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

ThatDonGuy
ThatDonGuy
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August 2nd, 2020 at 8:22:59 AM permalink
Quote: Ace2

To resolve it. To win or lose it. Must be all six points to win


I'll see if I can work up a computed solution later today. It's not quite as easy as it sounds, as you have to take into account the re-rolling of earlier points.
For example, if, e.g., E{6,8} means the expected number once you have already made the 6 and 8, then E{4,5,6} is the sum of:
1 -- the next roll
1/3 x E{4,5,6} -- rolling a natural or craps
1/12 x 2/3 x (the expected number of rolls to miss a point of 4)
1/12 x 1/3 x (the expected number of rolls to make a point of 4) x E{4,5,6}
1/9 x 3/4 x (the expected number of rolls to miss a point of 5)
1/9 x 1/4 x (the expected number of rolls to make a point of 5) x E{4,5,6}
5/36 x 6/11 x (the expected number of rolls to miss a point of 6)
5/36 x 5/11 x (the expected number of rolls to make a point of 6) x E{4,5,6}
5/36 x 6/11 x (the expected number of rolls to miss a point of 8)
5/36 x 5/11 x (the expected number of rolls to make a point of 8) x E{4,5,6,8}
1/9 x 3/4 x (the expected number of rolls to miss a point of 9)
1/9 x 1/4 x (the expected number of rolls to make a point of 9) x E{4,5,6.9}
1/12 x 2/3 x (the expected number of rolls to miss a point of 10)
1/12 x 1/3 x (the expected number of rolls to make a point of 10) x E{4,5,6,10}
Wizard
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Wizard
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August 2nd, 2020 at 9:50:03 AM permalink
Quote: Ace2

Starting with a new shooterís first roll, what the expected number of rolls to resolve the fire bet (all six points)?



I think this not an "easy" problem. That said, here is my ...


8.52412536496924

As a reality check, the expected rolls per shooter is 8.52551020408163, which should be a little more, as the player will still keep shooting after winning the Fire bet.

The difference is 0.00138483911240.

It can be found on my site the probability of winning the Fire Bet is 0.000162435.

0.00138483911240/0.000162435 = 8.52551020408163.

This makes sense as if the player wins the Fire Bet, he can expect to roll the same number of times from that point as a fresh shooter.

I probably should have done it that way to begin with, but I did it from scratch with a Markov Chain. Yes, next I'll do it with an integral, as soon as I figure out how to set it up properly.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
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August 2nd, 2020 at 10:32:04 AM permalink

As a rational number, I get 1,812,411,430,047,906,383,008,275,074,367,503,120,787,919,877 / 212,621,395,445,003,244,039,537,720,489,806,989,098,501,552

This is about 8.524, which is in line with simulation.

Wizard
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Wizard
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August 2nd, 2020 at 2:31:56 PM permalink
Quote: ThatDonGuy


As a rational number, I get 1,812,411,430,047,906,383,008,275,074,367,503,120,787,919,877 / 212,621,395,445,003,244,039,537,720,489,806,989,098,501,552

This is about 8.524, which is in line with simulation.



You can use my Calculator to solve that fraction to about 200 decimal places.
It gives and answer of ...
8.524125364969235306716226354263265894844054830893892380461387183547919960879226601357329869160257663444176437649024965837412598384033096488651707166597741954157645984860065445620151460834281272750147688402872739349839291833042135572671058657848177493067744
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
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August 2nd, 2020 at 3:17:31 PM permalink
What surprised me the most about this was, the expected number of rolls needed to resolve a point of 4 (or 10), the expected number needed to make a point of 4, and the expected number to miss a point of 4, are all the same (4), and that the same applies for points of 5/9 (3.6) and 6/8 (36/11).
Ace2
Ace2
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August 2nd, 2020 at 3:21:07 PM permalink
Congratulations Wizard. Thatís the correct answer

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
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
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August 2nd, 2020 at 4:05:02 PM permalink
Quote: Ace2

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.


So "the expected length of a shooter's run = the expected length of a fire bet + (the probability of winning a fire bet x the expected length of a shooter's run)."

That does not seem intuitive. Could you explain the reasoning?

It can also be expressed as "the expected length of a fire bet = the expected length of a shooter's run x the probability of losing a fire bet" (R = L - PL = L (1 - P)).
Ace2
Ace2
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August 2nd, 2020 at 5:13:49 PM permalink
Thinking in terms of average total rolls per shooter L:

New shooter rolls R times until thereís a resolution of the fire bet (sixth point won or seven-out, whichever comes first). Thereís a 1-P chance that the resolution was a seven-out and his turn is over...so no more rolls. Thereís a P chance that the resolution was a fire bet win. In that case the shooter starts another come out roll with L rolls.

So R + PL = L

Thatís not intuitive?
Itís all about making that GTA
Wizard
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Wizard
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August 2nd, 2020 at 5:32:20 PM permalink
Thinking out loud.

Probability fire bet wins

Integral: 1-(1-exp(-x/24))^2*(1-exp(-x/15))^2*(1-exp(-25*x/264))^2

Integral calculator

Expected rolls to achieve every point at least once = integral of 1-(1-exp(-x/36))^2*(1-exp(-2*x/45))^2*(1-exp(-25*x/396))^2 = 64.91860341630701 rolls. I'm not sure if that is right. Next step is to introduce the 7-out as a stopping event.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
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August 2nd, 2020 at 5:58:53 PM permalink
Quote: Ace2

Thinking in terms of average total rolls per shooter L:

New shooter rolls R times until thereís a resolution of the fire bet (sixth point won or seven-out, whichever comes first). Thereís a 1-P chance that the resolution was a seven-out and his turn is over...so no more rolls. Thereís a P chance that the resolution was a fire bet win. In that case the shooter starts another come out roll with L rolls.

So R + PL = L

Thatís not intuitive?


Well, when you put it that way...

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