Replay craps?
August 17th, 2020 at 8:14:55 PM
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Wiz tweeted:
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?
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?
Craps is paradise (Pair of dice).
Lets hear it for the SpeedCount Mathletes :)
August 17th, 2020 at 10:08:08 PM
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https://wizardofodds.com/wizfiles/pdf/replay.pdf
It’s all about making that GTA
August 18th, 2020 at 4:58:58 AM
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Main page: https://wizardofodds.com/games/craps/side-bets/replay/.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 18th, 2020 at 7:44:53 AM
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25% house edge?!Quote: Ace2https://wizardofodds.com/wizfiles/pdf/replay.pdf
jesus...
Craps is paradise (Pair of dice).
Lets hear it for the SpeedCount Mathletes :)
August 18th, 2020 at 7:58:20 PM
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I am trying to find a feasible way to calculate the edge for this bet using only algebra and inclusion-exclusion. Feasible meaning inclusion-exclusion formulas that don’t have 100 terms.
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
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
It’s all about making that GTA
August 19th, 2020 at 6:57:20 AM
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Cool alternative to the FireBet. Very interesting.
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)?
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)?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
August 19th, 2020 at 12:03:54 PM
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I assume he is paid for the highest win only.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 19th, 2020 at 1:06:55 PM
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Every side bet I can think of pays only the highest prize won, even if you necessarily qualify for lower prizes. Like the Fire bet, for instance
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
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
It’s all about making that GTA
August 19th, 2020 at 6:30:17 PM
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Oh, I totally get the concept of only paying the highest qualifier. But in all those other games mentioned, it's impossible to have the higher qualifier without first having the lower qualifier. I.E. You can't have a 6 point FireBet without already having the 5 point.
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.
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.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
August 21st, 2020 at 4:16:12 PM
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Quote: WizardMain page: https://wizardofodds.com/games/craps/side-bets/replay/.
Nice job Michael!
August 22nd, 2020 at 4:11:09 AM
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can 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
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
August 22nd, 2020 at 8:17:18 AM
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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?
Last edited by: Ace2 on Aug 22, 2020
It’s all about making that GTA
August 22nd, 2020 at 9:43:59 AM
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Quote: SteenNice job Michael!
Thank you!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 22nd, 2020 at 9:48:02 AM
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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.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 22nd, 2020 at 10:26:18 AM
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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.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 22nd, 2020 at 10:34:32 AM
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You’re better off playing the Britney Spears slots.
August 22nd, 2020 at 11:18:17 AM
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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
It’s all about making that GTA
August 22nd, 2020 at 12:21:13 PM
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Quote: Ace2I disagree.
I agree. I was thinking about this away from the computer and had another idea. Let me try again.
6564030407855551 / 21936950640377856
Approximation: 0.2992225544681486
Approximation: 0.2992225544681486
Integrate 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 6 advancing next (and winning).
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).
Last edited by: Wizard on Aug 22, 2020
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 22nd, 2020 at 12:26:26 PM
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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).
It’s all about making that GTA
August 22nd, 2020 at 12:43:21 PM
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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*
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 22nd, 2020 at 12:45:21 PM
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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*
It’s all about making that GTA
