## Poll

8 votes (53.33%) | |||

6 votes (40%) | |||

3 votes (20%) | |||

2 votes (13.33%) | |||

6 votes (40%) | |||

1 vote (6.66%) | |||

3 votes (20%) | |||

2 votes (13.33%) | |||

8 votes (53.33%) | |||

5 votes (33.33%) |

**15 members have voted**

Quote:Ace2To 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}

Quote:Ace2Starting with a new shooter’s 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.

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.

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 ...

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

Quote:Ace2Since 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)).

New shooter rolls R times until there’s a resolution of the fire bet (sixth point won or seven-out, whichever comes first). There’s a 1-P chance that the resolution was a seven-out and his turn is over...so no more rolls. There’s 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

That’s not intuitive?

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.

Quote:Ace2Thinking in terms of average total rolls per shooter L:

New shooter rolls R times until there’s a resolution of the fire bet (sixth point won or seven-out, whichever comes first). There’s a 1-P chance that the resolution was a seven-out and his turn is over...so no more rolls. There’s 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

That’s not intuitive?

Well, when you put it that way...