I moved off my analysis of the Replay craps side bet to a separate page and did a full math analysis via integral calculus, as opposed to the previous random simulation.

This page took at least a week to create and I consider probably the hardest side bet I've ever analyzed.

Google cant find it?

25% house edge?!Quote:Ace2https://wizardofodds.com/wizfiles/pdf/replay.pdf

jesus...

It’s fairly easy to algebraically calculate the odds of each event independent of other events. However, for all lines besides the top prize, you must calculate the probability of winning that prize AND not winning a higher prize, which makes the calculation considerably more complex and ideal for integral calculus, as the Wizard has done.

Perhaps one of the inclusion-exclusion gurus here would like to take a stab at this problem. I’ve thought about it from many angles without success

I have a question about the rules, which makes me also question the math.

What happens if a player is on a really hot roll and makes multiple numbers 3 or more times? Is the player paid for all qualifying wins?

I mean, I get that if the shooter makes four points of a certain number, then only the 4 or more gets paid - not both the 3 times and 4 or more times. But what about multiples?

I.E. If he makes the 5 three times, and the 8 four times, will the payout be 100 (for the four 8s), 95 (for the three 5s) or 195 (for both)?

This particular bet is resolved by a seven out or hitting the highest prize of four 4s or 10s. In the case of the highest prize being won (1 in 27,000 trials) it would be at the casino’s discretion to allow new bets to be made (same shooter). Same odds since it’s an entirely new bet. All other cases will be resolved by a seven-out and new shooter

In this Reply bet, it's possible (albeit rare) to have multiple numbers make 3 or more points.

I mean, unless I'm reading it wrong, this already has a less than 1% chance of paying. This question of multiple wins would be extremely rare.

And that rarity is the key. Perhaps they allow that multiple payout, since it will slightly reduce that outrageous house edge.

Quote:WizardMain page: https://wizardofodds.com/games/craps/side-bets/replay/.

Nice job Michael!

It’s not actually required, but you’d have to algebraically work through a staggering amount of states otherwise.Quote:odiousgambitcan someone distill for me why integral calculus is needed? to my 'gut' it just doesn't seem like a Craps analysis would ever require it

Here’s an example. There’s a 6 x 6 checkerboard and 6 checkers. In the first five columns there’s a checker in the first row and in the six column the checker is in the second row. We will roll a standard die...if a 4 is rolled we move the checker in column 4 up 1 row. We continue rolling until one of the checkers makes it to row 6. What is the probability that the checker in column 6 (with a 1 row head start) wins? Simple dice problem, right?

Solving this problem is quite similar to solving the Replay Bet since it’s about figuring which outcome will accumulate a certain amount of “wins” first. In other words, it’s a race, which is often ideal for a Poisson integration.

For the checkerboard problem you could list out all 5^5 * 4 = 12,500 states and tediously work through them with algebra. Or you could solve it quickly by integrating one short equation.

Extra credit: what’s the answer to the checkerboard problem?

Quote:SteenNice job Michael!

Thank you!

Quote:odiousgambitcan someone distill for me why integral calculus is needed? to my 'gut' it just doesn't seem like a Craps analysis would ever require it

You could do a Markov chain, but there are 5^6=15,625 states to the game. You would have to determine the probability of each state leading to each state. Very tedious.

I find it amazing that calculus can be used to solve a discrete problem like that.

Quote:Ace2Here’s an example. There’s a 6 x 6 checkerboard and 6 checkers. In the first five columns there’s a checker in the first row and in the six column the checker is in the second row. We will roll a standard die...if a 4 is rolled we move the checker in column 4 up 1 row. We continue rolling until one of the checkers makes it to row 6. What is the probability that the checker in column 6 (with a 1 row head start) wins? Simple dice problem, right?

Solving this problem is quite similar to solving the Replay Bet since it’s about figuring which outcome will accumulate a certain amount of “wins” first. In other words, it’s a race, which is often ideal for a Poisson integration.

For the checkerboard problem you could list out all 5^5 * 4 = 12,500 states and tediously work through them with algebra. Or you could solve it quickly by integrating one short equation.

Extra credit: what’s the answer to the checkerboard problem?

5426820388378871923714339111 / 16736565124800000000000000000 = Approximation: 0.3242493515194159

Integrate for x from 0 to infinity of (1-exp(-x/6)*(1+(x/6)+(x/6)^2/2+(x/6)^3/6))*(exp(-x/6)*(1+(x/6)+(x/6)^2/2+(x/6)^3/6+(x/6)^4/24))^5/6.

The function in the integral represents the probability at the moment the game ends the checker in column 6 will have advanced four or more times AND the other five checkers will have advanced 0 to 4 times each.

I disagree.Quote:Wizard

5426820388378871923714339111 / 16736565124800000000000000000 = Approximation: 0.3242493515194159

Integrate for x from 0 to infinity of (1-exp(-x/6)*(1+(x/6)+(x/6)^2/2+(x/6)^3/6))*(exp(-x/6)*(1+(x/6)+(x/6)^2/2+(x/6)^3/6+(x/6)^4/24))^5/6.

The function in the integral represents the probability at the moment the game ends the checker in column 6 will have advanced four or more times AND the other five checkers will have advanced 0 to 4 times each.

I believe you’ve calculated the probability that, at any given time:

1) column 6 has advanced >3

2) columns 1-5 have advanced <5

3) column 6 wins next roll

If column 6 has advanced >3, it’s already won and doesn’t need another win. Also, it can’t continue winning at every position above the fifth row, as your formula implies

Quote:Ace2I disagree.

I agree. I was thinking about this away from the computer and had another idea. Let me try again.

Approximation: 0.2992225544681486

exp(-x/6)*(x/6)^3/6*(exp(-x/6)*(1+(x/6)+(x/6)^2/2+(x/6)^3/6+(x/6)^4/24))^5*(1/6)

This should be the probability of the checker in column 6 advancing 3 more squares, the other five advancing 0 to 4, and the one in column 6 advancing next (and winning).

Well done Wizard. I owe you an O’DoulsQuote:WizardI agree. I was thinking about this away from the computer and had another idea. Let me try again.

6564030407855551 / 21936950640377856

Approximation: 0.2992225544681486Integrate for x from 0 to infinity of:

exp(-x/6)*(x/6)^3/6*(exp(-x/6)*(1+(x/6)+(x/6)^2/2+(x/6)^3/6+(x/6)^4/24))^5*(1/6)

This should be the probability of the checker in column 6 advancing 3 more squares, the other five advancing 0 to 4, and the one in column x advancing next (and winning).

Quote:Ace2Well done Wizard. I owe you an O’Douls

Thanks Ace! My first time winning a drink, if you can call O'Douls that.

I went to a soccer game in Barcelona and thought I was ordering a beer. Little did I know until after standing in a long line and taking a sip all they sold was non-alcoholic beer. *sheesh*

But they make up for that by having real beer in vending machinesQuote:WizardThanks Ace! My first time winning a drink, if you can call O'Douls that.

I went to a soccer game in Barcelona and thought I was ordering a beer. Little did I know until after standing in a long line and taking a sip all they sold was non-alcoholic beer. *sheesh*