Probability of rolling 6 when throwing k dice one by one vs all at once.

Question

I'm trying to wrap my head around combinations and permutations. Calculating the probability of rolling a 6 in these two senarios.

$$ P(\text {rolling a 6 in k throws}) = 1 - P(\text{not rolling 6 in k throws}) = $$

When rolling the dice one by one, using permutations with repetition ($n^k$), we get: $$ 1- \frac{5^k}{6^k} $$

when thrown all at once, since the dice are indistinguishable, we use combinations with repetition: $\binom{n+k-1}{k}$

$$ 1 -\frac{\binom{5+k-1}{k}}{\binom{6+k-1}{k}} = 1 - \frac{5}{5+k} $$

Shouldn't those two probabilities be the same? Where is the error in my reasoning?

Answer 1

Your combinations are not equiprobable. Try it with two coins. The possible combinations are $HH,TH,TT$ and the probabilities are $\frac 14,\frac 12,\frac 14$.

Answer 2

Take $2$ "normal" ($6-$sided, fair) indistinguishable dice, throw them and add their numbers. You get a value between $2$ and $12$. There was once a dispute between $2$ historically known people (forgot who) if the sums $11$ and $12$ should come up with the same frequency or not. The person who said they should come up with the same frequency argued that both $12$ and $11$ can occur as a sum of the two dice values in exactly one way: $12=6+6$ and $11=5+6$. Since the dice are indistinguishable, there is no difference between throwing a $5$ and a $6$ $(11=5+6)$ and throwing a $6$ and a $5$ $(11=6+5)$.

If you do this in reality, however, you will find that sum $11$ occurs roughly double as often as sum $12$. The reason is easy to understand if the dice are distinguishable (say one is red, the other blue). Then it is clear that the results of the red and blue dice are independent, so you get $36$ possible outcomes with equal probability. Now the sum $12$ is reached only in $1$ of those $36$ outcomes: the red die must show a $6$ as well as the blue die. To get sum $11$, there are now $2$ outcomes: red die shows $6$ and blue die shows $5$ and red die shows $5$ and blue die shows $6$.

The person who argued that sums $11$ and $12$ should come up with equal frequency made the error of assuming that the probability of coming up with a sum $x$ was proportional to the number of ways that number can be partitioned into the sum of two integers between $1$ and $6$. This assumption is incorrect, and weeding it (and other errors) about what constitutes an "equal probability" out is essential to getting these kinds of problems right.