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# 51(since there are 51 cards left of which 3 are jacks

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Unformatted text preview: 51 (since there are 51 cards left, of which 3 are jacks); the probability that the third card dealt is a jack, given that the first two are jacks, is 2 / 50. So the answer is 4 · 3 · 2 52 · 51 · 50 = 1 5525 7. (a) You roll a fair die four times. What is the probability that you see at least one 6? (b) You roll two fair dice twenty-four times. What is the probability that at least once, you see two 6s? (c) You bet someone, at even money, that upon rolling a fair die four times you will see at least one 6. You repeat this bet many times. Do you expect to win or lose money? (d) You bet someone, at even money, that upon rolling two fair dice twenty-four times you will, at least once, see two 6s. You repeat this bet many times. Do you expect to win or lose money? (Both bets mentioned here were bets that the Chevalier de Mere was interested in.) Solution. (a) The probability of not seeing a six on a single roll is 5 / 6. The probability of not seeing a six on any of four rolls is (5 / 6) 4 = 625 / 1296, or about 0 . 4822. The probability of seeing at least one six is 1- 625 / 1296, or about 0 . 5178. (b) The probability of not seeing two sixes when rolling two dice is 35 / 36. The proba- bility of not seeing two sixes on any of twenty-four rolls is (35 / 36) 24 , or about 0 . 5086. The probability of seeing at least one double-six is 1 minus this, or about 0 . 4914. (c) The probability of seeing at least one six is greater than 1 / 2, so in the long run you win. 2 (d) The probability of seeing at least one double-six is less than 1 / 2, so in the long run you lose. 3...
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51(since there are 51 cards left of which 3 are jacks the...

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