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Person A and person B throw three dice each. Score is the 28 May 2007, 19:52
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Person A and person B throw three dice each. Score is the sum of points on three dice. If person A scores 10 in his attempt. What is the probability that B will score more than A?
1. 24/64
2. 1/2
3. 9/16
4.10/16
5. 42/64



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gmatiscoming
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Person A and person B throw three dice each. Score is the 28 May 2007, 21:59
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i'm just starting my probabilty prep; how would i solve this problem without having to list out a sample space of 108 combinations?
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ricokevin
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Person A and person B throw three dice each. Score is the 28 May 2007, 23:30
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Possible score range is 3~18.

Since to win, B has to be in the range 11~18 and there are 8 different scores in that range. There are also 8 different scores in the range 3~10.

There is only one way to get 3 (1,1,1) as there is to get 18 (6,6,6).
Similarly, there are same number of ways to get 4 as there are to get 17, and so on.

Hence, the total number of combinations for the scores in both ranges are the same.

Thus, the answer is 1/2.

I'd go with B.
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ywilfred
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Person A and person B throw three dice each. Score is the 29 May 2007, 02:19
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Maximum score from 3 dies = 18
So B mus score 11,12,13,14,15,16,17,18

Number of possible scores = 3-18

So P = 8/16 = 1/2
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Person A and person B throw three dice each. Score is the 29 May 2007, 10:38
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Thanks for responses guys.
OA is 1/2
I like Ywilfred's solutions.
This question can NOT be solved without the trick in time pressure.
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ethan_casta
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Person A and person B throw three dice each. Score is the 29 May 2007, 11:08
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I wonder what the formal solution would be? I mean you'd have to just brute force it?

for ex., there is 1 way to get 18
there are 3 ways to get 17
there are 6 ways to get 16; 6,6,2 (this combo can come up 3 ways); 6,5,5 (this combo can come up 3 ways).
etc.

just sum up total ways to get a score of 11 - 18 and divide by 216?
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Person A and person B throw three dice each. Score is the 29 May 2007, 21:13
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ricokevin
Possible score range is 3~18.

Since to win, B has to be in the range 11~18 and there are 8 different scores in that range. There are also 8 different scores in the range 3~10.

There is only one way to get 3 (1,1,1) as there is to get 18 (6,6,6).
Similarly, there are same number of ways to get 4 as there are to get 17, and so on.

Hence, the total number of combinations for the scores in both ranges are the same.

Thus, the answer is 1/2.

I'd go with B.
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ricko...

very nice. i never thought to approach the data set as symmetrical... thank you.
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KillerSquirrel
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Person A and person B throw three dice each. Score is the 29 May 2007, 21:32
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ywilfred
Maximum score from 3 dies = 18
So B mus score 11,12,13,14,15,16,17,18

Number of possible scores = 3-18

So P = 8/16 = 1/2
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Perfect !

:-D
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LM
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Person A and person B throw three dice each. Score is the 29 May 2007, 21:37
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ywilfred
Maximum score from 3 dies = 18
So B mus score 11,12,13,14,15,16,17,18

Number of possible scores = 3-18

So P = 8/16 = 1/2
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AWESOME...VERY FAST AND QUICK METHOD..thanks a million!



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