March 23rd, 2020 at 6:04:53 PM
permalink

I must have reduced that formula. Here is a formula in expanded form:Quote:WizardCan you please expand a bit on why this works?

(((x/36)/e^(x/36)+1/e^(x/36)) * ((x/18)/e^(x/18)+1/e^(x/18)) * ((x/12)/e^(x/12)+1/e^(x/12)) * ((x/9)/e^(x/9)+1/e^(x/9)) * ((x*5/36)/e^(x*5/36)+1/e^(x*5/36)))^2 * 1/e^(x/6) * 1/6 dx

In this case it's much easier to calculate the chance of losing the bet since there is only one way (7) to lose and ten ways (pair of 2,3,4,5,6,8,9,10,11,12) to win.

We lose whenever a seven is thrown before a pair is rolled. For example, a 3 will be rolled every 36/2 = 18 rolls. Therefore the chance of it being rolled once is (x/18)/e^(x/18) and the chance of it being rolled zero times is 1/e^(x/18). The chance of it being rolled less than twice (no pair) is the sum of those: ((x/18)/e^(x/18)+1/e^(x/18)). Do that for all numbers 2-6, take the product, then square it since 8-12 have the same average waiting times. The term 1/e^(x/6) is the probability of 7 being rolled zero times.

The integral sums the probability over all time of being in the following state: numbers 2,3,4,5,6,8,9,10,11,12 have been rolled less than twice, a 7 has not been rolled, and then a 7 is rolled (multiply by 1/6 probability) to lose the bet.

The chance of losing is 1942942141 / 3673320192 =~ 0.529

It’s all about making that GTA

March 23rd, 2020 at 9:06:59 PM
permalink

Thank you. I think I see the light. For my own benefit, I put the following into integral-calculator.com/:

(exp(-x/36)*(1+x/36))^2*(exp(-x/18)*(1+x/18))^2*(exp(-x/12)*(1+x/12))^2*(exp(-x/9)*(1+x/9))^2*(exp(-5x/36)*(1+5x/36))^2*exp(-x/6)/6

Integrating from 0 to infinity, the probability of no winner is 1942942141 / 3673320192 = 0.5289335095893541.

This is the actual integral, before inserting the bounds of integration: (25*(-x^10-10*x^9-90*x^8-720*x^7-5040*x^6-30240*x^5-151200*x^4-604800*x^3-1814400*x^2-3628800*x-3628800)*e^(-x)+4110*(-x^9-9*x^8-72*x^7-504*x^6-3024*x^5-15120*x^4-60480*x^3-181440*x^2-362880*x-362880)*e^(-x)+290421*(-x^8-8*x^7-56*x^6-336*x^5-1680*x^4-6720*x^3-20160*x^2-40320*x-40320)*e^(-x)+11639700*(-x^7-7*x^6-42*x^5-210*x^4-840*x^3-2520*x^2-5040*x-5040)*e^(-x)+293947380*(-x^6-6*x^5-30*x^4-120*x^3-360*x^2-720*x-720)*e^(-x)+4903428960*(-x^5-5*x^4-20*x^3-60*x^2-120*x-120)*e^(-x)+54884392128*(-x^4-4*x^3-12*x^2-24*x-24)*e^(-x)+408146688000*(-x^3-3*x^2-6*x-6)*e^(-x)+1934615301120*(-x^2-2*x-2)*e^(-x)+5289581076480*(-x-1)*e^(-x)-6347497291776*e^(-x))/38084983750656.

Thanks again Ace!

(exp(-x/36)*(1+x/36))^2*(exp(-x/18)*(1+x/18))^2*(exp(-x/12)*(1+x/12))^2*(exp(-x/9)*(1+x/9))^2*(exp(-5x/36)*(1+5x/36))^2*exp(-x/6)/6

Integrating from 0 to infinity, the probability of no winner is 1942942141 / 3673320192 = 0.5289335095893541.

This is the actual integral, before inserting the bounds of integration: (25*(-x^10-10*x^9-90*x^8-720*x^7-5040*x^6-30240*x^5-151200*x^4-604800*x^3-1814400*x^2-3628800*x-3628800)*e^(-x)+4110*(-x^9-9*x^8-72*x^7-504*x^6-3024*x^5-15120*x^4-60480*x^3-181440*x^2-362880*x-362880)*e^(-x)+290421*(-x^8-8*x^7-56*x^6-336*x^5-1680*x^4-6720*x^3-20160*x^2-40320*x-40320)*e^(-x)+11639700*(-x^7-7*x^6-42*x^5-210*x^4-840*x^3-2520*x^2-5040*x-5040)*e^(-x)+293947380*(-x^6-6*x^5-30*x^4-120*x^3-360*x^2-720*x-720)*e^(-x)+4903428960*(-x^5-5*x^4-20*x^3-60*x^2-120*x-120)*e^(-x)+54884392128*(-x^4-4*x^3-12*x^2-24*x-24)*e^(-x)+408146688000*(-x^3-3*x^2-6*x-6)*e^(-x)+1934615301120*(-x^2-2*x-2)*e^(-x)+5289581076480*(-x-1)*e^(-x)-6347497291776*e^(-x))/38084983750656.

Thanks again Ace!

It's not whether you win or lose; it's whether or not you had a good bet.