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Manager  Joined: 17 Sep 2011
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A gambler rolls three fair six-sided dice. What is the probability  [#permalink]

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9 00:00

Difficulty:   55% (hard)

Question Stats: 48% (01:45) correct 52% (01:40) wrong based on 100 sessions

### HideShow timer Statistics A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

A) 1/8
B) 5/18
C) 1/3
D) 5/12
E) 5/6

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Re: A gambler rolls three fair six-sided dice. What is the probability  [#permalink]

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3
This is a tricky question because there is a final twist that it is easy to overlook.

First off, we want to find the probability of rolling a pair of any number. There are six
pairs and a total of 36 different ways to roll a pair of dices: 6/36 = 1/6.

Next, we want to account for the third die - it cannot be the same as the first two,
so it can be any of five numbers out of 6: 5/6.

Therefore probability is 5/6 x 1/6 = 5/36.

It is tempting here to think that we have solved the problem. But if we read carefully, the
question does not say that the first two dice have to be the same and the third die has
to be different.

the question says a gambler rolls 3 dice. So the first two do not have to be pairs.
As long as 2 of the 3 dice are pairs and the last die is a different number. To find the
number of way this can occur, we use combinations formula: 3!/2!1! = 3.

We multiply this to the numerator to give us 3 x 5/36 = 15/36.

For some even trickier dice problems, check out our Magoosh GMAT blog post (these ones are really nasty!)

GMAT Probability: Difficult Dice Questions
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A gambler rolls three fair six-sided dice. What is the probability  [#permalink]

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abhi47 wrote:
A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

A) 1/8
B) 5/18
C) 1/3
D) 5/12
E) 5/6

Responding to a pm.

Combinations approach:

Total # of outcomes is $$6^3$$;

Favourable outcomes are all possible scenarios of XXY:
$$C^1_6*C^1_5*\frac{3!}{2!}=6*5*3=90$$, where $$C^1_6$$ is # of ways to pick X (the number which shows twice), $$C^1_5$$ is # of ways to pick Y (out of 5 numbers left) and $$\frac{3!}{2!}$$ is # of permutation of 3 letters XXY out of which 2 X's are identical.

$$P=\frac{Favourable}{Total}=\frac{90}{6^3}=\frac{15}{36}=\frac{5}{12}$$.

Hope it's clear.
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Re: A gambler rolls three fair six-sided dice. What is the probability  [#permalink]

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Hi, can somebody please explain a fundamental aspect of this question: I understand that the probability of getting 2 die with the same number AND one dice with a different number is = 6/6 x 1/6 x 5/6 = 5/36

BUT, beyond this, I get confused whether I should multiply this by 3 for the other 3 other combinations OR 3!?

Could somebody explain while accounting for different combinations, when we multiply by the # of combi VS the factorial of the # of combi?
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Re: A gambler rolls three fair six-sided dice. What is the probability  [#permalink]

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Aximili85 wrote:
Hi, can somebody please explain a fundamental aspect of this question: I understand that the probability of getting 2 die with the same number AND one dice with a different number is = 6/6 x 1/6 x 5/6 = 5/36

BUT, beyond this, I get confused whether I should multiply this by 3 for the other 3 other combinations OR 3!?

Could somebody explain while accounting for different combinations, when we multiply by the # of combi VS the factorial of the # of combi?

In this case, you should multiply by 3 because you can get two identical results and the third different in three different scenarios: AAB, ABA and BAA. Each scenario has the same probability of 5/36.
You can look at the 3 as 3C1, meaning how many choices we have to place the different result, or you can interpret 3 as 3!/2! because you have all the permutations of the triplet A,A,B, with A repeated twice. Obviously, as they should, they really give the same result.
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Re: A gambler rolls three fair six-sided dice. What is the probability  [#permalink]

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Ways in which a given set can appear = 3 i.e. AAB ABA BAA
For case AAB
Probability of getting a number on first dice = 6/6
Probability of getting the same number on second dice = 1/6
Probability of getting a different number on third dice = 5/6

Thus probability = 6/6 * 1/6 * 5/6 * 3

5/12 (D)
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Re: A gambler rolls three fair six-sided dice. What is the probability  [#permalink]

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2
My approach:

Total number of possible way = 6*6*6 = 216
3 discs will appear in any one of the following arrangements: AAA, AAB, ABC
where, AAA=All are same, AAB=Two are same, ABC=all three different

Now, total number of AAA = 6
Total number of ABC = 6*5*4 = 120
therefore, total number of AAA, ABC = 126
So, total number of AAB = 216-126=90
Probability of AAB= 90/216=5/12
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Re: A gambler rolls three fair six-sided dice. What is the probability  [#permalink]

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Top Contributor
abhi47 wrote:
A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

A) 1/8
B) 5/18
C) 1/3
D) 5/12
E) 5/6

Let's first calculate P(same, same, different)
P(same, same, different) = P(1st roll is ANY value AND 2nd roll matches 1st roll AND 3rd roll is different from first 2 rolls)
= P(1st roll is ANY value) x P(2nd roll matches 1st roll) x P(3rd roll is different from first 2 rolls)
= 6/6 x 1/6 x 5/6
= 5/36
So, P(same, same, different) = 5/36

However, this is not the only way to get 2 sames and 1 different.
There's also: same, different, same as well as different, same, same

Applying similar logic, we know that P(same, different, same) = 5/36
And P(different, same, same) = 5/36

So, P(2 same rolls and 1 different) = P(same, same, different or different, same, same or same, different, same)
= P(same, same, different ) + P(different, same, same) + P(same, different, same)
= 5/36 + 5/36 + 5/36
= 15/36
= 5/12

Cheers,
Bret
_________________ Re: A gambler rolls three fair six-sided dice. What is the probability   [#permalink] 25 Aug 2018, 06:11
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