If I flip 10 dice, what's the probability I get 1 1, 2 2s, 3 3s, and 4 4s?

If I flip 10 dice, what's the probability I get 1 1, 2 2s, 3 3s, and 4 4s?

Here is what I tried: I realized that we can model this question in terms of bit strings. A 0 can represent a single dice roll. Since we roll 10 dice, we have 10 0s. A 1 can represent a separator between different values. Since a dice has 6 possible outcomes, we need 5 1s to separate these values. For instance, the bit string

010010001000011


represents rolling 1 1, 2 2s, 3 3s, and 4 4s. There are ${10 + 6 - 1}\choose{10}$ ways of forming bit strings with 10 0s and 6 - 1 = 5 1s, and the above string is only 1 of these ways. Thus, the answer to the original question is $\frac{1}{{10 + 6 - 1}\choose{10}} = \frac{1}{3003}$.

However, my concern is that my example with the bit strings does not perfectly biject to the original question. Each bit string is equally likely to occur. However, are all combinations of rolls of the 30 dice equally likely to appear? For instance, would rolling 10 1s be less likely than rolling 5 1s and 5 2s since the latter has more permutations?

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I think the proper term is roll. Coins are flipped. Dice are rolled. – Michael Chernick Sep 12 '12 at 20:37

Look at a smaller example: rolling one $1$ and two $2$’s with three dice. There are clearly $3$ ways to get this result: $122,212$, and $221$. Thus, the probability is $\dfrac3{6^3}=\dfrac1{72}$.

Now compute it using your bit-string method: your target is $01001111$, one string out of a possible $\dbinom83=56$, giving the incorrect result $\dfrac1{56}$. Thus, the method cannot be right, and your doubts are justified.

The problem is indeed the one that you identified at the end of the question. There is only one way to roll ten $1$’s; there are $\dbinom{10}5$ ways to roll five $1$’s and five $2$’s.

Added: To return to the larger problem, you want to count the ways of getting the desired outcome. There are $\dbinom{10}4$ ways to pick which $4$ dice come up $4$, $\dbinom63$ ways to choose which of the remaining $6$ dice come of $3$, and so on, so the total number of ways is

$$\binom{10}4\binom63\binom32\binom11=\frac{10!}{4!3!2!1!}=12,600\;,$$ the multinomial coefficient $$\binom{10}{4,3,2,1}\;.$$

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Let $X_i$ denote score of $i$-th dice. Let $N_k$ denote the number of digits $k$ appearing after 10 throws: $$N_k = \sum_{i=1}^{10} [ X_i = k ]$$ Obviously $N_1+N_2 +\cdots + N_6 =10$. Moreover, $N_k$ clearly follows binomial distribution with parameters $n=10$ and $p=\frac{1}{6}$, thus the tuple $(N_1, N_2, \ldots, N_6)$ follows the multinomial distribution with parameters $n=10$ and $p_k =\frac{1}{6}$.

The probability of the outcome of interest is then: $$\mathbb{P}(N_1=1,N_2=2,N_3=3,N_4=4,N_5=0,N_6=0) = \binom{10}{1,2,3,4} \frac{1}{6^{10}} \approx 0.02 \%$$

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Hmm, thanks, but doesn't ${10}\choose{1, 2, 3, 4}$ discount order, while $6^{10}$ considers order? – David Faux Sep 12 '12 at 20:17
@DavidFaux I am not sure what you mean. I am not basing my argument on combinatorial enumeration. But you can interpret it as follows: $6^{10}$ counts the total number of possible dice score configurations, i.e. 10 slots, 6 possibilities for each. Now one choose one such configuration, matching the description. Then there $\binom{10}{1,2,3,4}$ others, obtained through permutations. – Sasha Sep 12 '12 at 20:42
Binomial coefficients are often used to count strings (permutations). For a simple example, how many strings of length $10$, of $0$'s and/or $1$'s, have exactly three $0$'s? The $0$'s can be placed in $\binom{10}{3}$ ways. – André Nicolas Sep 12 '12 at 20:51
Thanks! That makes sense. – David Faux Sep 13 '12 at 2:04