In the game of Yahtzee, a "full house" occurs when a player

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?

The player has already rolled XX on the first two dice. To get a "full house" the next three dice must show YYY or XYY in any combination.

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*3-abb*3]

i.e. 30/36 * 6/216

Ans = [6/36 * 5/54 ] + [30/36 * 6/216 ] = 25/648

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?

Say, the player rolled two 6's on the first two rolls: 66

In YYY case he must roll something other than 6 (so, the probability is 5/6) and then that number (Y) twice, thus 5/6*1/6*1/6.

Hope it's clear.

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)

combination

XX YYY = 5/216
XX XYY (this y can be anything other than x ) so 5 (3 permutation) so 15

20/216

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

Hi All,

This is a quirky probability question. Here's one way to solve this problem:

We're told that the first 2 dice already match, so we'll call them AA. It's the other 3 dice we have to deal with. To get a "full house", we need a set of 3 and a pair (the pair we start with COULD be part of the set of 3 though). We should map out the possibilities:

AA BBB = (5/6)(1/6)(1/6) = 5/216

AA ABB = (1/6)(5/6)(1/6) = 5/216

AA BAB = (5/6)(1/6)(1/6) = 5/216

AA BBA = (5/6)(1/6)(1/6) = 5/216

Total = 20/216 = 5/54

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

so, we do not need to care about the six possibilities of the pair.
5/216 is a trap.
It took 2 min to do this question correctly at the first time.
This is my second attempt, and I make it wrong.

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