Four dice are tossed let $N$ be the number of distinct outcomes. Find $E[N]$

Question

Four dice are tossed let $N$ be the number of distinct outcomes. Find $E[N]$

If we let $X$ be the number of distinct outcomes in a given toss.

E.g. if we get the toss $1,2,2,4$ then $X=3$

Calculating some probabilities:

$P(X=1) = 6 \cdot 6^{-4}$ since we have six ways we can end up with all four dice having the same value face-up.

Now according to a solution I have for this:

$\displaystyle P(X = 2) = 6^{-4}\cdot 6 \cdot 5 \cdot \binom{4}{2}$

This is where I am not so sure; why $4$ choose $2$ as opposed to $6 $ choose $2$ ?

$P(X= 4) = 6^{-4}\cdot 6 \cdot 5 \cdot 4 \cdot 3$

and finally $P(X = 3) $ I haven't calculated as I am not clear on $P(X=2)$.

The issue for me seems to be the combinatorial part of the argument. i.e. The number of ways each event occurs.

Answer 1

Alternatively, split into indicator variables and use linearity of expectation.

For $1\leq i\leq 6$, let $X_i$ be $1$ if we roll $i$ at least once, and $0$ otherwise. Then $$\mathbb E[X]=\sum_{i=1}^6\mathbb E[X_i]=6\mathbb E[X_1].$$ But $\mathbb E[X_1]=\mathbb P(\text{roll a }1)=1-\left(\frac56\right)^4$, so we get $\mathbb E[X]=\frac{671}{216}$.

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This method has the benefit of generalising easily. For a $d$ sided die rolled $n$ times, the expected number of distinct outcomes is $$\mathbb E[X]=d\left[1-\left(\frac{d-1}{d}\right)^n\right].$$

Answer 2

Expected value

$$E[X] = \sum_{k=1}^4 [k \times p(k)]$$

where, for each $k, ~p(k)$ will represent the probability of having exactly $k$ distinct numbers, and will equal

$$\frac{N\text{(umerator)}_k}{D\text{(enominator)}}.$$

For all values of $k$,

$$D = 6^4.$$

As the OP indicated,

$N_1 = 6 \times \binom{4}{4} = 6.$

To compute $N_2$ note that you can have either [3 of one number, 1 of a different number] or [2 of one number, 2 of another number].

In the first case, there are $(6 \times 5)$ ways of choosing the two numbers that show and $\binom{4}{1}$ ways of choosing the odd die out. In the second case, superficially, one would imagine that there are [again] $(6 \times 5)$ ways of choosing the two numbers and $\binom{4}{2}$ ways of grouping the 4 dice in pairs. This is wrong, as has already been indicated, because of overcounting. You have to divide the second case by $2$.

A safer way of enumerating the second case is to pretend that one of the dice is green, and the others are red. There are 3 potential partner dice that can be joined with the green die. Once the partnership is made, there are $6$ choices for the [green die + partner] and then $5$ choices for the other partnership.

Therefore
$N_2 = [6 \times 5 \times 4] + [6 \times 5 \times 3] = 210.$

To compute $N_3$, there must be 2 dice that roll the same number and two other dice that roll different numbers. There are $\binom{4}{2}$ ways of deciding which two dice get partnered. Then, there are $6$ choices for which number this partnership will roll.

For the other two dice, although it (wrongly) appears that an overcounting trap is present, similar to the second case for $N_2$, that is false. The easiest way to see this is to pretend that the two single dice are two colors, blue and green. After one of the 6 numbers is assigned to the partnership, there are then $5$ choices for the blue die and then $4$ [remaining] choices for the green die.

Therefore
$N_3 = \binom{4}{2} \times 6 \times 5 \times 4 = 720.$

To compute $N_4$, simply assume that each die has a different color (e.g red, green, blue, yellow) and that there are 6 choices for the red die, then 5 choices for the green die, then 4 choices for the blue die, and then 3 choices for the yellow die.

Therefore
$N_4 = 6 \times 5 \times 4 \times 3 = 360.$

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Before proceeding further, there is a sanity check. There are $6^4$ ways that dice of 4 different colors may be rolled. Therefore, $N_1 + N_2 + N_3 + N_4$ must equal $6^4 = 1296.$

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Final answer:

$$\frac{1}{D} \times \sum_{k=1}^4 [k \times N_k] ~=~ \frac{(1 \times 6) + (2 \times 210) + (3 \times 720) + (4 \times 360)}{1296}$$

$$=~ \frac{4026}{1296} ~=~ \frac{671}{216}.$$

Answer 3

Your outcomes will be a quadruple ($X{=}1$), one triple and one single ($X{=}2$), two pairs (also $X{=}2$), one pair and two singles ($X{=}3$), or four singletons ($X{=}4$).

To evaluate the individual probabilities, select faces, then places.

$$\begin{align}\mathsf P(X{=}1)&=\frac 1{6^4}\binom 61\binom 44 &&\text{four of a kind}\\&=6/1296\\\mathsf P(X{=}2) &=\frac 1{6^4} \left(\binom 62\binom42\binom 22+\binom 61\binom43\,\binom 51\binom 11\right)&&\text{two pair or three of a kind}\\&=210/1296\\\mathsf P(X{=}3) &=\frac 1{6^4} \binom 61\binom 42\,\binom 52\binom 21\binom 11&&\text{one pair and two singletons}\\&=720/1296\\\mathsf P(X{=}4)&=\frac{1}{6^4}\binom 64\binom 41\binom 31\binom 21\binom 11&&\text{four singles}\\&=360/1296\end{align}$$

If you plot a histogram, there would be a strong peak at $X{=}3$ with a little skew to the right, so we should anticipate the expected value to lie just above $3$.   And indeed...

$$\mathsf E(X)=\dfrac{6+2\cdot210+3\cdot720+4\cdot360}{1296}=\dfrac{4026}{1296}=\dfrac{671}{216}$$

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However, as stated by other answers, its easier to evaluate the expected count of faces shown.$$\begin{align}\mathsf E(X)&=\sum_{k=1}^6\mathsf P(X_k{>}0)\\&=6\cdot \left(1-\frac {5^4}{6^4}\right)\\&=\dfrac{6^4-5^4}{6^3}\\&\approx 3.1\end{align}$$

Answer 4

You're counting the number of distinct ways to:

1. Select two numbers to show up ($6\times5$) with order
2. Then, you're selecting which two dice to show the first (vs the second) number, hence $\binom{4}{2}$
3. Which for example, double-counts 3,3,4,4. Select 3, 4, select the first two dice to show 3. Select 4, 3, select the second two dice to show 4.
4. Additionally, some ways to have $X=2$ have not been counted.