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Bill and Jane play a simple game involving two fair dice, each of
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30 Oct 2014, 08:52
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25% (02:40) correct 75% (01:59) wrong based on 211 sessions
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Re: Bill and Jane play a simple game involving two fair dice, each of
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08 Oct 2015, 21:53
Thoughtosphere wrote: Bunuel wrote: Tough and Tricky questions: Probability. Bill and Jane play a simple game involving two fair dice, each of which has six sides numbered from 1 to 6 (with an equal chance of landing on any side). Bill rolls the dice and his score is the total of the two dice. Jane then rolls the dice and her score is the total of her two dice. If Jane’s score is higher than Bill’s, she wins the game. What is the probability the Jane will win the game? (A) 15/36 (B) 175/432 (C) 575/1296 (D) 583/1296 (E) 1/2 I can't believe that this one is a 600 Level Question. It took me around 4 minutes to solve it. Here's my take: There are three possible outcomes in the game: 1. B Wins 2. J Wins 3. Draw So, \(1 = P(J) + P(B) + P(Draw)\) The probability of B winning or J winning the game is same. so, \(1 = 2*P(J) + P(Draw)\) So, we need to calculate the probability of a draw. A draw can happen when both will have the same sum. The possible sums are  2,3,4,5,6,7,8,9,10,11,12 Prob. of each sum is  2  1,1  1/36 3  1 2, 2 1  2/36 4  1 3, 2 2, 3 1  3/36 5  1 4, 2 3, 3 2, 41  4/36 6  1 5, 2 4, 3 3, 4 2, 5 1  5/36 7  1 6, 2 5, 3 4, 4 3, 5 2, 61  6/36 8  2 6, 3 5, 4 4, 5 3, 6 2,  5/36 9  3 6, 4 5, 5 4, 63  4/36 10  4 6, 5 5, 6 4  3/36 11  5 6, 6 5  2/36 12  6 6  1/36 Now, since both have to get the same sum, the probability will be same for both and the combined probability will be the square of the probabilities i.e. 2  (1/36)^2 3  (2/36)^2 etc. Squaring the probabilities and adding them will give 146/1296 So, P(D) = 146/1296 1 = 2*P(J) + P(D) 2*P(J) = 1150/1296 P(J) = 575 / 1296 Ans. C There are some calculations involved here but you can make them a bit more efficient: There are 11 possible sums and 7 is the middle sum. The probability of getting each sum equidistant from 7 will be the same. Probability of getting sum of 2 (= 1/36) will be the same as the probability of getting sum of 12 (= 1/36) because of the symmetry of outcomes. Probability of getting a sum of 3 = Probability of getting a sum of 11 = 2/36 and so on... Probability that both players get a sum of 2 = \((1/36)*(1/36) = 1^2/36^2\) Probability that both players get a sum of 3 = \((2/36)*(2/36) = 2^2/36^2\) and so on ... Note that the sum of squares of first n consecutive positive integers is given by n(n+1)(2n+1)/6 \(1^2/36^2 + 2^2/36^2 + ... + 5^2/36^2 = (1^2 + 2^2 + 3^2 + ..5^2)/36^2 = (5*6*11/6)/36^2 = 55/36^2\) \(6^2/36^2 = 36/36^2\) \(5^2/36^2 + 4^2/36^2 + ....1^2/36^2 = 55/36^2\) (from above) \(Total = (55 + 36 + 55)/36^2 = 146/36^2\) This is the probability of draw.
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Re: Bill and Jane play a simple game involving two fair dice, each of
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30 Oct 2014, 19:58
Bunuel wrote: Tough and Tricky questions: Probability. Bill and Jane play a simple game involving two fair dice, each of which has six sides numbered from 1 to 6 (with an equal chance of landing on any side). Bill rolls the dice and his score is the total of the two dice. Jane then rolls the dice and her score is the total of her two dice. If Jane’s score is higher than Bill’s, she wins the game. What is the probability the Jane will win the game? (A) 15/36 (B) 175/432 (C) 575/1296 (D) 583/1296 (E) 1/2 I can't believe that this one is a 600 Level Question. It took me around 4 minutes to solve it. Here's my take: There are three possible outcomes in the game: 1. B Wins 2. J Wins 3. Draw So, \(1 = P(J) + P(B) + P(Draw)\) The probability of B winning or J winning the game is same. so, \(1 = 2*P(J) + P(Draw)\) So, we need to calculate the probability of a draw. A draw can happen when both will have the same sum. The possible sums are  2,3,4,5,6,7,8,9,10,11,12 Prob. of each sum is  2  1,1  1/36 3  1 2, 2 1  2/36 4  1 3, 2 2, 3 1  3/36 5  1 4, 2 3, 3 2, 41  4/36 6  1 5, 2 4, 3 3, 4 2, 5 1  5/36 7  1 6, 2 5, 3 4, 4 3, 5 2, 61  6/36 8  2 6, 3 5, 4 4, 5 3, 6 2,  5/36 9  3 6, 4 5, 5 4, 63  4/36 10  4 6, 5 5, 6 4  3/36 11  5 6, 6 5  2/36 12  6 6  1/36 Now, since both have to get the same sum, the probability will be same for both and the combined probability will be the square of the probabilities i.e. 2  (1/36)^2 3  (2/36)^2 etc. Squaring the probabilities and adding them will give 146/1296 So, P(D) = 146/1296 1 = 2*P(J) + P(D) 2*P(J) = 1150/1296 P(J) = 575 / 1296 Ans. C
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Re: Bill and Jane play a simple game involving two fair dice, each of
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08 Oct 2015, 10:18
Bunuel wrote: Tough and Tricky questions: Probability. Bill and Jane play a simple game involving two fair dice, each of which has six sides numbered from 1 to 6 (with an equal chance of landing on any side). Bill rolls the dice and his score is the total of the two dice. Jane then rolls the dice and her score is the total of her two dice. If Jane’s score is higher than Bill’s, she wins the game. What is the probability the Jane will win the game? (A) 15/36 (B) 175/432 (C) 575/1296 (D) 583/1296 (E) 1/2 Hi Bunnel Can you please explain why the answer is not 1/2 here ? If we try to solve logically when both have fair chance of winning here why the probability of winning for each is not 1/2 ? Both have equal chance of winning here.



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Re: Bill and Jane play a simple game involving two fair dice, each of
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08 Oct 2015, 15:26
Hi Radhika11, According to the prompt, Jane only 'wins' if she scores HIGHER than Bill. She does NOT win if she TIES his score. While they would each have an equal chance of outscoring the other, the probability would NOT be 50/50. GMAT assassins aren't born, they're made, Rich
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Re: Bill and Jane play a simple game involving two fair dice, each of
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30 Mar 2016, 04:48
isn't there a faster method to calculate the answer in this case? p.s. it took me 5 minutes to get the correct answer



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Re: Bill and Jane play a simple game involving two fair dice, each of
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16 Jul 2016, 14:53
I am still unable to find a faster way of calculating the probability of a draw. All of the methods suggested above involve a lot of calculation and I strongly believe require more than 2 minutes. Bunuel do you have any suggestions ?



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Re: Bill and Jane play a simple game involving two fair dice, each of
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08 Jun 2018, 12:09
Consider the case of Bill He can have 6C1*6C1= 36 possibilities for the throw.
Same for Jane, She can have 6C1*6C1= 36 possibilities for the throw.
Thus total possible cases= 36*36= 1296
Number of ways both Bill and Jane get a 2: Bill: (1,1) and Jane: (1,1) = 1 This is same for the case when the both get 12: Bill: (6,6) and Jane: (6,6) = 1
Number of ways both Bill and Jane get a 3: Bill: (1,22,1) and Jane: (1,22,1) = 2C1*2C1= 4 This is same for the case when the both get 11: Bill: (5,66,5) and Jane: (5,66,5) = 4
Number of ways both Bill and Jane get a 4: Bill: (1,33,12,2) and Jane: (1,33,12,2) = 3C1*3C1= 9 This is same for the case when the both get 10: Bill: (4,66,45,5) and Jane: (4,66,45,5) = 9
Number of ways both Bill and Jane get a 5: Bill: (1,44,12,33,2) and Jane: (1,44,12,33,2) = 4C1*4C1= 16 This is same for the case when the both get 9: Bill: (3,66,35,44,5) and Jane: (3,66,35,44,5) = 16
Number of ways both Bill and Jane get a 6: Bill: (2,44,23,31,55,1) and Jane: (2,44,23,31,55,1) = 5C1*5C1= 25 This is same for the case when the both get 8: Bill: (2,66,25,33,54,4) and Jane: (2,66,25,33,54,4) = 25
Number of ways both Bill and Jane get a 7: Bill: (3,44,32,55,26,11,6) and Jane: (3,44,32,55,26,11,6) = 6C1*6C1= 36
Total number of ways for draw: 2*1 + 2*4 + 2*9 + 2*16 + 2*25 + 36= 146
Thus total ways in which either Bill or Jane wins: 1296146= 1150
Jane will win in half the cases whereas Bill will win in half the cases Hence Jane wins in 1150/2= 575 cases
Thus total probability 575/1296
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