A gambler rolls three fair six-sided dice. What is the probability

Question

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

ChrisLele · Accepted Answer

This is a tricky question :-D

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

Bunuel · Answer

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} .

Answer: D.

Hope it's clear.

Aximili85 · Answer

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?

EvaJager · Answer

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.

anujkch · Answer

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)

premnath · Answer

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

BrentGMATPrepNow · Answer

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

Answer: D

Cheers,
Bret

BrushMyQuant · Answer

Given that A gambler rolls three fair six-sided dice and We need to find What is the probability that two of the dice show the same number, but the third shows a different number?

As we are rolling three dice => Number of cases =  6^3  = 216

Now, out of the three rolls lets pick the two rolls which show the same number. We can do that in 3C2 ways
=  \frac{3!}{2!*(3-2)!}  =  \frac{3*2!}{2!*1!}  = 3 ways

Now, out of the 6 numbers the two dice which show the same number can show any of these 6 numbers in 6 ways

The third die can show any number apart from the number which these two dice are showing in 5 ways. (5 numbers out of 6 except the number which the two dice are showing)

=> Total number of ways = 3 * 6 * 5

=> Probability that two of the dice show the same number, but the third shows a different number =  \frac{3*6*5}{216}  =  \frac{5}{12}

So,  Answer will be D 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

IcanDoIt23 · Answer

Thanks for the solution, Bunuel!

Could you please help me spot what is going wrong in below approach?

My process of solving it was like:

There are three places for 3 dices _ _ _

Since 2 different numbers are required out of 6, I would select them as 6C2
Now, I need to arrange them in 3 places (since there are 3 dices with 2 same numbers), as 3!/2!

Above 2 gives me the numerator, and the total combinations are 6*6*6

Therefore: 6C2 * (3!/2!) / 6*6*6 = 5/24

pancakeprotein · Answer

Fresh thoughts after doing this. I was able to get to the correct answer D, but almost made a silly mistake.

Thought process:
1.) Okay what would the chances for two dice to be 1 and the other to be not 1.

(1/6 * 1/6 * 5/6 ) * 3 (3 because there are 3 ways this could end up)

2.) I also realized that we also need to take into account all the other chances, such as 22 1 or 33 2,

So the final calculation would be

1/6 * 1/6 * 5/6 * 3/1 * 6/1 = 5/12

Answer is D

Bunuel · Answer

6C2 = 15 gives the number of different pairs without order. For example, the pair (1, 2) will only be counted once. However, in the case of XXY, X = 1 and Y = 2 is different from X = 2 and Y = 1. To account for this distinction, you should use 6C1 * 5C1 = 30, which includes both (1, 2) and (2, 1) as separate cases.

canadianmonkey3 · Answer

Sharing my two cents, my approach was the following:

I was unable to figure out a method with combinatorics so I used the following logic:

a) The first number can be anything, thus, 6/6 options.

b) The second number has to be the same as the first number, thus, 1/6 options.

c) The third number cannot be whatever the first number is, thus, 5/6 options.

d) These 3 numbers can also be arranged in 3 different ways, therefore the math is:

Calculation) 6/6 * 1/6 * 5/6 * 3 = 5/12

A gambler rolls three fair six-sided dice. What is the probability

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A gambler rolls three fair six-sided dice. What is the probability

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