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A and B in turns, throw a dice. If A gets a sum of 8 before

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A and B in turns, throw a dice. If A gets a sum of 8 before [#permalink] New post 08 Aug 2013, 06:27
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A and B in turns, throw a dice. If A gets a sum of 8 before B gets a sum of 9, then A wins. But if B gets a sum on 9 before A gets a sum of 8, then B wins. Find the chances of winning of A.

A. 31/36
B. 31/76
C. 5/36
D. 45/76
E. 1/9
[Reveal] Spoiler: OA

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Re: A and B in turns, throw a dice. If A gets a sum of 8 before [#permalink] New post 09 Aug 2013, 09:22
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PiyushK wrote:
A and B in turns, throw a dice. If A gets a sum of 8 before B gets a sum of 9, then A wins. But if B gets a sum on 9 before A gets a sum of 8, then B wins. Find the chances of winning of A.

A. 31/36
B. 31/76
C. 5/36
D. 45/76
E. 1/9

Dear PiyushK,
I find a few things unclear about this particular question. I point out that a good GMAT Quant question is very tight, and leaves not a single ambiguity --- it can be difficult to write a Quant question to those standards.

Question #1: you say "a dice" ---- that's grammatically incorrect and, hence unclear. The word "die" is singular and "dice" is plural. When I first glanced at the question, I assumed you meant --- each person was throwing two dice, and on each new throw, the question was: is the current sum an 8 or 9. That game could go on for quite some time, and calculating that is quite challenging (in that game, I found A has a 5/9 chance of winning). Upon re-reading, it appeared to me that you meant: each was throwing a single die. If this is correct, that leads me to ....

Question #2: Suppose A rolls a 6 on the first roll, then another 6 on the second roll, for a sum of 12. Does this count as winning? In other words, it is a matter of getting to a sum of 8 or more, before B gets to 9 or more? Or, do you mean that the game continues until, say, A has a string of consecutive rolls that has the exact sum of eight? (This latter question would be one of the hardest probability questions I have seen or imagined!)

Question #3: It's not explicit, but I am inferring that the rolls are not simultaneous, but rather, that A rolls once first, then B rolls once, then A rolls again, then B rolls again, etc. This was my take on the rolling procedure, but it is not made explicit in the question.

I can't even begin to think about finding an answer until I know the answer to all of these questions.

All of these are ambiguities that must be explicitly address in the way the problem is formulated. Does all this make sense?
Mike :-)
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Re: A and B in turns, throw a dice. If A gets a sum of 8 before [#permalink] New post 09 Aug 2013, 11:54
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PiyushK wrote:
A and B in turns, throw a dice. If A gets a sum of 8 before B gets a sum of 9, then A wins. But if B gets a sum on 9 before A gets a sum of 8, then B wins. Find the chances of winning of A.

A. 31/36
B. 31/76
C. 5/36
D. 45/76
E. 1/9

........................
Solution:

Let, A throws the dice 1st.
For the following combinations we can get an 8.
6+2
5+3
4+4
3+5
2+6
So the probability of getting an 8 in consecutively thrown 2 dices by A= 5/36 = Probability of winning of A and losing of B (B will not get any chance here because A scored the 8 1st, that’s why B is out of calculation) Like this : A B A

Next if B throws the dice 1st then this combinations will give a 9.
3+6
4+5
5+4
6+3
So the probability of getting a 9 in consecutively thrown 2 dices by B = 4/36
And probability of losing A = 1 – 5/36 = 31/36
Again, the probability of winning of B and losing of A = 4/36 × 31/36 = 1/9 × 31/36

So the conditional probability of winning of A = (5/36) / (1/9 × 31/36 + 5/36)
= 45/76 (Answer D)
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Re: A and B in turns, throw a dice. If A gets a sum of 8 before [#permalink] New post 25 Oct 2013, 05:01
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PiyushK wrote:
A and B in turns, throw a dice. If A gets a sum of 8 before B gets a sum of 9, then A wins. But if B gets a sum on 9 before A gets a sum of 8, then B wins. Find the chances of winning of A.

A. 31/36
B. 31/76
C. 5/36
D. 45/76
E. 1/9


Since language of the question is ambiguous, i prefer making a logical guess. Since A needs to get a total of eight and B needs to get a total of nine, A's task is slightly easier than B. BUT ONLY SLIGHTLY

So A has slight advantage over B.

A. 31/36 => Very Close to 1, this is highly biased towards A winning but A needs to have a slight bias. Hence Reject.
B. 31/76 => Slightly less than 0.5 but we want A to have slightly more than half probability of winning. Hence Reject.
c. 5/36 => Significantly less than 0.5 and close to 0. Again, reject since A needs to have slight edge over B.
D. 45/76 => Hmmmmm.... Slightly more than 0.5, almost what we wanted. Could be...
E. 1/9 => Again, significantly less than 0.5 and close to 0. Reject for same reason as option C.

Hence, Answer is D.
Re: A and B in turns, throw a dice. If A gets a sum of 8 before   [#permalink] 25 Oct 2013, 05:01
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