Question 1191771: In the game of Yahtzee, five dice are thrown. On such a roll, Pat gets a 4,4,4,3, and 5. The rules allow her two more chances to roll any subset of the 5 dice. Hoping for a Yahtzee, she saves the first three dice and rerolls the dice that had a 3 and 5. Her strategy is:
a) if her 2nd roll is 4 and 4, then she stops and has achieved a Yahtzee
b) if her 2nd roll has a 4 and something else, she will keep the 4 and roll the other die one more time trying to get a 4
c) if her 2nd roll has no 4's then she will reroll those two dice one more time, trying to get both dice to be a 4
What is the probability she will end up with a successful Yahtzee?
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Sorry for the long question, I have no idea where to start and it's overwhelming... Thank you to whoever finds the time to help me!
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
She gets a Yahtzee in any of three ways; calculate the probability in each case.
(1) both dice show 4 on the second roll
P(4) on each die is 1/6, so P(4 on both dice) is (1/6)(1/6)=1/36
(2) one die shows 4 on the second roll and the remaining one die shows 4 on the third roll
P(4) on each die is 1/6; P(not 4) on each die is 5/6.
Either the first die can be a 4 and the other not, or the other way around.
P(4 then not 4) = (1/6)(5/6) = 5/36
P(not 4 then 4) = (5/6)(1/6) = 5/36
P(1 of 2 dice showing 4) = 10/36 = 5/18
P(4 on third roll) = 1/6
P(Yahtzee with one 4 on second roll and one on third roll) = (5/18)(1/6) = 5/108
(3) neither die shows 4 on the second roll and both show 4 on the third roll.
P(neither die shows 4 on second roll) = (5/6)(5/6) = 25/36
P(both dice show 4 on third roll) = (1/6)(1/6) = 1/36
P(Yahtzee with two 4's on third roll - (25/36)(1/36) = 25/1296
P(Yahtzee with three 4's on first roll) = (1/36)+(5/108)+(25/1296) = (36/1296)+(60/1296)+(25/1296) = 121/1296
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