## Poll

8 votes (47.05%) | |||

6 votes (35.29%) | |||

3 votes (17.64%) | |||

2 votes (11.76%) | |||

6 votes (35.29%) | |||

2 votes (11.76%) | |||

3 votes (17.64%) | |||

2 votes (11.76%) | |||

8 votes (47.05%) | |||

6 votes (35.29%) |

**17 members have voted**

Quote:Ace2Congratulations Wizard. That’s the correct answer

Thank you.

Quote:Since we already know the average roll length L of a shooters turn and the probability P of winning the fire bet, we can say that R + PL = L, where R is avg rolls to resolve the fire bet. Solve for R to get 8.524125 rolls. I think that qualifies as an easy puzzle.

Or take the integral over all time of this to get the same answer

e^(-196x/1671)*(1-((1-e^(-125x/6684))*(1-e^(-22x/1671))*(1-e^(-55x/6684)))^2) dx

1671/196 is average rolls for a seven out

6664/125 is average rolls for winning an 8

1671/22 is average rolls for winning a 9

6684/55 is average rolls for winning a 10

Exact answer

1812411430047906383008275074367503120787919877

/212621395445003244039537720489806989098501552

I just finished my calculus solution and agree with the above. Definitely not an "easy" problem. I prefer to express the integral as e^(-196x/1671)*(1-(1-e^(-125x/6684))^2*(1-e^(-22x/1671))^2*(1-e^(-55x/6684))^2).

Starting with a new shooter, what’s the expected number of rolls to win the fire bet ? The winner doesn’t have to be that shooter

Integral from 0 to infinity of (1-(1-e^(-125x/6684))^2*(1-e^(-22x/1671))^2*(1-e^(-55x/6684))^2) = apx. 219.1494672902

Quote:Wizard

Integral from 0 to infinity of (1-(1-e^(-125x/6684))^2*(1-e^(-22x/1671))^2*(1-e^(-55x/6684))^2) = apx. 219.1494672902

Quote:Ace2Here’s an easy one.

Starting with a new shooter, what’s the expected number of rolls to win the fire bet ? The winner doesn’t have to be that shooter

Somewhere around 52,477; it's the expected number of rolls to resolve a fire bet divided by the probability of winning

Let E be the number needed for the current shooter to win or lose the Fire Bet, and P the probability of a shooter winning the Fire Bet

First shooter wins: EP

Second shooter wins: (1-P) P x 2E

Third shooter wins: (1-P)^2 P x 3E

...

Sum = EP (1 + 2 (1-P) + 3 (1-P)^2 + ...)

= EP (1 + (1-P) + (1-P)^2 + ...)^2

= EP / P^2

= E / P

Quote:Ace2219 rolls for a bet that has about a 1 in 6,000 chance of winning ?

Never mind, I misunderstood the question. My answer would be appropriate for this question:

Assuming the shooter could never seven-out, how many rolls on average would it take to achieve the Fire Bet?

CorrectQuote:ThatDonGuy

Somewhere around 52,477; it's the expected number of rolls to resolve a fire bet divided by the probability of winning

Let E be the number needed for the current shooter to win or lose the Fire Bet, and P the probability of a shooter winning the Fire Bet

First shooter wins: EP

Second shooter wins: (1-P) P x 2E

Third shooter wins: (1-P)^2 P x 3E

...

Sum = EP (1 + 2 (1-P) + 3 (1-P)^2 + ...)

= EP (1 + (1-P) + (1-P)^2 + ...)^2

= EP / P^2

= E / P

It will take an average of 1/P trials to win and each trial will average E rolls. So E/P

Can also be expressed as (1/P - 1)L where L is average rolls per shooter

Can I solve it this way ?

(1-y) d2y = dx2

(y - y^2/2 + c1)dy = (x + c2)dx

y^2/2 - y^3/6 + c1(y) + c3 = (x^2/2) + c2(x) + c4

Correct ?

Quote:ssho88Anyone here can help to solve d2y/dx2 = 1/(1-y) ?

Quote:ChesterDogI cannot solve it, but at www.WolframAlpha.com, you can type in y''=1/(1-y) to get the general solution.

Can I solve it this way ?

(1-y) d2y = dx2

(y - y^2/2 + c1)dy = (x + c2)dx

y^2/2 - y^3/6 + c1(y) + c3 = (x^2/2) + c2(x) + c4

Correct ?