Probability that 7 dice show 4 of one face and 3 other faces

Question

I tried using $$1*\frac16*\frac16*\frac16*\frac56*\frac46*\frac36*\frac{6!}{3!1!1!1!}$$ to show the number of ways to divide the different cases, but this doesnt seem right. Another approach is $$\frac16*\frac16*\frac16*\frac16*\frac56*\frac46*\frac36*\frac{7!}{4!1!1!1!}*6$$ I don't know which of them is the right one, please help me understand which is the right way?

Answer 1

There are $6^7=279936$ 7-dice rolls in total, and all the rolls that satisfy the required condition can be generated as follows:

• $\binom74=35$ ways to pick the four dice showing the same number $A$
• 6 ways to pick what $A$ is
• $5×4×3=60$ ways to pick the remaining three numbers $B,C,D$

Therefore we have $35×6×60=12600$ rolls with four of one number and three other numbers. The probability of such a combination occurring is $$\frac{12600}{279936}=\frac{175}{3888}=4.50\%$$ So neither of your two expressions is correct; the first one gives a probability of $15.43\%$ and the second one $27.01\%$. For problems like these it's better to work from a combinatorial viewpoint rather than use probabilites directly.

Answer 2

Choose a number so that exactly four of the dice have that number. This can be done in $6$ ways. Let this number be $A$.

Now choose three numbers so that the remaining three dice have these three distinct numbers. Supposing order does not matter for now, this can be done in ${{5}\choose{3}} = 10$ ways. Notice we choose out of five numbers, since one of the numbers is already reserved for the four dice we mentioned earlier. Let these three numbers be $B$, $C$, and $D$.

Then we see that there are $6^7 = 279936$ ways to roll seven dice, assuming order does matter.

Now we note that when we roll the seven dice, a successful outcome (i.e. four dice having same number, the other three having different numbers) will look like $AAAABCD$, ignoring order. Now is the time to account for order. There are $7 \times 6 \times 5 = 210$ ways to order these dice ($B$ can be in any of the seven spots, $B$ can be in any of the remaining six spots, and $C$ must take one of the final five spots).

Therefore, there are $6 \times 10 \times 210 = 12600$ successful outcomes out of $279936$ total outcomes. The probability is then $\frac{12600}{279936} = \frac{175}{3888}$.