A Wide Range of Subjects

Good day, fellow gaming math enthusiasts! Even for those of you who would prefer not to do the math, but are interested in the subject, good day to you, as well!

Everyone has been keeping me busy with questions, at least recently, so it’s time for me to share some of my favorite answers, once again. Here is the most recent installment prior to this one if you missed that and want to catch up!

If you’ve got a question of your own, a great place to visit would be our sister site Wizard of Odds message board as, much like The Beatles, I sometimes do this, “With a Little Help From My Friends.”

More importantly than that, sometimes the phrasing of some questions isn’t quite clear to me, so either that or Friending me on Facebook at Brandon James would be another way that we can engage in a back and forth so I can understand the specific question that I’m supposed to be answering. It’s usually my fault more than it is the other person’s because I often communicate in a somewhat technical way and can occasionally have trouble discerning the meaning of communication that I consider less precise. 

QUESTIONS AND ANSWERS

ALTERED DICE PROBABILITIES

Our first question from WoV comes from Member, Zcore13, who has a question related, I presume, to electronic craps machines, with one of the more popular ones being, “Bubble Craps.” These are machines that have large magnetized dice inside of a bubble that, ‘Pops,’ them to simulate a random roll.

But…what if it weren’t so random after all? Perhaps some of these machines aren’t as random as a pseudo-RNG that powers game such as the one that you can find on our site here. 

ZCore13 Asks:

I'm guessing it would be a player advantage, but someone with better math skills can probably figure it out pretty essy...

What if in craps, every time the dice were rolled, you knew that the number on each die that just hit, would not repeat on the next roll? For example, the dice end up 6 4 on the comeout roll. On the next roll, the first die could be anything, except a 6. The second die could be anything except a 4. So, I bet no 10 on the next roll, but then change it to no whateverr the next roll is after that.

Mission146 Answers

ADDED: For clarity, please note that this question does NOT mean that neither a four nor a six can come up again. What the questions means is that specific die will not be a six and the other specific die will not be a four. For that reason, in my answer, I have opted to name the dice, “Blue,” and, “Red,” to illustrate that this only relates to the face that showed up on the previous roll for each specific die and also to make this easier for me to figure out. 

For me, it's easier just to create new dice and go from there, so here you go:

Die 1 (Blue): 1, 2, 3, 4, 5

Die 2 (Red): 1, 2, 3, 5, 6

Possible combinations:

BR

1 1

1 2

1 3

1 5

1 6

2 1

2 2

2 3

2 5

2 6

3 1

3 2

3 3

3 5

3 6

4 1

4 2

4 3

4 5

4 6

5 1

5 2

5 3

5 5

5 6

All 25 possible combinations are covered. We know that there are 25 combinations because combinations for any two dice are simply (# of Faces) * (# of Faces), for comparison, if you had a game that involved rolling a six-sided die and flipping a coin, then there would be 12 possible combinations as follows:

1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T

With that, we will now look at the probabilities for these combinations:

Snake Eyes: 1/25

Three: 2/25

Hard Four: 1/25

Soft Four: 2/25

Five: 3/25

Hard Six: 1/25

Soft Six: 3/25

Seven: 4/25

Hard Eight: Impossible 0/25

Soft Eight: 3/25

Nine: 2/25

Hard Ten: 1/25

Soft Ten: 1/25

Yo: 1/25

Midnight: Impossible 0/25

Sum: 25

Okay, really, you can do any bet you want just by multiplying the new probability (found above) by the payouts. 

Snake Eyes: 1/25*****4% compared to 2.778% (!)*

Three: 2/25*****8% compared to 5.556% (!)

Hard Four: 1/25*****4% compared to 2.778% (!)

Soft Four: 2/25*****8% compared to 5.556% (!)

Five: 3/25*****12% compared to 11.1111% (!)

Hard Six: 1/25*****4% compared to 2.778% (!)

Soft Six: 3/25*****12% compared to 11.1111% (!)

Seven: 4/25***** 16% compared to 16.667%

Hard Eight: Impossible 0/25*****0% compared to 2.778%

Soft Eight: 3/25*****12% compared to 11.1111% (!)

Nine: 2/25:*****8% compared to 8.333%

Hard Ten: 1/25*****4% compared to 2.778%

Soft Ten: 1/25*****4% compared to 5.556%

Yo: 1/25*****4% compared to 5.556%

Midnight: Impossible 0/25*****0% compared to 2.778%

*(!) denotes more likely than before. 

With that out of the way, before we do our Place Bets on a one roll basis, we are going to go ahead and look at our single roll combination bets to determine whether they are more likely (warrants further examination) or less likely.

Crap Check: Okay, so our any Any Craps bet would normally be 11.1111% to occur, but it is now 12% to occur despite the fact that Midnight is impossible. 

Field: Um...are we tripling the two as opposed to 12? Are we in Reno? Either way, Field is 44% to win compared to 56% to lose. This would normally be 44.444% to win compared to 55.556% to lose, so the Field has become a little worse on the face of it. 

On the other hand, if Snake Eyes pays triple on the Field, then the Field has become better by virtue of eliminating a double and keeping a triple with the probabilities otherwise being the same. Single ways to make the Field have also been eliminated, so it's a good trade off for us.

Oh, screw it, let's do the Crap Check and Field first:

Crap Check

What sounds good? 7:1? I think so. All bets are $5:

(7 * 5 * .12) - (5 * .88) = -0.2

Okay, so this bet loses twenty cents every time and is, therefore, bad; it's just not as bad as it would normally be.

Field:

Triple Snake Eyes:

(5 * 3 * .04) + (5 * .4) - (5 * .56) = -0.2

Once again, this loses twenty cents every time and is, therefore, bad. It's actually worse than usual. 

However, suppose the previous roll had been 3-3, we are still left with 25 combinations that are possible, though all combinations that involve a three are now gone. In this scenario, we lose three possible Field rolls, but, perhaps crucially, we keep our doubler and our triple. 

Also, the previous scenario only left us with 11 possible Fields, but now we have twelve. 

( 5 * 3 * .04) + (5 * 2 * .04) + (5 * .4) - (5 * .52) = $0.40

Under this scenario, we are winning $0.40 on every $5.00 bet and are, therefore, at an 8% advantage. 

With that, the questions are simply:

1.) How many Fields become disqualified?

AND:

2.) Which Fields become disqualified?

In the above, even if 2 and 12 only doubled, you would still be winning $0.20 on every $5 bet and would be, therefore, at a 4% advantage. 

Hard Hops

That bring us to Hard Hops, of which we have four that are possible and all four of which that are equally likely. 

All four are now 4% to happen and 96% not to happen. Let's say they pay 30:1:

(30 * 5 * .04) - (5 * .96) = $1.20

In other words, we are expected to win $1.20 on every $5.00 bet and are now at a 24% advantage.

Hard Ways

For Hard Ways, we will DEFINITELY want to potentially pull these back after one roll; the reason why is because, suppose we bet 2-2 under the current conditions, if the next result is 2/5, then our Hardway becomes impossible on the roll after that one. 

Since we are talking about the Hard Four, I'm just going to go ahead and do that one. 

Hard Four: 4% Easy Four (Loses) 8% Seven (Loses) 16% Nothing Happens (72%)

Okay, so we are getting 7:1 if this hits:

(7 * 5 * .04) - (5 * .24) = $0.20

In other words, we expect to make $0.20 for every $5 bet and are at a 4% advantage accordingly. 

Hard Six pays better and we have actually made one Easy Way six impossible in the current scenario, which we did not do to a Easy Four, so let's see what happens:

Hard Six: 4% Easy Six (Loses) 12% Seven (Loses) 16% Nothing Happens (68%)

(9 * 5 * .04) - (5 * .28) = $0.40 for every $5 bet and reflects an 8% advantage. 

As we established, a Hard Eight is not possible. Hard Ten has eliminated an Easy Ten, which Hard Four had not, so let's look at it:

Hard Ten: 4% Easy Ten (Loses) 4% Seven (Loses) 16% Nothing Happens (76%)

(7 * 5 * .04) - (5 * .2) = $0.40 which reflects an 8% advantage for one roll on these.

Hard Ways v. Hard Hops

Given the choice, you would want Hard Hops as the advantage is through the roof. I guess if this were an actual certainty you might load the layout with everything possible, but assuming an overall total bet limit with a cap that could, in theory, be reached, you would prefer Hard Hops to Hard Ways easily.

Soft Hops

This depends on the soft hop in question. 

Any Soft Hop that is more likely than it was before is now automatically 8% to happen as compared to 5.556% to happen. My thinking is that any Soft Hop that is now more likely immediately becomes good, so let's see how good:

(15 * 5 * .08) - (5 * .92) = $1.40, which reflects an advantage of 28% every time. 

An example of a less likely Soft Hop is 10, because B6R4 cannot happen, only B4R6 can. Or, that might be the other way around. I'm sure I mixed the colors up at some point in this description. It doesn't matter. Only one soft ten can happen, the invert of whatever the first one was.

(15 * 5 * .04) - (5 * .96) = -$1.80 or a 36% disadvantage. This is obviously much worse than normal. 

Place Bets

Place Bets are effectively going to become one roll propositions because they can become bad bets based on what the dice do on the next roll, so you might end up picking some of them up. 

I am not going to bother with calculating all Place Bets because they are going to potentially change with every roll following this one depending on what ways, if any, are made impossible and what ways are made more likely. I will, however, do a best case scenario. If the eliminated dice are a B1R2, then every way to win a Place Ten is possible and the probability of doing so on the next roll goes from 8.333% to 12%. We only care about tens and sevens. Sevens will always be 4/25 (16%) because there is no way to eliminate two faces without eliminating two sevens. 

(.12 * 9) - (5 * .16) = $0.28

Okay, this is the best case scenario for a Place Ten bet because the number of sevens eliminated will always be two and every possible ten still exists. This represents a 5.6% advantage and is nothing compared to our advantage of just hopping all possible tens, so you would just want to hop all possible tens and do this only if you did not hit the bet cap yet and there was nothing else (such as other Hops) that are better. 

If you want to calculate the other Place Bets, on a one roll basis, just do the correct variation of the above for whatever dice you want to have eliminated. 

Come/Don't Come---:Spoiler Not Doing It:

Come bets can be calculated for the next roll as far as the initial bet goes. You would win on the very next roll 20% of the time (7 or 11) as compared to the normal 22.222%, but I'm not going to calculate it beyond that because I don't know if B-6 is always illegal and R-4 is always illegal. In any event, since you now only lose on Snake Eyes and Three, there is a 12% probability of instantly losing, resulting in a +8% probability of winning. Normally, the probability of losing would be 11.11%, for a difference of +11.111% in favor of winning the very next roll, so as far as initial roll Come Bet goes, it is worse. 

I assume that Blue die and Red die make illegal whatever happens next, and, yeah, not doing it. The initial Come Bet is worse; good enough for me. 

The initial Don't Come bet would normally be 1/36 push, 2/3 is win (3/36) and 7/11 loses (8/36). The difference is 5/36 which means it is 13.8888% more likely to lose as win. 

With our new Don't Come bet, pushes are gone, but we are 3/25 to win instantly, which is 12%, though we remain 5/25, or 20% to lose instantly. The difference between the two is 8% more likely to lose instantly as win. 

Once again, I don't know the impact going forward of whatever Blue is being illegal and whatever Red is being illegal on the roll after this one. It can be figured out because we know Blue-6 and Red-4 cannot happen, and therefore, will not be illegal the roll after this one, but I'm not even getting into it. 

If all else remained equal for the Don't Come bet, and I don't immediately recognize why it wouldn't (though, maybe it does) then we just say that we immediately lose 5.888% less frequently. 

Of course, the fact that seven will continue to be less likely that normal is bad for us after we have survived the Come Out roll. Normally, a seven would be 16.667% to appear and now it is 4/25 = 16%. It will always be 16% because every roll disqualifies two possible sevens from the roll after that. Blue 3 Red 3, for example, would still disqualify Blue 3 Red 4 and Blue 4 Red 3 from being possible the roll after this one. 

I don't know whether or not it ends up being a wash and I am not going to figure it out. Everything relies upon whatever the next roll after the Don't Come bet is made is, though we do know, if it is a point number, it immediately becomes less likely the following roll as it has immediately eliminated one of the ways to roll itself. 

Conclusion:

I'm going to say that's all I am doing absent very specific questions that only require a single step/decision. For those, the above should contain enough information that anyone should be able to figure out how to calculate it anyway. 

The short answer is that the correct one roll Hop Bets are always going to be the way to go. For example, the following ways to make six are typically possible:

5/1, 4/2, 3/3, 2/4, 1/5

But, we have eliminated 2/4, (or was it 4/2?) whatever the case, one of those is gone, so you would not want to Hop that because:

(15 * 5 * .04) - (.96 * 5) = -1.8

So, that's terrible. You do want to Hop 1/5 because:

(15 * 5 * .08) - (.92 * 5) = $1.40

Is awesome. 

I conclude that the best possible bets in this scenario are whatever Hard Hops or Soft Hops bets become more likely to win compared to normal. The worst bets to make are results that have become literally impossible.

ADDED NOTE: All Hard Hops that remain possible are always good and the math is always the same because there will always be one way to roll those and 24 ways to roll anything that is not that. See above for the math on that.

***Eventually, I decided to go ahead and hammer out the math on the full layout of bets as I realized that all Hard Hops and all Soft Hops not involving the dice faces seen would be at an advantage given the hypothetical of the dice faces that showed up on the previous roll not being possible for those specific dice. With that, here is the information that I added:

ANOTHER ADDED NOTE:

Okay, the last thing that I am going to do is make every Hop bet that has become better. We are only going to look at the case of one Soft Hop Winning and one Hard Hop winning, so let's do that:

Snake Eyes: 4% ($5)

Threes: 8% ($5)

Hard Four: 4% ($5)

Soft Four: 8% ($5)

Soft Five 3/2: 8% ($5)

Soft Five 1/4: There is only one way to do this, so you would not bet it.

Hard Six: 4% ($5)

Soft 6 1/5: 8% ($5)

Soft 6 2/4: There is only one way to do this, so you would not bet it.

Seven 2/5: 8% ($5)

Hard Eight: Impossible

Soft Eight 3/5: 8% ($5)

Soft Eight 6/2: There is only one way to do this, so you would not bet it.

Nines: There is only one way to roll each Hop Nine, so you would not bet these.

Hard Ten: 4% ($5)

Soft Ten: There is only one way to do this, so you would not bet it.

Eleven: There is only one way to do this, so you would not bet it.

Midnight: Impossible.

Okay, so we would be making $50 in bets in this scenario. 36% of the time, everything just loses.

Hard Ways to Win: 1/1, 2/2, 3/3, 5/5 (4)---4/25

Soft Ways to Win (Being Bet): 1/2, 2/1, 3/1, 1/3, 2/3, 3/2, 1/5, 5/1, 2/5, 5/2, 3/5, 5/3 (12)---12/25

Soft Ways to Lose (NOT Being Bet): 1/4, 4/2, 4/3, 6/1, 6/2, 4/5, 3/6, 4/6, 6/5 (9)---9/25

Okay, so there are three net results possible:

ALL LOSS: There are nine ways to just lose $50.

Win Hard: If we win on one of the four Hard Hops, then we are paid $150 and have lost $45 on the other bets for total gains of $105.

Win Soft: If we win on one of the twelve Soft Hops (six bets), then we are paid $75 and have lost $45 on the other bets for total gains of $30.

With that:

(12/25 * 30) + (4/25 * 105) - (9/25 * 50) = $13.20

Therefore, given the present situation, we will expect to gain $13.20 for every $50 bet, thereby reflecting an advantage of 26.4% on every single roll under the present circumstances.

Mechanically, this would also be the easiest thing to do as it would be fast. Eventually, someone would immediately recognize what ways they should bet and what ways they should not, so this is going to be much easier than keeping track of multi-roll bets and/or whether the Field is good this roll or not.

Another thing that would make it quicker is that Hard Hops are really easy. If you didn't see the face on the previous roll, then bet that Hard Hop. If you didn't see either of the faces on the previous roll, then bet those Soft Hops. Anything involving a face you saw on the previous roll you do not want.

ADDED (FOR LCB): The reason that I saw no point in doing this for the other bets is because Electronic Craps machines offer a limited amount of time for players to make bets and the Hop Bets are going to be, by far, the biggest advantage that you could possibly achieve in this hypothetical. If someone got really fast, then they could maybe cover all of the applicable Hops Bets and look to add bets later if they wouldn’t run into the machine’s Table Limit. 

Needless to say, if one could find an E-Craps machine where this was always the case, then it would be an absolute goldmine. If you want to look into it, then an easy way to disprove it would be to wait for two dice to show the same face several times, such as 1/1, 2/2, 3/3, 4/4, 5/5 or 6/6, after that, if you see any rolls where at least one of those numbers is showing, then it is not the case. I’d probably want to see about twenty rolls without that ever happening before I started to think, “...maybe,” and then would want to see forty more applicable rolls to be sure. 

That is, unless you’re really good at keeping your eyes on which die is which, in which case, you would be able to disprove that the machine was operating this way faster. 

MORE CRAPS

With that, we move on to a procedural question wherein Member Poll2k01 wonders why a particular combination of bets can not be made and if it is because that combination would offer an advantage. 

Of course, the readers here will already know that Wizard, one of the best game analysts to ever do it, will tell you that there is no way to combine negative expectation bets such that a positive expectation is created, but people persist in trying to do just that. 

Poll2k01 Asks:

Hello fellow Craps enthusiasts - I've been looking at rules on the Craps and was wondering why when playing DON'T PASS, players are not allowed to bet on PLACE or BUY bets ON THE POINT NUMBER once a point has been established. If this was allowed, would this be Advantage Play for players?

Example:

1. Comeout roll - Bet $10 on Don't Pass
2. Point 4 was established
3. Player Bets

$5 Buy on 4

$1 Each on other numbers ($1 Buy on 5, $1 Place on 6, $1 Place on 8, $1 buy on 9, $1 buy on 10)

1. Expected outcome:

7-out: Don't Pass wins $10, all number bets loses (-$10), net result = 0 (wash)

4 rolls point wins - $5 Buy wins $9.75, but Don't Pass losses $10, net result = - $0.25

I would like to see some math to validate if this is Advantage Play for player or not with the following assumptions:

1. Given the house edge at #1 at $10
2. The average rolls of 4 rolls before point number gets rolled OR a 7 out

I think the question would boil down to is: Assume if there is a player's edge POST-COME OUT given the conditions above, would it be greater than the COME-OUT disadvantage when betting don't pass.

Again, thank you for enlightening me.

Mission146 Answers:

Here's the thing: We can step-by-step this, which we will, but we really don't have to.

The most fundamental principle there is when it comes to advantage (or lack thereof) is that there is simply no way for multiple negative expectation bets to become a positive expectation bet. It is simply impossible absent something external to just the normal rules of the game.

Okay, so immediately we know that a DP $10 bet has an expected loss of $0.14 per bet resolved; we also know that the $5 Buy 4 has an expected loss of $0.0835 per bet resolved, as it seems that you are using the 39:20.

Therefore, no advantage is possible because both are negative expectation bets. Procedurally, they don't allow this on the Point Number; I have no insight into their reasons why, but what I do know is that you could make a DC bet, have it travel, then make a Buy bet on whatever the DC traveled to. In other words, you can do this same concept in a different way and the house is absolutely not afraid of it.

Point of 4 or 10 is the best possible scenario for a DP pass short of an immediate win. Your expectation:

(6/9 * 10) - (3/9 * 10) = $3.33~

In other words, you find yourself at a 33.33% advantage, so what you are talking is to hedge that advantage against a new bet which would be at a disadvantage. The math on the possibilities works as follows:

Don't Pass Bet Wins: +$5

Buy Bet Wins: -$0.25

(5 * 6/9) - (.25 * 3/9) = $3.25

You will notice that your net expectation went down by the expected loss that the Buy 4 bet has in the first place.

Superficially, this seems like the DP + BUY is in an advantageous situation, because it is, but the DP had an even bigger advantage prior to you making the Buy Bet. It's also important to remember that, while the DP is at an advantage NOW (which is true on its own) it's because a point was established, therefore, you overcame the only roll (Come Out) which represents a disadvantage for DP. The entire House Edge of the DP bet stems from the huge disadvantage that it's at on the CO roll.

You'll forgive me for not including the Place Bets, but they are irrelevant. The House Edge is just the House Edge, as far as those go. If the DP wins, then the Place bets lose, but you're just adding more disadvantageous bets to the advantageous situation (DP surviving CO) that you find yourself in.

Is the, "Post Come Out Advantage," greater than the Come Out disadvantage when betting DP?

When it comes to the percentage, absolutely, but that's simply because a CO of 4 or 10 (aside from just winning on the CO) is the most favorable situation for the DP bet. The DP bet ceases to have a disadvantage any time it survives the CO.

Now, does the, "Post Come Out Advantage," ameliorate the disadvantage of making the DP bet in the first place? Absolutely not. Factored into the 1.36% (bet made) House Edge and 1.4% (bet resolved) House Edge of the DP bet is the fact that a DP surviving a CO will be at a huge advantage. You have made the bet where you were expected to lose; you have survived the only disadvantageous phase that the bet has and now you are wanting to make more bets that are expected to lose---every single one of which will subtract from what would otherwise be your expected profit on the DP bet having survived the CO.

ADDED (FOR LCB): Later on in the applicable thread, someone expressed doubt that casinos even have the rule that the OP asked about. I know of at least one casino that has that rule, though I have no idea why they have it, because you could do it on a Don’t Come bet that travels, though I don’t know why you would. Perhaps it is to encourage people to make a, “Put Bet,” which is an a bet that you make after the Point has been established that pays Even Money. Wizard defines it here:

In craps, the player may skip the come out roll on a pass or come bet. Such a late bet on the pass and come is known as a "put bet." Much of the value in pass and come bets is in the come out roll, so skipping it carries a high house edge. To be specific, 33.33% on the 4 and 10, 20.00% on the 5 and 9, and 9.09% on the 6 and 8, on a per bet resolved basis. However, you can combine a put bet with an odds bet, bringing down the overall house edge. Here are the breakeven points, according to point.

6 and 8: Bettor must combine a put bet with 5X odds to have the same overall house edge of 1.52% as a place bet.

5 and 9: Bettor must combine a put bet with 4X odds to have the same overall house edge of 4.00% as a place bet.

4 and 10: Bettor must combine a put bet with 19X odds to have the same overall house edge of 1.67% as a buy bet, assuming the commission is paid on a win only. If the commission is always paid then the bettor must combine a put with with 6X odds to match the 4.76% house edge.

At casinos that offer 3-4-5X odds, or worse, there is no reason to make a put bet, because you will do equal or bettor to make a place or buy bet.

That equates to 66.67% RTP, 80% RTP and 90.91% RTP, respectively, on Put Bets with no Odds. That is obviously NOT something that you would ever want to do when the House Edge of a Pass Line bet is 1.41%. Why are there no Put Bets on the Don’t Pass Line? The answer is because that would put the player at a ridiculously huge advantage. 

NEW ROULETTE STYLE GAME?

WoV Member ThomasandMiguel inquires about a game similar to Roulette with an interesting set of rules. Specifically, the player makes a bet, but that one bet lasts for three spins and can win increasing amounts of money as that player’s numbers come up. 

Since new games are always interesting, and this one was fairly simple to figure out, I went ahead and solved the House Edge for him.

ThomasandMiguel Asks:

Would you be able to work out the following house edge, probability and Return to Player etc for the following bet on a 23 number roulette wheel with a neighbour bet that covers three of the numbers. i.e a £100 wager would cover all three numbers on a roulette wheel with 23 pockets and play for up to three spins.

Neighbour bet - This wager will cover three numbers and is resolved across one, two or three spins. If the first spin matches any of the three wagered numbers, the player will win 5 times their wager, otherwise the wager is lost. The wager is not returned to the player if they win the first spin, instead it plays for a second spin.

If the second spin again matches the three wagered numbers then the player will win an additional 10 times their wager, otherwise the wager is lost. The wager is not returned to the player if they win the second spin, instead it plays for a third spin.

If the third spin again matches the the three wagered numbers then the player will win an additional 60 times their wager and the wager will be lost.

Many Thanks!

Mission146 Answers:

The first thing that we establish is that there are 23 numbers and the player is essentially betting on three of them at once. We're going to make the bet amount $5 for this.

I'm going to do the best I can with this, but I will say that I don't quite know what I get from your wording. For the second and third spins, when you say, "Additional," do you mean on top of the results from previous spins, or do you mean that is the total amount won?

Also, it seems that you lose your wager no matter what happens, so we start from a basis of -$5 and just have to go from there:

First Step Win and then Loss: (3/23) * (20/23) * 5 * 5 = 2.83553875236

First Step Win, Second Step Win, then Loss if 15x total: (3/23) * (3/23) * (20/23) * 5 * 15 = 1.10955864223

First Step Win, Second Step Win, then Loss if 10x total: (3/23) * (3/23) * (20/23) * 5 * 10 = 0.73970576148

Win/Win/Win if 75x total: (3/23)^3 * 75 * 5 = 0.83216898167

Win/Win/Win if 60x total: (3/23)^3 * 60 * 5 = 0.66573518533

SUM (If Wins Stack): 2.83553875236 + 1.10955864223 + .83216898167 = 4.77726637626

SUM (If Wins DO NOT Stack): 2.83553875236 + 0.73970576148 + 0.66573518533 = 4.24097969917

Expected Loss if Stack: 5 - 4.77726637626 = 0.22273362374

Expected Loss if Not Stack: 5 - 4.24097969917 = 0.75902030083

House Edge if Stack: .22273362374/5 = 0.04454672474 or 4.454672474%

House Edge if Not Stack: .75902030083/5 = 0.15180406016 or 15.180406016%

Again, this all assumes that the $5 wager is taken regardless of what happens with the spins, which is how you seemed to have phrased it. If you want to know it for the player retaining the wager upon three consecutive winning spins, then just replace '75' with '80' in the Win/Win/Win if stack and '60' with '65' if no stack and go from there. You'll also have to determine the probability of the player losing the $5 wager, which you can basically do the same way I did above.

Return to Player is just the inverse of the House Edge relative to one. If you want the RTP, just subtract the decimal form of the House Edges above from 1.

For the probabilities, again, your phrasing leads me to the impression that the original $5 bet is not to be returned no matter what. With that, here are the probabilities:

Spin One Loss: 20/23 = 0.86956521739

Spin One Win, then Loss = 3/23 * 20/23 = 0.11342155009

Spin One Win, then Spin Two Win, then Loss = 3/23 * 3/23 * 20/23 = 0.01479411522

All Three Win: (3/23)^3 = 0.00221911728

ADDED (FOR LCB): The only thing that I have to add to the above, since this game does not actually exist yet (as far as I know) is that I wasn’t sure whether or not future amounts won were added to previous amounts won or the multiplier reflected the total amount won at various points. For that reason, since this was pretty easy anyway, I solved it for both. 

ULTIMATE TEXAS HOLD ‘EM BARRAGE

Recently, I was barraged by a wide variety of Ultimate Texas Hold ‘Em questions. You can read the general rules of the game here if you are unfamiliar with them, but I would recommend looking at the Wizard of Odds site for playing strategies as this is a fairly complicated game. 

Anyway, I answered one question on it and then a great many, “What would this do to the House Edge?”, type of questions followed.

KSDJDJ Asks:

Hi, I noticed dealer errors the last few months playing UTH (Ultimate Texas Holdem) and was wondering what the advantage would be for the following, assuming "it happened every applicable hand" (see table below):

Situation Advantage

Miss Pays:

Paid ante on ALL wining hands

Paid 3/1 on the blind bet for a flush

Paid 1/1 blind bet, with less than a straight

Paid on a dealer flush ###(player has losing hand)

***

"Flashing dealer, one flop card" situations:

Can see the card perfectly

Can only see color, and if it is a picture card or not

###: Two dealer pocket flush cards , and all combos of exactly three community flush cards.

Cell ***: Means, same as ### above, except NOT miss-paid if the community cards are: FL-FL-FL-N-N, N-FL-FL-FL-N or N-N-FL-FL-FL

Note: FL = Flush Card, and N = not a flush card (so, all other suits).

----

Thanks in advance for any help

ksdjdj

Mission146 Answers:

Paying Ante on ALL Winning Hands:

Okay, so the first thing that we have to do is determine what percentage of hands are winning, but would normally not have the ante paid on them. We can go here:

If we go down to the return table for the game, then what we want are hands where the player wins and the dealer does not qualify. These probabilities are given in the return table, so we would just add those together.

0.055972 + 0.002927 + 0.001829 + 0.000008 + 0.000003 + 0.038965 + 0.001620 + 0.001396 + 0.000013 + 0.025907 + 0.004890 + 0.001471 + 0.000010 = 0.135011

In other words, this results in an expected gain of 0.135011 units every hand.

We typically expect to wager 4.152252 and lose .02185, or 2.185% so that represents an expected loss of 0.0907267062 units.

Instead, we are going to gain an additional .135011 units, so when we subtract from that the .0907267062 units that we would normally expect to lose, we end up with an expected positive outcome of 0.0442842938 units, which reflects a player advantage of 0.0442842938/4.152252 = 0.01066512673 or 1.0665%.

One thing that is probably worth noting is that there may be a few strategy changes that would come about as a result of this, particularly when it comes to an unpaired (and, obviously not better than a pair) board on the final decision to call or fold. In this instance, the player might also win the ante bet even if the dealer does not qualify.

I remember thinking the advantage was much larger before...hopefully I was wrong then and am right now.***

Paying 3:1 on the Blind Bet for a Flush

We have the following probabilities for a Flush in this game:

0.008666 + 0.001829 + 0.005487 + 0.001396 + 0.006866 + 0.001471 = 0.025715

Okay, so that represents the total probability of all winning flushes.

Typically, the Blind Bet would pay 3:2 on a Flush, so for $5 units, this would be $7.50. Instead, you are getting paid at 3:1, which would be $15.

(.025715 * 3) - (.025715 * 3/2) = 0.0385725

With that, the expectation seems to be that we would gain an expected .0385725 units every hand.

Normally, we would expect to lose 0.0907267062 units every hand, so when we subtract our gains from that we end up with 0.0521542062 units that we would still expect to lose every hand. We expect to bet a total of 4.152252 units, so the House Edge becomes 0.0521542062/4.152252 = 0.01256046265

With that, it does reduce the House Edge to 1.256% and then you can divide by the expected total bet, in units, to get EoR if you like.

Paid 1/1 Blind Bet w/Less Than Straight

We're going to the Return Table again and this is just a function of how often does the player win with less than a straight.

0.055972 + 0.076036 + 0.131987 + 0.038965 + 0.049437 + 0.025907 = 0.378304

Once again, the expected units gained is the same as the probability because this pays us one unit that we would not otherwise be getting. With that, we take our normal expected loss, in units, of 0.0907267062 instead becomes an expected gain of 0.2875772938. With that:

0.2875772938/4.152252 = 0.06925815046 or a 6.9258% player advantage.

It's also important to note that this is not going to be exactly right as there would be strategy changes involved, as well. The player would fold significantly less often, but how much less often is above my pay grade.

Paid on a Dealer Flush

Paid what on a dealer Flush? Do you mean just the blind as if the player had the flush, or totally across the board paid as if the player had won the hand?

***I think I remember and the other question was the dealer always paying the Blind Bets, even if the dealer had won.

As far as the other stuff, Teliot handled most hole-carding UTH questions in his A.P. Heat articles on 888. 

***After that, KSDJDJ wondered about situations in which the dealer did not recognize that he had the flush, of course, for the math, I had to assume that happens all the time. 

KSDJDJ Asks:

1) Yes, the dealer thought he had nothing / didn't qualify.

2) Sorry about the wording, the only other times it has happened to me so far, are when the dealer's "next best hand" was a 2 pair or pair (where I had either a better pair or better two-pair, both times)

MIssion146 Answers:

Dealer Has a Flush, but We Do Not Lose:

Okay, the first one is going to be a little tough. This is very much going to be an approximation.

The first thing that we have to look at is that I'm going to go down to, "Player Plays 4x Blind," and get the two probabilities for the player getting a flush, which is both with and without the dealer qualifying:

0.004696 + 0.021019 = 0.025715

In other words, this is the probability of the player getting any flush and winning, which is automatically also true of the dealer.

Obviously, if the dealer has a flush, then the dealer should have qualified, even if the dealer doesn't seem to think so. That being the case, the only results that are being flipped are ones in which the player has a flush, or worse, and loses to a dealer who qualifies with a flush.

The probability of the dealer qualifying and beating the player is:

0.141353 (Loss of Six Units) + 0.066196 (Loss of Four Units) + 0.097079 (Loss of Three Units)

I'm also not going to count folding because you just lose anyway, regardless if the dealer knows he has a flush, or not. Although, this might change the decision to call or fold against certain flush rich boards if perhaps you have a high(ish) card and there are four suited on the board; you'd no longer count the four-suited towards the 21 outs for the call/fold because the dealer doesn't know what a flush is.

Anyway, let's get our total expected loss from these. If it seems like it should be more, remember, we are only taking the losses in which the dealer qualifies, because he normally would with a flush. If the dealer did not qualify, then he did not have a flush anyway.

0.848116 + 0.264783 + 0.291238 = 1.404137 (Units)

Okay, so when the dealer qualifies and we lose, that represents a total of 1.404137 expected units lost.

The probability of the player (betting blind) having a Flush that is the superior hand should be the same as that for the dealer, so we multiply this by that probability:

0.025715 * 1.404137 = 0.03610738295

This reflects a swing of about 0.03610738295 units...but don't take that as gospel because I really just did the best I could with this one. My logic could be flawed somewhere. In any event, this would reflect an expected gain of .03610738295/4.152252 = 0.00869585539 or 0.869585539% Which is a swing of less than one percent and merely reduces the House Edge from 2.1850% to 1.31541446%****

Again, I could be off, so would appreciate any corrections. That said, this would impact only hands where the dealer would qualify anyway AND flushes are not that common, so it doesn't have a huge swing on anything.

When a flush beats another flush and the dealer's would have beaten the player's, but is treated as if the opposite happened, this also swings that the player will be paid for the Flush on the blind bet, as well, but it's an unlikely enough scenario (and is already accounted for on Ante/Play) that any change in EV from that, specifically, will be miniscule. I guess the same is true with dealer flush otherwise normally beating a player straight, but again, has already been accounted for on Ante + Play and is going to be a pretty negligible swing on the blind bet just because it happens so infrequently.

In general, I am going to OPINE that this alone would not be enough to create an advantage, even without accounting for what happens with the blinds when a dealer flush would otherwise normally beat an inferior player flush or a dealer flush would otherwise beat a player straight. I say opine because I didn't actually do it, but we know that the difference just on how the Blind bets are resolved is not going to be greater than what the Ante/Play bets have done as a result of not losing to dealer flushes.

I still don't quite get what #2 means, but if it relates back to when the dealer has a flush, I don't think the player winning including all situations in which the dealer has a flush and the player would have otherwise lost is, by itself, advantageous***. That being the case, the situation needing to be more specific than that would not be an advantage, either.

****Also, this assumes that the dealer would never just win some other way, which is not the case. For example:

Dealer: Ah, 8h

Board: Qh, 10h, 5h, Ac, 2d

Player: Qd Kd

Even if the dealer fails to recognize that he has a flush, the player still loses to a dealer pair of Aces v. a player's pair of Queens, so the fact that the dealer had a flush wasn't even needed for the player to still lose this hand. There is no quick way to account for all of these situations that I can think of, but suffice it to say, the dealer doesn't always need the flush that he has to beat the player, so the change in House Edge is not even as good as when we assume the player always wins this.

***Another potential dealer mistake that has been seen, evidently, is of a dealer allowing the player to make a 4x Raise AFTER seeing the flop. Normally, the max Raise allowed would be 2x, and pursuant to the strategy, the player would only make this raise if he had the advantage. That being the case, you woulds always do 4x if you could. 

JackSpade Asks:

I had a session at NYNY with two really dumb dealers and a nearly as dumb pit boss. One dealer told me I could bet 4x AFTER the flop! I did so twice when having a pair (winning once) before the next dealer came in. On the hand I lost, I knew it was strong enough to bet the standard 2x but hesitated for a moment wondering whether it was also good enough for a 4x raise. I think the answer is that I should never hesitate to raise as large as possible when I have a raise-worthy hand.

Assuming a player was at a table that allowed a 4x raise after the flop, what statistical advantage would the player enjoy?

Mission146 Answers:

That's the correct line of reasoning. If you're making the 2x Raise at all, then it's because you have the mathematical advantage on that bet, so you would want to raise ANYx that they would allow you to raise in that situation.

In terms of the advantage that a player would have if that player could make the 4x raise after the flop, all you would have to do is go to the Wizard of Odds return table, add the probabilities of every possible way for the player to win with the, "Middle Raise," then multiply that by two to get your expected units gained. After that, you would want to sum the probabilities of the possible ways to lose making the middle raise, then multiply those by two to get the expected number of units lost on those potential outcomes.

(0.076036 + 0.004785 + 0.005487 + 0.010405 + 0.000666 + 0.000097 + 0.000004 + 0.038965 + .001620 + 0.001396 + 0.000013) * 2 = 0.278948*

*A few events have a return, but a probability that was so low as to not be expressed on the return table, so are not included.

(0.066196 + 0.000287) * 2 = .132966

In total, this advantage would cause the player to win an additional expected .145982 units, per hand.

Generally speaking, the expected loss of 2.185% relative to the initial two units being bet, so an initial expected loss of 0.0437 units.

If we take our expected gain of .145982 units per hand and subtract the .0437, then we would expect to win .102282 units per initial bet of two units, this would give us a player advantage of 5.1141%.

In theory, this should change nothing about the playing strategy otherwise, all else being equal.

ADDED (FOR LCB): There was later some discussion to the effect of NEVER taking the 4x Raise until after you have seen the flop, but could you imagine what the casino would think if they saw you playing UTH and you did not make the 4x raise until after the flop? Personally, I tend to think that NEVER making the preflop raise is going to get the error detected by casino staff more quickly as it is almost certain that someone in the place knows how the game is actually supposed to be dealt. My tendency, overall, would be just to try to ride out the sizable advantage that just being able to bet 4x post-flop gives me and otherwise play normally. 

Also, all of these things (individually) are errors that people have seen. It’s a tough game to deal and not all dealers are properly trained, other times, they just get lackadaisical and have a memory lapse. UTH is one of those games that I would only want my UTH dealers to also be on easier games, or common games, like Blackjack, rather than other card games that are unique and have unique rules...I think part of the issue for the dealers is mixing the rules for two different games up with each other.

DICE SETTING

A recent WizardofVegas discussion concerned the shift in probabilities and House Edge/RTP if dice influencing were viable and we assumed a change in probabilities. GenoDRPh was the first to inquire:

GenoDRPh Asks:

So as not to hijack an existing thread, here's a hypothetical:

A DI practitioner claims to be able to throw the dice and reduce a probability of a seven rolling from 1 in 6 to 1 in 8.

How does that change the HE on a 1 unit pass line bet with no odds if there is no point established? How does the HE change with a decrease in chance of come out roll wins, but increase in chance of come out roll loses, and increased chance of points being established.

How does that change the HE on a 1 unit pass line bet with no odds if there is a point established? How does the HE change due to a decrease in chance of seven out, an increase in chance of making the point, and an increase in chance of a push?

Feel free to assume either the probability of the other number combinations increase equally, or increase in proportion of their existing probabilities. Ideally, an analyses of both assumptions would be ideal.

Gene

ADDED (FOR LCB): Some people believe that influencing the dice to a certain degree is possible and some don’t. I think the debate will go on for ages and I even wrote an entire WoV article about what I would take for statistical proof it is more likely than not. For my part, I am Agnostic and leaning no. In any event, I am willing to discuss hypotheticals, and so I did: Also, I REALLY screwed up my analysis on the first try out of trying to go too fast, so that’s why I call myself an idiot here. That, and I also happen to be kind of an idiot. 

Mission146 Answers:

DISCLAIMER: THIS POST HAS BEEN SUBSTANTIALLY EDITED TO REMOVE MISTAKES DUE TO GOING TOO FAST AND ALSO THE FACT THAT I AM AN IDIOT.

First of all, I apologize for the delay, but my big project took a little longer than anticipated.

Secondly, I'm going to answer this, but I do want to somewhat modify the parameters. The key to being a good snake oil salesman is that the product you are offering must superficially sound viable, with that, I think 1 in 8 sevens is entirely too high. It wouldn't even take a very large sample of tosses to prove it is mathematically unlikely that the claimant does that long-term, so I seriously doubt anyone would even make that claim.

Superficially, it looks like just a 4.1% change of outcome, but look at it over 100 rolls with 12.5 (actual) compared to 16.6 (normally expected) and that example would be saying that it effectively reduces the frequency of sevens by about 25%.

What I am going to do is reduce this to a more reasonable 1 in 6.5, which would change the expected percentage from 16.6667% to about 15.3846%.

The second modification I am going to make is that we are just going to assume that the CO roll is thrown such as to not change anything vis-a-vis the long-term probabilities. After all, a PL bet enjoys its only advantage (in terms of probability of winning as opposed to losing) on the CO anyway. Now, one might argue, "But, I try to roll more sevens on the CO, but I don't care," since this is all hypothetical anyway.

The one thing that I am keeping is that we will treat the increase to other combinations as equal. I get that dice influencers say they are trying to hit certain other numbers with greater frequency, but in the non-hypothetical world, we have yet to see any actual mathematical evidence to back the assertion that they can even reduce sevens, much less increase the probability of other numbers.

With that, the probabilities on the CO are:

Snake Eyes: 2.7778%

Three: 5.5556%

Four: 8.3333%

Five: 11.1111%

Six: 13.8889%

Seven: 16.6667%

Eight: 13.8889%

Nine: 11.1111%

Ten: 8.3333%

Yo: 5.5556%

Midnight: 2.7778%

Okay, so what we have are the following:

Immediate Loss: 11.1111%

Immediate Win: 22.2222%

Point Established (4, 10): 16.6667%

Points Established (5,9): 22.2222%

Point Established (6, 8) 27.7778%

Any Point Established: 66.6667%

Therefore, our expectation on the Pass Line bet (taken alone) is simply:

.222222 - .111111 = .111111

Thus, on a bet of one unit, we have an expected gain of 11.1111% looking at the Come Out roll in isolation.

The point being established, we only care about sevens and that point number. With that, we have the following expected losses based on the probability of arriving at that Point Number (via the CO) to begin with and then comparing to sevens:

Point of Four or Ten: 3/36 * (3/9 - 6/9) = -0.02777777777 * 2 = -0.05555555555

Point of Five or Nine: 4/36 * (4/10 - 6/10) = -0.02222222222 * 2 = -.00444444444

Point of Six or Eight: 5/36 * (5/11 - 6/11) = -0.01262626262 * 2 = -0.02525252525

Now, we just total these losses and then subtract the result from our Come Out advantage:

.111111 - (.0555555555555 + .0444444444444 + .02525252525) = -0.01414152524

Thus, reflecting the House Edge of the Pass Line of about 1.414%.

That's the Pass Line bet in a nutshell.

Okay, so what we want to do is reduce the odds of a seven to 1 in 6.5 rather than 1 in 6, which will result in the following probability shift:

(1/6) - (1/6.5) = 0.01282051282

Okay, so we want sevens to decrease by that and everything else to increase proportionately. It's important to note that, given the intended impact of dice setting, these things are not meant to increase proportionately to one another, but there's really no way to measure the success rate of something that has not been demonstrated to be successful in the first place. For that reason, we can't know the exact probabilities of what should shift away from sevens to other numbers.

Also, we have taken away from sevens, so we do not want to add to sevens. For that reason, what we will do is take this change in percentage and multiply it by the fact that the sevens must also be something else and then add those together.

0.01282051282 + (0.01282051282 * 1/6) = 0.01495726495

With that out of the way, we simply multiply our other results' normal probabilities by the above, then add what the probability would normally be into that, and hopefully, the sum of all probabilities after we have done so will be something very close to 1.

Snake Eyes: 2.7778%---> (1/36 * 0.01495726495) + 1/36 = 0.02819325735

Three: 5.5556%---> (2/36 * 0.01495726495) + 2/36 = 0.05638651471

Four: 8.3333%---> (3/36 * 0.01495726495) + 3/36 = 0.08457977207

Five: 11.1111%---> (4/36 * 0.01495726495) + 4/36 = 0.11277302943

Six: 13.8889%---> (5/36 * 0.01495726495) + 5/36 = 0.14096628679

Seven: 16.6667--->0.15384615384

Eight: 13.8889%---> 0.14096628679

Nine: 11.1111%---> 0.11277302943

Ten: 8.3333%---> 0.08457977207

Yo: 5.5556%---> 0.05638651471

Midnight: 2.7778%---> 0.02819325735

SUM: 0.02819325735 + 0.02819325735 + 0.05638651471 + 0.05638651471 + 0.08457977207 + 0.08457977207 + 0.11277302943 + 0.11277302943 + 0.14096628679 + 0.14096628679 + 0.15384615384 = 0.99964387454

I'd assume this is slightly off of one due to rounding. It's certainly a hell of a lot closer than it was last time.

Okay, so now what we will do is look at our situations the same way, so we only care about those numbers and seven.

Four or Ten + Seven Combined Probability:

0.08457977207 + 0.15384615384 = 0.23842592591

Four and Ten Solved:

(0.08457977207/0.23842592591) - (0.15384615384/0.23842592591) = -0.29051530996 * 6/36 = -0.04841921832

Five and Nine Combined Probability:

0.15384615384 + 0.11277302943 = 0.26661918327

Five and Nine Solved:

(0.11277302943/.26661918327) - (0.15384615384/.26661918327) = -0.15405164739 * 8/36 = -0.03423369942

Six and Eight Combined Probability:

0.15384615384 + 0.14096628679 = 0.29481244063

Six and Eight Solved:

(.14096628679/.29481244063) - (.15384615384/.29481244063) = -0.04368834307 * 10/36 = -0.01213565085

As we can see, the disadvantage of these situations has been reduced due to the reduced frequency of sevens thus necessitating an increased frequency of these other numbers. Will it change our results overall to an expected positive, let's find out:

.111111 - 0.04841921832 - 0.03423369942 - 0.0121356085 = 0.01632247376

The result is a player advantage of 1.632247376%, which would be even more insurmountable than the, 'Normal,' disadvantage on the Pass Line bet is over the long-term.

Also, keep in mind we made the following other assumptions:

1.) We roll, 'Normally,' on the Come Out.

2.) While we reduce the frequency of sevens to 1 in 6.5, we are not disproportionately increasing the frequency of other numbers. This is important to note because people who purport to be dice influencers will often state that they are trying for sixes and eights. Since nobody has ever demonstrated that they can reduce the frequency of sevens to 1 in 6.5, much less demonstrated what doing so will increase their expected sixes and eights to, I really don't care.

3.) This is Pass Line only. If you want to know what this does to Odds Bets, it would be trivial to incorporate the Odds into what I have already done. Alternatively, I can always do so later.*

*Obviously, given the above assumptions, Odds bets would be made at an advantage if you could reduce the frequency of sevens (without reducing the frequency of the desired number) at all.

ADDED (FOR LCB): After this, I would go on to calculate what this would do to the Odds bets, as follows:

Player Edge for Odds

With the Pass Line out of the way and our methodology (I hope) down, we can calculate the Player Edge or remaining House Edge for Odds.

ODDS BETS

This is easy as it just requires changing one part of our formulas to reflects the Odds Payout:

Odds for Four and Ten:

((0.08457977207/0.23842592591)*2) - (0.15384615384/0.23842592591) = 0.06422703504

The above reflects the advantage on the Odds Bet taken alone. For its influence on a, "Per Come Out," basis, simply multiply the above by the overall probability of occurrence based on the CO, which is 6/36:

0.06422703504 * 6/36 = 0.01070450584

That is the advantage per Come Out roll given the potential for being able to make the Odds Bet. As we have previously established, the player is at an overall advantage anyway. The new Expected Value of a Point of Four Established that we determined is -0.04841921832, so our expectation on Single Odds is insufficient to offset that, 3x/4x/5x odds also wouldn't, because this would be the 3x, but it becomes offset and overall profitable EVEN WITH THIS POINT HAVING BEEN ESTABLISHED at 5x Odds.

Odds for Five and Nine:

((0.11277302943/.26661918327)*3/2) - (0.15384615384/.26661918327) = 0.05743544075

The above reflects the advantage on the Odds Bet taken alone. For its influence on a, "Per Come Out," basis, simply multiply the above by the overall probability of occurrence based on the CO, which is 8/36:

0.05743544075 * 8/36 = 0.01276343127

As we can see, our expectation for Single Odds is slightly greater in this scenario, which makes sense, because trying to Make a Point of 5 or 9 is not as bad of a situation relative to 4/10 to begin with. With regard to the Pass Line bet, -0.03423369942 is the new expected loss of a Point of Five or Nine established given the adjusted probabilities. With that, any Odds greater than 2x Odds would turn a point of five or nine established profitable.

Odds for Six and Eight:

((.14096628679/.29481244063)*6/5) - (.15384615384/.29481244063) = 0.05194282261

With that, we multiply that by the probability of even being the Point Number and get:

0.05194282261 * 10/36 = 0.01442856183

Six and Eight becomes positive even with single odds as our new expected loss of these points established is -0.01213565085, which is less than our positive expected outcome from the Odds Bet. Therefore, in addition to just winning on the Come Out roll, having a Come Out roll of six or eight has positive expected value if we are able to take any Odds whatsoever.

The reason that 6/8 becomes better (with odds) than 5/9 followed by 5/9 being better than 4/10 is because of these changed probabilities in the first place.

Once again, dice influencing proponents will argue that sixes and eights will increase out of proportion to other numbers given the reduced amount of sevens, but once again, nobody has ever demonstrated that they can knock the frequency of sevens down to 1 in 6.5 rolls, so I really don't care.

However, when we assume that the frequency of sevens has dropped so, then because numbers such as six and eight are more likely to begin with, they ultimately end up with a greater percentage of the roll frequency that was taken away from sevens. For that reason, single odds would produce an advantage on 6/8, at least 3x odds to produce an advantage on 5/9 and starting at 5x Odds for Points of four and ten.

With that all considered, if you were at a table upon which you could take 5x Odds, 10x Odds, or something even more than that...and again assuming you could roll sevens with a frequency of 1 in 6.5, or less often, then every single roll of the dice would see you at an advantage. Recall, just rolling the dice, 'Randomly,' you have an advantage on the Come Out roll to begin with.

Do you think dice influencing is possible? Let me know in the comments!

HOME RUN PROMOTION

An online casino that won’t be named has recently been offering profit boosts, with the most recent being 61%, on Aaron Judge of the New York Yankees to hit a home run in whatever game they happen to be playing that day. If you had this offer, then the Odds with the boost are extremely strong and you would expect to profit. I think that is why:

AitchtheLetter Asks:

What do yall think about Aaron Judge's chances on hitting a homer tonight?

Mission146 Answers:

Aaron Judge has appeared (to this point in the season) in 143 games and has started in 139 of those games. I do not know if there is a stipulation that cancels the bet if Aaron Judge does not start, but we would assume that he would start anyway.

It would appear that he has played no fewer than seven innings in a game, so I am going to go with the 143 number.

Aaron Judge has 60 Home Runs in the season, but has eleven games with multiple Home Runs; as a result, he has hit a Home Run in 49 unique games this season.

The probability of Aaron Judge, based on this season alone, hitting a Home Run is therefore:

(49/143) = 0.34265734265*

I would say that was a great estimate by SOOPOO.

With that, we can calculate a fair Moneyline for this event. The first thing that we want to do is index this to losing $100 or winning x.

(.34265734265 * x) - (.65734265735 * 100) = 0

x = 191.836735

We can confirm this here by placing a Moneyline of 191.836:

Which yields an Implied Probability of 34.3% and confirms our findings.

The current odds of this event happening are +195 with the straight bet on (site omitted), so that would imply that we are at a slight advantage straight up as:

(.34265734265 * 195) - (.65734265735 * 100) = 1.08391608175

Thereby yielding an expected profit of $1.08391608175 on this bet just based on Aaron Judge. However, the expected starting pitcher for the Red Sox is Michael Wacha, and as this website notes:

Quoting in part:

“Michael Wacha has been quite good for the Red Sox, because of course. His 2.61 ERA in 114 innings jumps off the page, and even though he’s been considerably worse by FIP or xERA, its not like he’s been the luckiest guy in the world. He’s seen a three-year decline in his strikeouts, but like Taillon, has traded that for better command in the zone, fewer walks and home runs.”

With that, it sounds as though this pitcher gives up fewer home runs than the average pitcher does. However, with the Profit Boosts that I have seen recently being given out, if one of those is offered, then it's pretty much going to be an expected profit regardless of who the pitcher is...as far as I can tell. For example, yesterday's boost on this event was 60%, so if that remains the same, you would want this bet all day.

*Of course, this ignores the matchups against whoever the starting pitcher is against whoever they are playing, which I wouldn't know anything about and wouldn't really know how to analyze particularly well. I guess I would start with what percentage of Home Runs that guy gives up and go from there, but I can't even begin to know where from there would go. Secondly, this wouldn't account for matchups against potential relief pitchers and the probability of those relief pitchers coming in, etc. etc.

ADDED (FOR LCB): Anyway, the bets were enormously advantageous for players with this particular profit boost and bettors would have done well on September 20th, 2022 when this bet with the boost hit. Actually, the website in question cut the amount that you could bet on the proposition in half for the following game, so I think they realized they had made it too juicy for the bets that they were allowing. 

We might want a more specific analysis to decide the value on a bet like this, but I don’t really know enough about baseball this season to figure out a precise value, given all the variables, but after a general analysis could easily conclude, “There is no way this bet is not an advantage with the Odds Boost.”

CONCLUSION

Typically, I would want to put more questions in one of these, but it’s already running pretty long because of the math. I’d also like to include some questions asked here, but nobody has asked me any recently. I did get some questions asked directly (in-person) and a few via Facebook Messenger, but because of how long this has already gone, I will have to save some of those for the next time we do this. 

If you have any questions that you would like answered, please feel free to PM me here at, “Mission146,” at WizardofVegas at, “Mission146,” or send, “Brandon James,” a Friend request on Facebook and Direct Message me your questions. 

Even though I might not have been able to field as many questions as I would usually like in this series, we certainly did manage to cover a wide array of games and questions!