AhrZee
AhrZee
Joined: Nov 12, 2016
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November 12th, 2016 at 9:13:42 PM permalink
Hi,

Does anyone have a derivation of the ALL bet probability?

My Brother-in-law saw someone win the All bet playing craps in Las Vegas. Since I'm taking a probability class at the university, he asked me "What are the chances of winning that bet?" The Wizard says the House Edge is ~7.99% and the payout is 174 to 1. This implies the probability of making the ALL roll before a 7 roll is about 0.0053. However, when I calculate the probability I get a different answer.

My calculation strategy is to consider the probability of finishing the ALL sequence on roll n. Then I use infinite series to sum over all possible n.

For example, say 2 is the last point needed. if you roll 2 on the n-th roll, then you had n-10 rolls without a 7 and without a 2. And since you're completing the ALL sequence, you had 10 rolls that were the remaining numbers in the sequence: i.e., 3, 4, 5, 6, 8, 9, 10, 11, 12. Furthermore, there are 9! ways to arrange these remaining rolls. So the chance of getting the ALL sequence in n rolls and finishing with a 2 is: 9! P(2) P(3) ... P(6) P(8) ... P(11) P(12) [P(no 7 and no 2)]^n-10.

I sum over all n (to infinity) and then add the probabilities for 3 being last, 4 being last, etc. My calculation yields 1.47E-5. Much different than the stated 0.0053.

Do you know where I can see a derivation of the ALL bet probability? Or can you suggest where my logic is wrong?

Thanks for the help!
Ahr
DJTeddyBear
DJTeddyBear
Joined: Nov 2, 2009
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November 13th, 2016 at 3:55:36 AM permalink
The Wiz may have done it thru simulation.

I'm almost positive simulation is the only method for the Fire Bet.
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁 Note that the same could be said for Religion. I.E. Religion is nothing more than organized superstition. 🤗
ThatDonGuy
ThatDonGuy
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November 13th, 2016 at 4:05:03 PM permalink
Quote: AhrZee

For example, say 2 is the last point needed. if you roll 2 on the n-th roll, then you had n-10 rolls without a 7 and without a 2. And since you're completing the ALL sequence, you had 10 rolls that were the remaining numbers in the sequence: i.e., 3, 4, 5, 6, 8, 9, 10, 11, 12. Furthermore, there are 9! ways to arrange these remaining rolls. So the chance of getting the ALL sequence in n rolls and finishing with a 2 is: 9! P(2) P(3) ... P(6) P(8) ... P(11) P(12) [P(no 7 and no 2)]^n-10.


This is incorrect, as it assumes that the first 9 rolls have to be 9 different numbers. I am under the impression that any numbers (besides 7, of course) can be rolled multiple times, as long as each number from 2 to 6 and from 8 to 12 is rolled at least once before a 7.

There is at least one brute force way of doing it. Here's what I came up with off the top of my head:
First, take a permutation of the numbers 2-6 and 8-12 - in this case, the numbers 2, 3, 4, 5, 6, 8, 9, 10, 11, 12.
The probability here is P(2 before any numbers after it and before 7) x P(3 before any numbers after it and before 7) x P(4 before any numbers after it and before 7) x ... x P(11 before 12 or 7) x P(12 before 7) = 1/36 x 2/35 x 3/33 x 4/30 x 5/26 x 5/21 x 4/16 x 3/12 x 2/9 x 1/7.
Calculate this for all 3,628,800 permutations of the 12 numbers, and add them up.

And then, I would do a Monte Carlo simulation to confirm the math...
AhrZee
AhrZee
Joined: Nov 12, 2016
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November 13th, 2016 at 10:15:18 PM permalink
Quote:

This is incorrect, as it assumes that the first 9 rolls have to be 9 different numbers.



Just to clarify: there is no assumption about where the 9 numbers appear, so long as they occur sometime before the n-th roll. They could be first, last or anywhere in between. And by including the 9! factor, we also allow them to be in any order. All we assume is that the 2 is last (and the player hasn't yet rolled a 7).

And then we assume the 3 is last... then the 4... etc. And then we add these 10 probabilities to get the total probability.

Thanks for your suggestion about how to calculate this probability. Your strategy seems to assume that the player will roll the needed 10 numbers in exactly 10 rolls. At least that's how I interpret "P(2 before any numbers after it and before 7)..." This seems to mean that 2 is the first number rolled, 3 is the second, etc. Then you suggest accounting for all the different arrangements of the 10 numbers, i.e., the 10! or 3,628,800 permutations.

In that case (and only that case): P(All) = (10!) P2 P3 P4 P5 P6 P8 P9 P10 P11 P12 = (3628800) (0.00000000000394) = 0.0000143.

This is still not equal to the stated probability of 0.005... because it assumes the All set is observed in the first 10 rolls. It doesn't take into account rolling them by, say, the 57th roll, or the 5007th roll. The total probability must include all these other ways to roll the All numbers without rolling a 7. (My derivation includes an infinite geometric series (that converges) to account for all these other ways.)

Hopefully there are more ideas out there... Because my strategy isn't producing the internet answer either!

Thanks again!
miplet
miplet
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November 14th, 2016 at 4:11:13 AM permalink
Read this thread to see how it was done for the Firebet. For the All bet, there are 2^10 states instead of 2^6. My spreadsheet is in Google Docs here.
“Man Babes” #AxelFabulous
ThatDonGuy
ThatDonGuy
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November 14th, 2016 at 10:14:27 AM permalink
Quote: AhrZee

Thanks for your suggestion about how to calculate this probability. Your strategy seems to assume that the player will roll the needed 10 numbers in exactly 10 rolls. At least that's how I interpret "P(2 before any numbers after it and before 7)..." This seems to mean that 2 is the first number rolled, 3 is the second, etc. Then you suggest accounting for all the different arrangements of the 10 numbers, i.e., the 10! or 3,628,800 permutations.


Not quite. What I say "3 before any numbers after it and before 7," I am saying, "3 before 4, 5, 6, 8, 9, 10, 11, 12 (the numbers that come after it in that permutation; otherwise it would belong to a different permutation, and you would be calculating some groups of rolls multiple times) or 7 (which would end the bet)" - there could be any number of 2s rolled (including zero) before that 3. Similarly, say, "8 before any numbers after it and before 7" means "8 before the next 9, 10, 11, 12, or 7," regardless of how many 2s, 3s, 4s, 5s, or 6s are rolled before the 8.

However, Miplet's state-based solution is probably the best way to calculate it.
Let N = 512 if you have rolled a 2 + 256 if you have rolled a 3 + 128 if you have rolled a 4 + ... + 2 if you have rolled an 11 + 1 if you have rolled a 12.
To make this easier to describe, N will be expressed as a binary number, and to avoid confusion, a lowercase "b" will be placed at the end of each binary number (so if you have rolled a 2, a 6, and an 11, N = 10001000010b)

Let P(N) be the probability of winning with that particular value N

P(1111111111b) = 1, since you have rolled all 12 numbers

If you have rolled every number except 12, N = 1111111110b; you have a 1/36 chance of rolling a 12 and getting to 1111111111b, a 1/6 chance of rolling a 7 and losing (so your probability of winning = 0), and a 29/36 chance of rolling a number you already rolled, leaving your N unchanged.
P(1111111110b) = 1/36 x P(1111111111b) + 1/6 x 0 + 29/36 x P(11111110b)
7/36 P(1111111110b) = 1/36 x 1
P(1111111110b) = 1/7

If you have rolled every number except 11, N = 1111111101b; you have a 2/36 chance of rolling a 11 and getting to 1111111111b, a 1/6 chance of rolling a 7 and losing (so your probability of winning = 0), and a 28/36 chance of rolling a number you already rolled, leaving your N unchanged.
P(1111111101b) = 2/36 x P(1111111111b) + 1/6 x 0 + 28/36 x P(11111101b)
8/36 P(1111111101b) = 2/36 x 1
P(1111111101b) = 1/4

If you have rolled every number except 11 and 12, N = 1111111100b; you have a 2/36 chance of rolling a 11 and getting to 1111111110b, a 1/36 chance of rolling a 12 and getting to 1111111101b, a 1/6 chance of rolling a 7 and losing (so your probability of winning = 0), and a 27/36 chance of rolling a number you already rolled, leaving your N unchanged.
P(1111111100b) = 2/36 x P(1111111110b) + 1/36 x P(1111111101b) + 1/6 x 0 + 27/36 x P(1111111100b)
9/36 P(1111111100b) = 2/36 x 1/7 + 1/36 x 1/4
P(1111111100b) = 15 / (36 x 7) = 5/84

Keep working backward until you get to P(0000000000b), which is the probability of winning after having rolled none of the numbers (which is where the bet begins).
P(0000000000b) = 1/36 x P(1000000000b) + 2/36 x P(0100000000b) + 3/36 x P(0010000000b) + 4/36 x P(0001000000b) + 5/36 x P(0000100000) + 1/6 x 0 + 5/36 x P(0000010000b) + 4/36 x P(0000001000b) + 3/36 x P(0000000100b( + 2/36 x P(0000000010b) + 1/36 x P(0000000001b)

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