# Pair of dice probability

Gold Member

## Homework Statement

A and B alternate rolling a pair of dice, stopping either when A rolls the sum 9 or when B rolls the sum 6. Assuming that A rolls first, find the probability that the final roll is made by A.

## The Attempt at a Solution

A rolls a sum 9 on each roll with prob 1/9
B rolls a sum 6 on each roll with prob 5/36

Given that A wins, he will win on an odd number of turns. (since A starts)

Let E be the event that the game finishes on an odd number of turns
Then P(E) = (1/9)(1-5/36)(1/9)(1-5/36)....

Where do I go from here?

Related Precalculus Mathematics Homework Help News on Phys.org
tiny-tim
Homework Helper
Hi CAF123!
Let E be the event that the game finishes on an odd number of turns
Then P(E) = (1/9)(1-5/36)(1/9)(1-5/36)....
No, that's the probability that A wins on every throw, but generously pretends that he didn't, because he wants to let B win.

Try again.

Gold Member
Hi tinytim.
Can you give me a hint to start?

tiny-tim
Homework Helper
Try ∑ P(Wn)

where Wn is the event of A winning on his nth throw.

Gold Member
Try ∑ P(Wn)

where Wn is the event of A winning on his nth throw.
The event that A wins on his turn is just 1/9. What does this sum represent?

tiny-tim
Homework Helper
The event that A wins on his turn is just 1/9.
That's P(W1).

What's P(W2) ?​

Gold Member
That's P(W1).

What's P(W2) ?​
I thought that the probability of A winning on any of his turns is 1/9. Is this not correct?
If not, why not? Surely whether A wins is dependent only on what he throws and not B's result.
Or did I misunderstand something?

tiny-tim
Homework Helper
I thought that the probability of A winning on any of his turns is 1/9.
no, that's the probability of A winning on say his 10th turn, given that neither A nor B has already won

1/9 = P(Wn | neither A nor B has already won before the nth turn)

you want P(Wn)

for example, P(W2) is 1/9 times the probability that neither A nor B won on their first throws …

that's obviosuly less than 1/9 !

Gold Member
no, that's the probability of A winning on say his 10th turn, given that neither A nor B has already won

1/9 = P(Wn | neither A nor B has already won before the nth turn)

you want P(Wn)

for example, P(W2) is 1/9 times the probability that neither A nor B won on their first throws …

that's obviosuly less than 1/9 !
Ok, I think I understand now.
So P(W2) = (1/9)(1-1/9)(1-5/36)
P(W3) = (1/9)(1-1/9)^2 (1-5/36)^2..

Can I write this as $$\frac{1}{9} \sum_{i}^{∞} (\frac{8}{9})^i \sum_{i}^{∞} (\frac{31}{36})^i$$

HallsofIvy
Homework Helper
In order that P win on the first turn, he must roll a 9. The probability of that is 1/9. In order that P win on the second turn, he must roll anything except a 9 on the first roll, B must roll anything but a 6, and P must roll a 9 on. The probability of that is (8/9)(31/36)(1/9)= (2/9)(31/9)(1/9)= 62/729.

On any one turn, the probability that P does NOT roll a 9 and B does NOT roll a 6 is (8/9)(31/36)= 62/81. In order that P win on the nth turn both P and B must NOT roll the correct number the previous n- 1 turns and P must roll a 9 on the last turn- the probability of that is (62/81)n-1(1/9).

tiny-tim
Homework Helper
(type "\left(" and "\right)", and they come out the correct size )

almost

that's not the same as $\frac{1}{9} \sum_{i}^{∞} \left(\frac{8}{9}\frac{31}{36}\right)^i$ is it?

(and starting at i = … ?)

Gold Member
(type "\left(" and "\right)", and they come out the correct size )

almost

that's not the same as $\frac{1}{9} \sum_{i}^{∞} \left(\frac{8}{9}\frac{31}{36}\right)^i$ is it?

(and starting at i = … ?)
i would start at 1 here. Many thanks again for your help. Can I ask, in general, what advice you would offer when tackling probability problems. It is by far the hardest course I am doing this semester and I feel I have difficulties starting the problems, what area of probability to apply etc.. Any advice would be appreciated - thanks.

tiny-tim
Homework Helper
(isn't it i = 0?)

Probability problems are usually solved by rewriting the events in English first, so that you know clearly what the events are.

Then you can start translating them into maths.

Ray Vickson
Homework Helper
Dearly Missed
i would start at 1 here. Many thanks again for your help. Can I ask, in general, what advice you would offer when tackling probability problems. It is by far the hardest course I am doing this semester and I feel I have difficulties starting the problems, what area of probability to apply etc.. Any advice would be appreciated - thanks.
Often it is best to forget formulas for a while and concentrate on understanding the nature of the "sample space" underlying the problem. In this case, it would help to write down the first few instances where A wins:
Step 1: A wins---stop
Step 1: A does not win; go to step 2
Step 2: B does not win; go to step 3
Step 3: A wins---stop
Step 3: A does not win; go to step 4
Step 4: B does not win; go to step 5
Step 5: A wins--stop
Step 5: A does not win; go to step 6
etc., etc.

For these first few steps it is easy enough to write out the probabilities associated with the outcomes "A wins", and you can use the revealed pattern to develop a formula for the entire game.

RGV