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In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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25 Jan 2014, 06:04
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In the game of Yahtzee, a "full house" occurs when a player rolls five fair dice and the result is two dice showing one number and the other three all showing another (three of a kind). What is the probability that a player will roll a full house on one roll of each of the three remaining dice if that player has already rolled a pair on the first two? A. 1/3 B. 5/54 C. 5/72 D. 5/216 E. 1/1296
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Re: In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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25 Jan 2014, 09:21
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guerrero25 wrote: In the game of Yahtzee, a "full house" occurs when a player rolls five fair dice and the result is two dice showing one number and the other three all showing another (three of a kind). What is the probability that a player will roll a full house on one roll of each of the three remaining dice if that player has already rolled a pair on the first two?
A. 1/3 B. 5/54 C. 5/72 D. 5/216 E. 1/1296 We have that the player rolled XX on the first two dice. To get a "full house" the next three dice must show YYY or XYY in any combination. The probability of YYY is 5/6*1/6*1/6 = 5/216. The probability of XYY is 3*1/6*5/6*1/6 = 15/216 (multiply by 3 because XYY can occur in three ways: XYY, YXY, YYX). The overall probability is 5/216 + 15/216 = 20/216 = 5/54. Answer: B.
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Re: In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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07 Jun 2014, 00:44
HOw can it be XX? Five are rolled , so XXXYY something like that .. Not clear



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Re: In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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07 Jun 2014, 03:18



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In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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04 Oct 2014, 07:57
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Bunuel If the same question were to say that 2 dice have been rolled already but we do not know the result/outcome [or simply what those two numbers are], now when the three dice are rolled what is the probability for a full house ? For the above case, is my calculation below right?1. If the first two numbers were similar , then we need to get any other number on all the three dice [aa / bbb] or [aa/abb  in any order] i.e 6/36 * 5 / 54 2. If the first two numbers were dissimilar, then we need to get one of those numbers on one die and the other number on the other two dice in any order [i.e. ab / aab*3abb*3] i.e. 30/36 * 6/216 Ans = [6/36 * 5/54 ] + [30/36 * 6/216 ] = 25/648



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Re: In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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02 Jul 2015, 01:08
Bunuel kindly enlighten me on the statement The probability of YYY is 5/6*1/6*1/6 = 5/216. how do you get 5/6 at the beginning?



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Re: In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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02 Jul 2015, 01:44



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Re: In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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22 Feb 2016, 22:26
i think its a conditional probability qustion
where first condition is a pair with 2 dice so we also should keep it in dinomenator
20/216 / 1/6(6/6*1/6)



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Re: In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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29 Aug 2016, 02:25
combination
XX YYY = 5/216 XX XYY (this y can be anything other than x ) so 5 (3 permutation) so 15
20/216



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Re: In the game of Yahtzee, a "full house" occurs when a player [#permalink]
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31 Dec 2017, 13:21
guerrero25 wrote: In the game of Yahtzee, a "full house" occurs when a player rolls five fair dice and the result is two dice showing one number and the other three all showing another (three of a kind). What is the probability that a player will roll a full house on one roll of each of the three remaining dice if that player has already rolled a pair on the first two?
A. 1/3 B. 5/54 C. 5/72 D. 5/216 E. 1/1296 VERITAS PREP OFFICIAL SOLUTION:Correct Answer: B If the player has already rolled a pair, there are two different scenarios under which the full house could be completed. Either the player rolls a third die that matches the first two rolls and then another pair, or the player rolls three matching die. If the player starts with AA, we'll call these scenarios ABB and BBB. Let's consider the ABB scenario first: the probability of rolling A (and matching the existing pair) is 1/6 The probability of rolling a different number, B, on the second die is 5/6 Finally the probability of matching B on the third roll is again 1/6 The probability of achieving this combination is 1/6 x 5/6 x 1/6, or 5/216 There are three different permutations through which the ABB combination could be achieved: ABB, BAB, and BBA. 5/216 x 3 = 15/216, the number of possibilities that make the first scenario true. The second scenario is simpler. The probability of rolling one die, B, that is different from the original AA pair is 5/6 The probability of rolling each of the remaining dice and matching B is 1/6 and then 1/6 again. So the probability of rolling BBB is 5/6 x 1/6 x 1/6 = 5/216 The two scenarios are not mutually exclusive, so add their probabilities together. 15/216 + 5/216 = 20/216, which reduces to 5/54
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