When rolling 5 dice what is the probability (percentage of chance) or rolling a full house in 3 rolls?

Question

In three rolls, removing the dice you want to keep in order to make your full house what is the probabilty of getting the full house on the a) first roll, b) the second roll, c) the third roll.

Answer 1

"Removing the dice you want to keep" leaves a lot of scope for the player. Do we assume that the player chooses most wisely which dice to keep? Your question can be answered unambiguously for the first roll.

For example, if the player rolls 3 sixes, 1 five, and 1 four, the player may choose to keep just the sixes, or may choose to keep the sixes and the five. The probabilities of winning would be different in the two cases. If the player keeps just the sixes, the next roll would win if it came up doubles 1 through 5, with probability 5/36. If the player keeps the sixes and the five, the next roll would win if it comes up 5, with probability 1/6, slightly better.

For the first roll, we can argue this way:

The first die can come up anything, and the next two must match it, with probability (1/6)^2. Then the fourth die must be something different, with probability 5/6, and the last die must match that, with probability 1/6. Then we multiply the whole thing by the combinatoric formula for 5 things taken 3 at a time:

[(5 * 4) / 2] * (1/6)^2 * (5/6) * (1/6) = 50 / 6^4

Answer 2

your odds are 14-to-1 to roll a full house in 3 rolls

Answer 3

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