Eight identical dice are rolled simultaneously. In how many possible outcomes each of the six numbers appears at least once?
I got result as $\left(\dfrac 16\right)^8.$ And I think I missed some part. Could you please help me?
[Math] Probability with 8 number of dice
dicediscrete mathematicsprobability
Related Solutions
This is a revision of my previous answer (see below).
Well, I was being stupid. Or at least, I guessed wrong about the definitions of $R_{6,7,8}$ and $R_{7,6,8}.$ It looks like we are meant to define events as follows:
- the sequence ends in $8$ with no $7$ having been rolled. Call this event $R_{6,8}.$
- the sequence ends in $8$ with a $7$ having been rolled after the first $6.$ Call this event $R_{6,7,8}.$
- the sequence ends in $8$ with a $7$ having been rolled before the first $6.$ Call this event $R_{7,6,8}.$
These definitions differ from my previous ones in the replacement of "last $6$" by "first $6.$"
An outcome in $R_{6,8}$ consists of
- a sequence of rolls containing no $6,$ $7,$ or $8$,
- a $6$,
- a sequence of rolls containing no $7$ or $8$,
- an $8$.
The probability that the first two stages will unfold in this way is $$ \sum_{n=0}^\infty \left(\frac{20}{36}\right)^n\cdot\frac{5}{36}=\frac{1}{16/36}\cdot\frac{5}{36}=\frac{5}{16}. $$ This can be reformulated in terms of conditional probability by considering that a sequence containing no $6,$ $7,$ or $8$ has zero probability of continuing forever. So a $6,$ $7,$ or $8$ eventually comes up with probability $1.$ Given that a $6,$ $7,$ or $8$ comes up, the probability that it is a $6$ is $5/16.$ Applying this reasoning to the last two stages as well, we get $$ \mathbb{P}(R_{6,8})=\mathbb{P}(6\mid6,7,8)\cdot\mathbb{P}(8\mid7,8)=\frac{5}{16}\cdot\frac{5}{11}=\frac{25}{176}. $$
Now an outcome in $R_{6,7,8}$ consists of
- a sequence of rolls containing no $6,$ $7,$ or $8$,
- a $6$,
- a sequence of rolls containing no $7$ or $8$,
- a $7$,
- a sequence of rolls containing no $7$ or $8$,
- an $8$.
We get $$ \mathbb{P}(R_{6,7,8})=\mathbb{P}(6\mid6,7,8)\cdot\mathbb{P}(7\mid7,8)\cdot\mathbb{P}(8\mid7,8)=\frac{5}{16}\cdot\frac{6}{11}\cdot\frac{5}{11}=\frac{150}{1936}. $$
Finally, an outcome in $R_{7,6,8}$ consists of
- a sequence of rolls containing no $6,$ $7,$ or $8$,
- a $7$,
- a sequence of rolls containing no $6,$ $7,$ or $8$,
- a $6$,
- a sequence of rolls containing no $7$ or $8$,
- an $8$.
We get $$ \mathbb{P}(R_{7,6,8})=\mathbb{P}(7\mid6,7,8)\cdot\mathbb{P}(6\mid6,7,8)\cdot\mathbb{P}(8\mid7,8)=\frac{5}{16}\cdot\frac{6}{16}\cdot\frac{5}{11}=\frac{150}{2816}. $$
Previous answer
[This answer uses different definitions of $R_{6,7,8}$ and $R_{7,6,8}$ than does the answer above, leading to the interchange of the values of these probabilities. Interestingly, this difference also seems to result in a formulation that does not lend itself to a conditional probability interpretation. Obviously the definitions in the answer above are the ones intended by the book, so I retract my statement that the book is wrong.]
The answer your book gives is right, but, unless I'm missing something, the solution method is incorrect. In particular, the probabilities they obtain for $\mathbb{P}(R_{6,7,8})$ and $\mathbb{P}(R_{7,6,8})$ are switched. Intuitively, $\mathbb{P}(R_{7,6,8})$ should be larger than $\mathbb{P}(R_{6,7,8})$ because in the event $R_{7,6,8}$ you may continue to roll $6$s after a $7$ has been rolled whereas in the event $R_{6,7,8}$ you may not. Unfortunately, the book's solution says that $\mathbb{P}(R_{7,6,8})$ is the smaller of the two. The solution I give below verifies that the intuition is correct, and I have checked this by simulation as well. I don't actually understand how to use conditional probability to solve this problem in the manner that the book was attempting, and would be interested if anyone could shed light on that.
Here's my approach: by symmetry, the probability of ending in $8$ is equal to the probability of ending in $6.$ So we compute the probability of ending in $8$, and then double it. To end in $8,$ an $8$ must be rolled exactly once, a $7$ may be rolled at most once, and a $6$ may be rolled any number of times as long as it is rolled at least once. There are three cases:
- end in $8$ with no $7$ having been rolled. Call this event $R_{6,8}.$
- end in $8$ with a $7$ having been rolled after the last $6.$ Call this event $R_{6,7,8}.$
- end in $8$ with a $7$ having been rolled before the last $6.$ Call this event $R_{7,6,8}.$
Now $$\begin{aligned} \mathbb{P}(R_{6,8})=&\sum_{m=0}^\infty\left[\mathbb{P}(\text{not 7 or 8})\right]^m\cdot\mathbb{P}(6)\cdot\sum_{n=0}^\infty\left[\mathbb{P}(\text{not 6, 7, or 8})\right]^n\cdot\mathbb{P}(8)\\ =&\sum_{m=0}^\infty\left(\frac{25}{36}\right)^m\cdot\frac{5}{36}\cdot\sum_{n=0}^\infty\left(\frac{20}{36}\right)^n\cdot\frac{5}{36}\\ =&\frac{1}{11/36}\cdot\frac{5}{36}\cdot\frac{1}{16/36}\cdot\frac{5}{36}\\ =&\frac{5}{11}\cdot\frac{5}{16}=\frac{25}{176}. \end{aligned}$$ Similarly, $$\begin{aligned} \mathbb{P}(R_{6,7,8})=&\sum_{\ell=0}^\infty\left[\mathbb{P}(\text{not 7 or 8})\right]^\ell\cdot\mathbb{P}(6)\cdot\sum_{m=0}^\infty\left[\mathbb{P}(\text{not 6, 7, or 8})\right]^m\cdot\mathbb{P}(7)\\&\cdot\sum_{n=0}^\infty\left[\mathbb{P}(\text{not 6, 7, or 8})\right]^n\cdot\mathbb{P}(8)\\ =&\sum_{\ell=0}^\infty\left(\frac{25}{36}\right)^\ell\cdot\frac{5}{36}\cdot\sum_{m=0}^\infty\left(\frac{20}{36}\right)^m\cdot\frac{6}{36}\cdot\sum_{n=0}^\infty\left(\frac{20}{36}\right)^n\cdot\frac{5}{36}\\ =&\frac{1}{11/36}\cdot\frac{5}{36}\cdot\frac{1}{16/36}\cdot\frac{6}{36}\cdot\frac{1}{16/36}\cdot\frac{5}{36}\\ =&\frac{5}{11}\cdot\frac{6}{16}\cdot\frac{5}{16}=\frac{150}{2816}. \end{aligned}$$ Finally, $$\begin{aligned} \mathbb{P}(R_{7,6,8})=&\sum_{\ell=0}^\infty\left[\mathbb{P}(\text{not 7 or 8})\right]^\ell\cdot\mathbb{P}(7)\cdot\sum_{m=0}^\infty\left[\mathbb{P}(\text{not 7 or 8})\right]^m\cdot\mathbb{P}(6)\\ &\cdot\sum_{n=0}^\infty\left[\mathbb{P}(\text{not 6, 7, or 8})\right]^n\cdot\mathbb{P}(8)\\ =&\sum_{\ell=0}^\infty\left(\frac{25}{36}\right)^\ell\cdot\frac{6}{36}\cdot\sum_{m=0}^\infty\left(\frac{25}{36}\right)^m\cdot\frac{5}{36}\cdot\sum_{n=0}^\infty\left(\frac{20}{36}\right)^n\cdot\frac{5}{36}\\ =&\frac{1}{11/36}\cdot\frac{6}{36}\cdot\frac{1}{11/36}\cdot\frac{5}{36}\cdot\frac{1}{16/36}\cdot\frac{5}{36}\\ =&\frac{6}{11}\cdot\frac{5}{11}\cdot\frac{5}{16}=\frac{150}{1936}. \end{aligned}$$
The probabilities of $R_{6,7,8}$ and $R_{7,6,8}$ are the reverse of those stated by the book. Of course, the book does not actually define $R_{6,7,8}$ and $R_{7,6,8}$ (or $R_6,$ $R_7,$ $R_8$), at least not in the excerpt provided. So maybe under some interpretation, the book's method is correct.
This probability is $1$ minus the probability that all outcomes are different, just like in the birthday problem.
The first throw can be anything (probability 1), the second must be unequal the first so probability $\frac{5}{6}$, the third unequal to the previous 2, so $\frac{4}{6}$, and so on till the fifth, which has two options left, so $\frac{2}{6}$. Multiply these, and take one minus that.
This comes to (if I computed correctly) to $\frac{49}{54} = 0.9074074074\ldots$, so I think your answer was correct in the first place (maybe an exact answer was asked for instead of a decimal approximation?) It's also the same as the answer by ADG.
Best Answer
The overall number of possible outcomes is of course $6^8$. We want to subtract off the number of outcomes that have at most five different numbers.
Now let's start with an easier problem: How many outcomes are there with no sixes? That is obviously $5^8$. Similarly there are $5^8$ outcomes with no fives, with no fours, and so forth. So it looks like the answer will be $$6^8 - 6\cdot 5^8$$ But this is not quite right. Consider any roll with no sixes and no fives. We have subtracted that roll off twice -- once because it has no sixes, once because it has no fives. We should have subtracted it only once. So we will add those rolls back in. There are $\frac{6\cdot 5}{2}=15$ ways to lack 2 numbers, and each such pair leads to $4^8$ possible outcomes. So our answer so far is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8$$ But consider any roll that contains no sixes, no fives, and no fours. We will have subtract this roll three times, and we will have added it three times (once because it has no sixes or fives, once because it has no sixes or fours, once because it has no fives or fours). So this roll thus far has not been accounted for; we have to subtract it. Now there are $\frac{6\cdot 5 \cdot 4}{6}=20$ ways to choose $3$ numbers out of $6$. There are $3^8$ rolls for each combination of $3$ missing numbers, o our answer thus far is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8 - 20 \cdot 3^8$$ And now we come to rolls with no sixes, fives, fours, or threes. The same sort of reasoning shows that each of these was subtracted twice in total, so we need to add these back. And finally, consider wolls that are just six of the same number. We will have subtracted and added these the same number of times, so we will need to subtract these once more.
The final number of outcomes is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8 - 20 \cdot 3^8 + 15 \cdot 2^8 -6\cdot 1^8 = 191520$$
And the probability of rolling 8 dice and getting at least one of each number is $$\frac{191520}{6^8} = \frac{665}{5832} \approx 0.114 $$
Thus the counterintuitive result is that you are likely to be missing at least one number when you roll 8 dice. In fact, in order to have better than a 50% chance of getting one of each number, you would have to roll 13 dice!