Probablity of three-of-a-kind or better in a roll of four dice

So. I took a math competition and one of the questions seemed simple enough.

"Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value"

Hm. Easy. Tried solving it. I couldn't get it. How can I solve this? I know you can probably use Combination, and if so, can you step by step explain how (never learned combinations yet). Or if it's possible without combination, can you explain how to solve it logically? thanks.

Imagine the four dice are rolled one after another. There are $6\times6\times6\times6=1296$ different possible outcomes. Of these, $6$ have all four dice showing the same value and $4\times6\times5=120$ have three dice showing the same value and the other die showing a different value. So the probability that at least three of the four dice show the same value is

$${6+120\over1296}={126\over1296}={7\over72}$$

The $4\times6\times5$ can be understood as follows: If exactly three dice have the same value, then the odd die can be any one of the $4$ dice, it can have any of $6$ values, leaving any of $5$ values for the three equal dice.

• @JoelReyesNoche, good grief, yes. I'll fix it. Thank you! – Barry Cipra Feb 28 '14 at 13:11

We want at least three of the four dice to have the same value. So there are two favourable conditions.

1. All the dice have the same value. Let this be event $A$.
2. Three out of the four dice have the same value. Let this be event $B$.

Let us calculate the probabilities for each of these events.

For event $A$, let's say the first die gets a number $x$. This could be any number from $\{1,2,3,4,5,6\}$, we don't care. The probability for this to happen is $1$. Now, we want the rest of the three dice to have exactly the same number. The probability for this to happen is $1 \times \dfrac{1}{6} \times \dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{216}$.

Fine, now let's calculuate the probability for event $B$. One of the dice can have any number $x$. Probability for this to happen is $1$. Now, two out of the three dice should have this same number, and the third dice shouldn't. The probability for this to happen is $1 \times \dfrac{1}{6} \times \dfrac{1}{6} \times \dfrac{5}{6}$. Now, remember that we considered this for only one tuple, say dice $(1,2,3)$ gets the same number. We want this for all tuples. There are a total of $\dbinom{4}{3}$ tuples. Thus, the total probability for event $B$ is

$$\dbinom{4}{3} \times 1 \times \dfrac{1}{6} \times \dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{20}{216}$$

Thus, our final answer is the sum of the probabilities of events $A$ and $B$. We get $\dfrac{1}{216} + \dfrac{20}{216} = \boxed{\dfrac{21}{216}} \approx 9.72\%$.

The number of combinations of rolls of the 4 die is, correctly, $6*6*6*6 = 6^4$.

The number of ways in which all 4 die roll the same value is 6 (either they all roll 1 or 2 or 3 and so on...)

The number of ways in which Exactly 3 die roll the same value is more complicated. Consider first the number of ways in which the first 3 die roll the same value but the fourth die rolls another. Label the die $W_1 , W_2, W_3, W_4$ for convenience. If $W_1 , W_2, W_3$ all roll 1's for example, then the number of combinations in which $W_4$ is not 1 is just 5. (The combinations are just (1,1,1,2), (1,1,1,3), (1,1,1,4) (1,1,1,5) and (1,1,1,6)). Since there are 6 possible values for the first 3 die, there are 6*5= 30 ways in which $W_1 , W_2, W_3$ are all the same and $W_4$ is different.

There are also many ways to choose 3 die out of the 4 namely, ${ 4 \choose 3} = 4$.

Putting it all together:

P(at least 3 dice are the same) = P(exactly 4 dice are the same) + P(exactly 3 dice are the same)

$$= \frac{6}{6^4} + \frac{4*5*6}{6^4} = \frac{1+4*5}{6^3} = \frac{21}{6^3} = \frac{7}{72} = 0.09722$$

Good Question!