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Mary and Joe are to throw three dice each. The score is the

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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 28 Oct 2018, 05:49
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 28 Oct 2018, 06:11
Bunuel wrote:
imania wrote:
would you love to see how attacked it?
if Joe is expected to outscore his friend, he should get these sums, 11,12,13...18
all possibilities are from 3 to 18
so : prob =8/16 equal to 1/2
PS. If you are wondering how I came to 3 as min because 1+1+1
and likewise 18 is max (6+6+6)


Unfortunately this approach is not right though for this particular case it gave a correct answer.

Consider this: if it were that Mary scored not 10 but 17 then Joe to outscore Mary should get only 18 and according to your approach as there are total of 16 scores possible then the probability of Joe getting 18 would be 1/16. But this is not correct, probability of 18 is (1/6)^3=1/216 not 1/16.

This is because not all scores from 3 to 18 have equal # of ways to occur: you can get 10 in many ways but 3 or 18 only in one way (3=1+1+1 and 18=6+6+6).

Hope it's clear.



Hi Bunuel,

Thanks for the clarification. However if the question would have asked for "number of ways Joe can get a sum greater than 13" What would have been your approach?

Please explain.
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Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 18 Apr 2019, 09:13
Bunuel wrote:
noboru wrote:
Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?


Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.
Expected value of three dice is 3*3.5=10.5.

Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.

P=1/2.


Dear Bunuel,

Pls help me where I have made mistake.

The total number of ways of score 10 or less than 10
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 22 May 2019, 02:08
Hey Bunuel

Just need a little clarification.

Assuming that the sum of outcomes on 3 dices follows a normal distribution curve with 10.5 as an average, so from this i can easily say that probability of having sum>10 will be 1/2. But if i use the same concept for another problem ( for eg. sum>6), i am not getting correct result.
Since for normal distribution the area b/w +sigma & -sigma (sigma=standard deviation) is 68% and i can calculate sigma, giving me the probability of an event.

Can you tell where did i go wrong?
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Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 10 Jun 2019, 22:07
Another simple way to address this is

For 3 dices we can get a sum ranging from3(all 1s) to 18 (all 6s).
Hence ..Total possible outcomes=16

Favorable outcomes= 11,12,13,14,15,16,17,18 ( total 8 possibilities)

Probability= 8/16 = 32/64 , Hence B
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Mary and Joe are to throw three dice each. The score is the   [#permalink] 10 Jun 2019, 22:07

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