Probability Generating Functions Question

[Tim 1234]

Homework Statement

In playing a certain game, your ability scores are determined by six independent rolls of three dice. After each set of six rolls, you are given the choice of keeping your scores or starting over.

(a) How many times should you expect to start over in order to get a set of ability scores with at least two scores that are 18?

Homework Equations

Binomial Probability Formula = (N choose K)P^kq^n-k

E(X)=1/p
From the PGF for Geometric Distribution

The Attempt at a Solution

Probability of rolling three sixes (18) is 1/216.

P(Rolling 18 ≥ 2 in 6 trials) = 1 - P(0 18s) - P(1 18)
= 1 - (215/216)⁶ - (6)(1/216)(215/216)⁵ ≈ .00031755
(Binomial Probability Formula)

Using the Geometric (.00031755) Distribution, E(X) = 1/p = 1/.00031755 ≈ 3149

Where E(X) is the expected number of repeats before getting ≥ 2 18s.

[/B]

 

[Ray Vickson]

Homework Statement

In playing a certain game, your ability scores are determined by six independent rolls of three dice. After each set of six rolls, you are given the choice of keeping your scores or starting over.

(a) How many times should you expect to start over in order to get a set of ability scores with at least two scores that are 18?

Homework Equations

Binomial Probability Formula = (N choose K)P^kq^n-k

E(X)=1/p
From the PGF for Geometric Distribution

The Attempt at a Solution

Probability of rolling three sixes (18) is 1/216.

P(Rolling 18 ≥ 2 in 6 trials) = 1 - P(0 18s) - P(1 18)
= 1 - (215/216)⁶ - (6)(1/216)(215/216)⁵ ≈ .00031755
(Binomial Probability Formula)

Using the Geometric (.00031755) Distribution, E(X) = 1/p = 1/.00031755 ≈ 3149

Where E(X) is the expected number of repeats before getting ≥ 2 18s.
[/B]

Please clarify: (1) If you decide to start over, do you discard your current scores and go back to 0? (It sound like you do.) (2) If on one roll of 6 dice you get a total of 36 , does that automatically count as two '18s'? (3) Are we allowed to make six rolls (of 6 dice in each roll), getting 6 each time, for a total of 36 = 18 ×2? (4) What has any of this to do with probability generating functions?

 

[Tim 1234]

If you start over, all of your current scores are discarded.

The trials consist of 6 independent rolls of three dice. So, you have a set of 6 sums of the roll of three dice.
So to get two "18s" you would need to roll three sixes twice in 6 attempts at rolling three dice.

 

[Ray Vickson]

If you start over, all of your current scores are discarded.

The trials consist of 6 independent rolls of three dice. So, you have a set of 6 sums of the roll of three dice.
So to get two "18s" you would need to roll three sixes twice in 6 attempts at rolling three dice.

Of course: you did say that, more or less.