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*