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.

This reads as each of the remaining three dice are thrown individually, which isn't the way I've played the game in the past.

If the three dice are thrown as a group, I wonder what the equivalent answer is ? I came up with 5/12

regor60 wrote:This reads as each of the remaining three dice are thrown individually, which isn't the way I've played the game in the past.

If the three dice are thrown as a group, I wonder what the equivalent answer is ? I came up with 5/18

Whether the remaining dice are rolled one at a time or all at once, the probability is the same.

To illustrate:
If two dice are rolled, what is the probability of getting exactly one 3?

One at a time:
P(1st roll is 3 and 2nd roll is not 3) = 1/6 * 5/6 = 5/36.
P(1st roll is not 3 and 2nd roll is 3) = 5/6 * 1/6 = 5/36.
Since either case is a favorable outcome, we ADD the fractions:
5/36 + 5/36 = 10/36 = 5/18.

Both at once:
Total number of possible rolls = 6*6 = 36.
Ways that the pair formed by the die on the left and the die on the right will yield exactly one 3:
3-1, 3-2, 3-4, 3-5, 3-6, 1-3, 2-3, 4-3, 5-3, 6-3 --> 10 ways.
Of 36 possible rolls, 10 have exactly one 3.
Thus, P(exactly one 3) = 10/36 = 5/18.

Whether the two dice are rolled one at a time or both at once, the result is the same:
P(exactly one 3) = 5/18.

regor60 wrote:This reads as each of the remaining three dice are thrown individually, which isn't the way I've played the game in the past.

If the three dice are thrown as a group, I wonder what the equivalent answer is ? I came up with 5/12

It doesn't actually matter: throwing the dice together doesn't affect the result of each individual die. (They're independent.) Source: I used to play (and lose!) a lot of Yahtzee as a kid myself.