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Full House

by manik11 » Mon Jan 18, 2016 5:04 am
In the game of Yahtzee, a "full house" occurs when a player rolls five fair dice and the result is two dice showing one number and the other three all showing another (three of a kind). 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?

A) 1/3

B) 5/54

C) 5/72

D) 5/216

E) 1/1296

OA : B
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by [email protected] » Mon Jan 18, 2016 9:43 am
Hi manik11,

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

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by manik11 » Mon Jan 18, 2016 10:11 am
Thanks a lot! Rich.
I have a follow up question. If the probability to get a A or a B i.e P(A) and P(B) is 1/6 I can't understand how 'AA BBB' would equal [(5/6)(1/6)(1/6)].
Could you please explain this in detail?
[email protected] wrote:
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
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by [email protected] » Mon Jan 18, 2016 3:54 pm
Hi manik11,

You have to be very careful to keep in mind what each calculation represents.

In "AA BBB"...

The two As represent the pair of numbers that you already have (so there's nothing to calculate there). Since you're dealing with two dice, the pair could be any of the following: 11, 22, 33, 44, 55, or 66..
The three Bs represent a number that appears 3 times that is NOT the number that "A" represents.

eg.
You could have 22 555, but you CANNOT have 22 222

So whatever number is represented by "A", the first "B" could be ANY of the remaining 5 numbers (that's 5/6). Since the second and third "Bs" have to match the first B though, they can each only be 1 possible number (that's 1/6 and 1/6).

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by manik11 » Mon Jan 18, 2016 11:31 pm
[email protected] wrote:Hi manik11,

You have to be very careful to keep in mind what each calculation represents.

In "AA BBB"...

The two As represent the pair of numbers that you already have (so there's nothing to calculate there). Since you're dealing with two dice, the pair could be any of the following: 11, 22, 33, 44, 55, or 66..
The three Bs represent a number that appears 3 times that is NOT the number that "A" represents.

eg.
You could have 22 555, but you CANNOT have 22 222

So whatever number is represented by "A", the first "B" could be ANY of the remaining 5 numbers (that's 5/6). Since the second and third "Bs" have to match the first B though, they can each only be 1 possible number (that's 1/6 and 1/6).

GMAT assassins aren't born, they're made,
Rich
Thanks for the explanation Rich...This one sure is really quirky!. Really challenging to understand the propmpt and get it correct within 2 minutes.
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by GMATGuruNY » Tue Jan 19, 2016 8:29 am
manik11 wrote:In the game of Yahtzee, a "full house" occurs when a player rolls five fair dice and the result is two dice showing one number and the other three all showing another (three of a kind). 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?

A) 1/3

B) 5/54

C) 5/72

D) 5/216

E) 1/1296
Alternate approach:
P = (good outcomes)/(all possible outcomes).

It is given that the first 2 rolls are THE SAME.
Thus, we must consider outcomes only for the LAST 3 ROLLS.

All possible outcomes:
Numerical options for the 3rd roll = 6. (Any of the 6 numbers on the die.)
Numerical options for the 4th roll = 6. (Any of the 6 numbers on the die.)
Numerical options for the 5th roll = 6. (Any of the 6 numbers on the die.)
To combine these options, we multiply:
6*6*6 = 216.

Good outcome 1: Among the last 3 rolls, exactly one is the same as the first 2 rolls
Number of options for this roll = 3. (Any of the remaining 3 rolls could be the same as the first 2 rolls.)
Numerical options for the remaining two rolls = 5. (Of the 6 numbers on the die, any number but that of the other 3 rolls.)
To combine these options, we multiply:
3*5 = 15.

Good outcome 2: The last 3 rolls are the same, but a different number from the first 2 rolls
Numerical options for the last 3 rolls = 5. (Of the 6 numbers on the die, any number but that of the first 2 rolls.).

Thus:
P(full house) = (15 + 5)/216 = 20/216 = 10/108 = 5/54.

The correct answer is B.
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