Homework Help: Pair of dice probability

1. Oct 20, 2012

CAF123

1. The problem statement, all variables and given/known data
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.

3. 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?

2. Oct 20, 2012

tiny-tim

Hi CAF123!
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.

3. Oct 20, 2012

CAF123

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

4. Oct 20, 2012

tiny-tim

Try ∑ P(Wn)

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

5. Oct 20, 2012

CAF123

The event that A wins on his turn is just 1/9. What does this sum represent?

6. Oct 20, 2012

tiny-tim

That's P(W1).

What's P(W2) ?​

7. Oct 20, 2012

CAF123

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?

8. Oct 20, 2012

tiny-tim

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 !

9. Oct 20, 2012

CAF123

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 $$\frac{1}{9} \sum_{i}^{∞} (\frac{8}{9})^i \sum_{i}^{∞} (\frac{31}{36})^i$$

10. Oct 20, 2012

HallsofIvy

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).

11. Oct 20, 2012

tiny-tim

(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 = … ?)

12. Oct 20, 2012

CAF123

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.

13. Oct 20, 2012

tiny-tim

(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.

14. Oct 20, 2012

Ray Vickson

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