# Probability of exactly three of a kind in a roll of 5 dice

The answer according to edx course HarvardX: FC1x Fat Chance: Probability from the Ground Up

is $$\frac{6*5*5* \left(^5_2\right)}{6^5} =\frac{1500}{6^5}$$

The remaining two dice can be same say this is valid favourable outcome 4,4,4,5,5.

But according to my reasoning the probability is much more than that,

My reasoning:

6 options for 3 of a kind

5 options for 4th dice

4 options for 5th dice(let us consider only the cases where 4th and 5th dice are different, for sake of showing that even with excluding certain favourable outcomes, namely in which 4th and 5th dice are same, my probability is higher than the course answer.)

therefore total number of ways

$$\frac{ 6*5*4*5!}{3!} = 2400$$

5! ways of arranging 5 items, divided by 3!, since 3 are of a kind.

$$p=\frac{2400}{6^5}$$

Which is greater than the probability calculated by the edx course, and I have not even considered the case when we allow 4th and 5th dice to have same number.

what is wrong with my reasoning?

• There are also two ways of arranging the dice that are not part of the three of a kind. Thus you should use $\frac{5!}{3!2!}$ (which is equal to ${5 \choose 3})$ instead of only $\frac{5!}{3!}$. May 8, 2018 at 14:48
• Also you need to consider all 5 options for the 5th dice to get $6*5*5*5!$ in the numerator. May 8, 2018 at 14:50

Assume the dice are distinguishable. Then there are $6^5$ possible outcomes since there are six possible outcomes for each of the five dice.

Three of a kind: There are $\binom{5}{3}$ ways for three of the five dice to display the same outcome and six possible outcomes those three dice could display. There are $\binom{5}{2}2!$ ways for the two remaining dice to display two of the five other possible values (as there are $\binom{5}{2}$ ways to select two of the remaining five values and $2!$ ways to arrange those values on the remaining two distinct dice), giving $$\binom{5}{3}\binom{6}{1}\binom{5}{2}2! = 1200$$ favorable outcomes.

Thus, the probability that three of a kind is obtained is
$$\Pr(\text{three of a kind}) = \frac{1200}{6^5}$$

Edit: Evidently, we were also supposed to consider all cases in which exactly three of the dice show the same outcome, so we must add the results for a full house.

Full house: There are $\binom{5}{3}$ ways for three of the five dice to display the same outcome and six outcomes those dice could display. There are five possible outcomes the other two dice could both display. Hence, there are $$\binom{5}{3}\binom{6}{1}\binom{5}{1} = 300$$ ways to obtain a full house. Thus, the probability of obtaining a full house is $$\frac{300}{6^5}$$

Total: The probability that exactly three of the dice display the same number is found by adding the probabilities for three of a kind and a full house, which yields $$\frac{1200}{6^5} + \frac{300}{6^5} = \frac{1500}{6^5}$$ as the given answer states.

Check: We know that the total number of outcomes is $6^5 = 7776$.

All different: There are $\binom{6}{5}$ ways of selecting five different outcomes and $5!$ arrangements of those outcomes on the dice. Thus, there are $$\binom{6}{5}5! = 720$$ ways to obtain five different numbers.

One pair: There are $\binom{5}{2}$ ways for two of the dice to display the same outcome and six possible outcomes those two dice could display. There are $\binom{5}{3}3!$ ways for the three remaining dice to display three of the remaining five values. Hence, there are $$\binom{5}{2}\binom{6}{1}\binom{5}{3}3! = 3600$$ ways to obtain a pair.

Two pairs: There are $\binom{6}{2}$ possible outcomes for the pairs. There are $\binom{5}{2}$ ways for two of the five dice to show the smaller of those outcomes and $\binom{3}{2}$ ways for two of the other three dice to show the larger of those outcomes. There are four possible outcomes for the remaining die. Hence, there are $$\binom{6}{2}\binom{5}{2}\binom{3}{2}\binom{4}{1} = 1800$$ ways to obtain two pairs.

Three of a kind: We showed above that there are $1200$ ways to obtain three of a kind.

Full house: We showed above that there are $300$ ways to obtain a full house.

Four of a kind: There are $\binom{5}{4}$ ways for four of the five dice to display the same outcome and six outcomes those dice could display. There are five possible outcomes for the remaining die. Hence, there are $$\binom{5}{4}\binom{6}{1}\binom{5}{1} = 150$$ ways to obtain four of a kind.

Five of a kind: All the dice must show the same outcome. There are six possible outcomes. Hence, there are $6$ ways to obtain five of a kind.

Total: The above cases are mutually exclusive and exhaustive. Observe that $$720 + 3600 + 1800 + 1200 + 300 + 150 + 6 = 7776 = 6^5$$

• It seems I was not clear enough in framing my question. Sorry for that. 4th and 5th dice can have same value. I have edited the question to be more clear. May 8, 2018 at 16:12
• I have edited my answer accordingly. May 8, 2018 at 16:34
• Thanks for the answer. I have selected the other answer, as it explained what was wrong with my reasoning. May 9, 2018 at 5:52

As pointed out in a comment, the main problem with your reasoning is that you're overcounting the possibilities for the two leftover dice: for instance, you're counting "three $1$s and a $2$ and a $3$" and "three $1$s and a $3$ and a $2$" among the sets of dice that can each be arranged in $\frac{5!}{3!}$ ways... hence double-counting arranged cases like $[1,3,2,1,1]$. For the cases where those dice are distinct (only), you need to divide by $2$. So the probability of three-of-a-kind with the two leftover dice distinct is half what you stated: $\frac{6\cdot 5\cdot 4 \cdot 5!}{2\cdot 3!}=1200$ cases out of $6^5$. Three-of-a-kind plus a pair leads to $\frac{6\cdot 5\cdot 5!}{2!\cdot 3!}=300$ additional cases out of $6^5$, for a total probability of $\frac{1500}{6^5}$.

It is simpler to solve it without distinguishing between the two possibilities for the leftovers: there are ${{5}\choose{3}}=10$ ways to select the dice participating in the three-of-a-kind, $6$ possibilities for their value, and $5^2$ possibilities for the values of the leftovers, for a net count of $10\cdot6\cdot 5^2=1500$.

• "you're overcounting the possibilities for the two leftover dice:" But when we consider the total sample space 6^5, we are considering,[1,1,1,2,3] as distinct from [1,1,1,3,2], aren't we? This is not rhetorical question, I am genuinely curious. Then why are we not considering them different in counting favourable outcomes? May 8, 2018 at 16:42
• I edited that sentence... hopefully it's more clear. Yes, $[1,1,1,2,3]$ and $[1,1,1,3,2]$ are distinct in the total sample space of $6^5$ cases; but your (correct) reasoning is that each set of values with three-of-a-kind and two distinct leftovers, like $\{1,1,1,2,3\}$, constitutes $5!/3!$ cases, and so you had better not count the set $\{1,1,1,2,3\}$ twice. May 8, 2018 at 17:33
• There are P(5,2) instead of $5^2$ ways for the leftovers, since in three of a kinda, the leftover two dices can't be duplicate.
– L Q
Aug 30, 2021 at 18:01