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

Wizard
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Wizard
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August 3rd, 2020 at 8:35:53 AM permalink
Quote: Ace2

Congratulations Wizard. Thatís 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).
Last edited by: Wizard on Aug 3, 2020
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2
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August 3rd, 2020 at 10:00:07 AM permalink
Hereís an easy one.

Starting with a new shooter, whatís the expected number of rolls to win the fire bet ? The winner doesnít have to be that shooter
Itís all about making that GTA
Wizard
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Wizard
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August 3rd, 2020 at 10:20:28 AM permalink

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
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2
Ace2
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August 3rd, 2020 at 10:34:25 AM permalink
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

219 rolls for a bet that has about a 1 in 6,000 chance of winning ?
Itís all about making that GTA
ThatDonGuy
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August 3rd, 2020 at 10:52:22 AM permalink
Quote: Ace2

Hereís an easy one.

Starting with a new shooter, whatís the expected number of rolls to win the fire bet ? The winner doesnít 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

Wizard
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Wizard
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August 3rd, 2020 at 10:55:55 AM permalink
Quote: Ace2

219 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?
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2
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August 3rd, 2020 at 12:26:49 PM permalink
Quote: 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

Correct

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
Itís all about making that GTA
ssho88
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August 3rd, 2020 at 7:43:18 PM permalink
Anyone here can help to solve d2y/dx2 = 1/(1-y) ?

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 ?
Last edited by: ssho88 on Aug 3, 2020
ChesterDog
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August 3rd, 2020 at 8:09:05 PM permalink
Quote: ssho88

Anyone here can help to solve d2y/dx2 = 1/(1-y) ?



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

ssho88
ssho88
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August 3rd, 2020 at 8:10:06 PM permalink
Quote: ChesterDog

I 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 ?

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