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Paul
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2 players each throw one fair die. In order to win, player A 11 Nov 2004, 18:53
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2 players each throw one fair die. In order to win, player A has to have a score greater than that of player B. What is the probability of player A winning?



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2 players each throw one fair die. In order to win, player A 11 Nov 2004, 20:50
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15/36 [1 min]

Total possible outcomes = 36
Same numbers (11, 22 etc) = 6
Difference = 36-6 = 30

In 50% of cases A will be greater then B. So, A will be greater then B in 15 outcomes.

Ans = 15/36
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pjk
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2 players each throw one fair die. In order to win, player A 12 Nov 2004, 00:56
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15/36 it it. Did it the long way, the above one is smarter though!
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Dharmin
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2 players each throw one fair die. In order to win, player A 12 Nov 2004, 02:38
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nice way to calculate buddy

15/36 should be the answer
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ninomoi
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2 players each throw one fair die. In order to win, player A 12 Nov 2004, 09:48
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Not sure I completely understand can someone please explain in itsy bitsy bits? :cry:
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2 players each throw one fair die. In order to win, player A 12 Nov 2004, 11:53
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ninomoi
Not sure I completely understand can someone please explain in itsy bitsy bits? :cry:
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Player will win if the combination of the numbers on the fair dice are as follows:
2,1; 3,1; 4,1;5,1; 6,1;
3,2; 4,2; 5,2; 6,2;
4,3; 5,3; 6,3;
5,4; 6,4;
6,5.

total numebr of outcomes= 6*6

favorable outcomes= 15.
hence , the probability of A winning= 15/36= 5/12.

hope this helps.
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2 players each throw one fair die. In order to win, player A 12 Nov 2004, 13:41
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OA is 15/36. Good job guys.
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ninomoi
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2 players each throw one fair die. In order to win, player A 12 Nov 2004, 18:48
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Thanks ruhi161084 a lot! :lol:



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