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5/36......(1,5)(5,1)(2,4)(4,2)(3,3)
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To win, she must roll a 6 on her second turn.

Pr(x+y = 6) = 2*Pr(1 + 5) + 2*Pr(2+4) + Pr(3+3) = 2(1/6)^2 + 2(1/6)^2 + (1/6)^2 = 5/36

There are 2 ways to roll 1 and 5, and 2 ways to roll 2 and 4 because order matters.
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To win, she must roll a 6 on her second turn.

Pr(x+y = 6) = 2*Pr(1 + 5) + 2*Pr(2+4) + Pr(3+3) = 2(1/6)^2 + 2(1/6)^2 + (1/6)^2 = 5/36

There are 2 ways to roll 1 and 5, and 2 ways to roll 2 and 4 because order matters.
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Bunuel
In a dice game, the only way that a player wins on her second turn is if the sum of the dice she rolls on her second turn matches the sum from her first. On each turn, a player simultaneously rolls two, fair, six-sided dice. If Lindsey's total from her first turn is 6, what is the probability that she will win on her second turn?

A. 1/12
B. 5/36
C. 1/6
D. 7/36
E. 17/36

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Note that the question has been written to feel more complicated than it actually is. We have no relevance of first turn or second turn. The question simply says that two dice will be rolled and what is the probability that the sum rolled is 6?

We obtain 6 in 3 ways - 1 and 5, 2 and 4, 3 and 3.

Now assume that the dice are distinct i.e. one red and the other yellow. So you can get the sum of 6 in the following ways
1 on Red, 5 on Yellow or the other way around - 2 ways
2 on Red, 4 on Yellow or the other way around - 2 ways
3 on Red, 3 on Yellow - 1 way

Total 5 ways.

There are a total of 6*6 = 36 ways in which we can get different numbers on the roll of 2 distinct dice.

Hence, probability = 5/36
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Bunuel
In a dice game, the only way that a player wins on her second turn is if the sum of the dice she rolls on her second turn matches the sum from her first. On each turn, a player simultaneously rolls two, fair, six-sided dice. If Lindsey's total from her first turn is 6, what is the probability that she will win on her second turn?

A. 1/12
B. 5/36
C. 1/6
D. 7/36
E. 17/36

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VERITAS PREP OFFICIAL SOLUTION:

Since six things can happen on each individual roll, and she'll roll two dice, there are 6(6) = 36 total possibilities for the denominator. Then you need to count how many ways she can win, which are:

1 and 5

2 and 4

3 and 3

4 and 2

5 and 1

Accordingly, 5 of the 36 outcomes allow Lindsey to win, so her probability is 5/36.
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Can someone please explain why we count the outcome [3,3] only once? There are two die, [3,3] can occur two different ways, it seems to me.
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Hi joejackson,

It helps to think about the possible outcomes in the format "first die, second die":

To get a total of 6, you need to have one of the following outcomes (presented in 'first, second' format):

1 and 5
2 and 4
3 and 3
4 and 2
5 and 1

That's a total of 5 'winning' outcomes out of 36 possible outcomes.

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joejackson
Can someone please explain why we count the outcome [3,3] only once? There are two die, [3,3] can occur two different ways, it seems to me.

Yes, say the two dice are distinct in colour - say one is red and the other is yellow.
You can get 2 on the Red one and 4 on the Yellow one. This is different from 2 on the Yellow one and 4 on the Red one - fine
How about this? You get 3 on the Red one and 3 on the Yellow one. How is it any different from 3 on the Red one and 3 on the Yellow one? They are the same case, right? So you should count it only once.
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