0

My work so far:

Let $X$ be a random variable representing the number of "missing" faces from 10 dice rolls. For each $X = x$, we have $6 \choose x$ ways of picking the specific faces that do not appear in our 10 rolls, for example if $x=2$ then missing faces could be $\{3,6\}$ or $\{4,5\}$. For each of these choices, the probability that these faces don't appear is $ \left( \frac{6-x}{6} \right)^{10}$. Since the faces that don't appear are mutually exclusive, I conclude that $P(X = x) = \left( 6 \choose x \right) \left( \frac{6-x}{6} \right)^{10}$. This formula breaks down for x=0 and after summing from x=1 to x=5 on wolfram alpha I ended up with a a probability of 1.29. I am not sure where my logic went wrong, and any help would be greatly appreciated

Misha
  • 35
  • You are overcounting. A roll of $10\ 1$'s is counted in $X=1$ five times for the five missing faces, in $X=2$ six times for the six pairs of non-$1$ faces, and so on. – Ross Millikan Mar 12 '23 at 01:21
  • I am not sure I follow, $X=1$ means that 1 face is missing. which is one of the numbers from 1 to 6. For each of these cases, there is a 5/6 chance per dice roll that this face is missing, for a total probability of $6 * (5/6)^{10}$. Assuming you meant $X=5$, then I don't see how a roll of 10 ones is counted five times, could you please explain? – Misha Mar 12 '23 at 01:31
  • 1
    Your formula does not ensure that all the other numbers do appear; e.g. if the missing faces are ${3,6}$, the probability $\left( \frac{4}{6} \right)^n$ still allows for the possibility that some of the other four numbers may also not appear. – A.J. Mar 12 '23 at 01:54
  • If $x=2$ and the set of missing faces is ${3, 6}$, then the corresponding probability will be $m/6^{10}$, where $m$ is the number of all possible outcomes, consisting of $1, 2, 4, 5$'s, that must contain each of $1, 2, 4, 5$ at least once. This number is smaller than $(6-2)^{10} = 4^{10}$, which counts the number of possible outcomes consisting of $1,2,4,5$'s but not necessarily containing all the four letters simultaneously. (For example, $m$ does not consider the outcome $(1,2,2,1,4,4,1,1,2,2)$ since $5$ is missing, but $4^{10}$ does.) – Sangchul Lee Mar 12 '23 at 04:29
  • $X=1$ means that one face is missing, but if you select as $2$ the face that is missing you just make sure that all the rolls are among $1,3,4,5,6$. A roll of $10\ 1$'s satisfies this and gets counted. It will get counted again when you count the cases where $3$ is missing, and again $4,5,6$. This is a common oversight. When you demand that $2$ is missing you don't make sure that all the other faces are represented. – Ross Millikan Mar 12 '23 at 05:24
  • The question that I closed this as a duplicate of assumes specific values, but I posted a general answer that also applies to your question here. Note that exactly $k$ out of $6$ faces not appearing is equivalent to exactly $6-k$ out of $6$ faces appearing. – joriki Mar 13 '23 at 09:03

2 Answers2

0

The table below shows the counts of the equally likely rolls with various numbers of faces seen and unseen. With $n$ rolls, divide by $6^n$ to get probabilities, so divide the bottom row by $6^{10}$ to answer your question.

If $x$ is the number of unseen faces seen from $n$ rolls, and $p(x,n)$ is the probability, than you can use a recurrence or Stirling numbers of a second kind or inclusion-exclusion to find $p(x,n)$.

In your example of $x=2, n=10$, inclusion-exclusion to take account of fewer faces seen would give ${6 \choose 2}{4 \choose 0}(\frac46)^{10}-{6 \choose 2}{4 \choose 1}(\frac36)^{10}+{6 \choose 2}{4 \choose 2}(\frac26)^{10}-{6 \choose 2}{4 \choose 3}(\frac16)^{10}+{6 \choose 2}{4 \choose 4}(\frac06)^{10} = \frac{12277800}{6^{10}}$ about $0.203$, compared to your attempt of ${6 \choose 2}(\frac46)^{10}$ (just the first term) about $0.260$ which would be too high as it counts cases with more unseen faces many times.

 seen   0   1       2         3        4        5        6
 unseen 6   5       4         3        2        1        0
rolls                               
0       1   0       0         0        0        0        0
1       0   6       0         0        0        0        0
2       0   6      30         0        0        0        0
3       0   6      90       120        0        0        0
4       0   6     210       720      360        0        0
5       0   6     450      3000     3600      720        0
6       0   6     930     10800    23400    10800      720
7       0   6    1890     36120   126000   100800    15120
8       0   6    3810    115920   612360   756000   191520
9       0   6    7650    363000  2797200  5004720  1905120
10      0   6   15330   1119600 12277800 30618000 16435440
Henry
  • 157,058
0

Generalized Principle of Inclusion-Exclusion: Suppose that $E_1,\dots,E_n$ are events in a common probability space. Let $X=\sum_{i=1}^n {\bf 1}(E_i)$ be the random variable equal to the number of events that occur. Then $$ P(X=k)=\sum_{m=k}^n (-1)^{m-k}\binom{m}{k}\sum_{1\le i_1<\dots<i_m\le n}P(E_{i_1}\cap \dots \cap E_{i_m}) $$

See Generalised inclusion-exclusion principle for references and proofs.

Applying this, we see that the probability that there are exactly $x$ unseen faces is $$ \sum_{m=x}^{6}(-1)^{m-x}\binom{m}x\binom{6}m\left(\frac{6-m}{6}\right)^{10} $$

Mike Earnest
  • 75,930