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Pair of dice probability

  • Thread starter CAF123
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  • #1
CAF123
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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?
 

Answers and Replies

  • #2
tiny-tim
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Hi CAF123! :smile:
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. o:)

Try again. :smile:
 
  • #3
CAF123
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Hi tinytim.
Can you give me a hint to start?
 
  • #4
tiny-tim
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Try ∑ P(Wn)

where Wn is the event of A winning on his nth throw.
 
  • #5
CAF123
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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?
 
  • #6
tiny-tim
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The event that A wins on his turn is just 1/9.
That's P(W1).

What's P(W2) ?​
 
  • #7
CAF123
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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?
 
  • #8
tiny-tim
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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 ! :wink:
 
  • #9
CAF123
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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 ! :wink:
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 [tex] \frac{1}{9} \sum_{i}^{∞} (\frac{8}{9})^i \sum_{i}^{∞} (\frac{31}{36})^i [/tex]
 
  • #10
HallsofIvy
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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
tiny-tim
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(type "\left(" and "\right)", and they come out the correct size :wink:)

almost :smile:

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

(and starting at i = … ?)
 
  • #12
CAF123
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(type "\left(" and "\right)", and they come out the correct size :wink:)

almost :smile:

that's not the same as [itex] \frac{1}{9} \sum_{i}^{∞} \left(\frac{8}{9}\frac{31}{36}\right)^i [/itex] 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.
 
  • #13
tiny-tim
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(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. :smile:
 
  • #14
Ray Vickson
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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
 

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