Roll a dice infinitely many times, what is the probability of getting a 5 before a 6

Question

My proof: For all possible events, based on the order of appearance between 5 and 6.l, they can be categorised into 2 groups.

$A:$ event with 5 appear before 6
$A^{c}:$ event with 6 appear before 5

And the union of $A$ and $A^{c}$ should be all possible events.

So $P(A) + P(A^{c}) = 1$

The last step: by symmetry, number of events that have 5 before 6 is the same as that for 6 before 5.

So $P(A) = P(A^{c}) = 0.5$

However, this last step I am not sure how to write out a formal proof to support my intuition for the symmetry. I think I need to evaluate the cardinality $A$ and $A^{c}$ but not sure how to. So I need help here.

Answer 1

to summarize the discussion in the comments:

For each $n\in \mathbb N$, let $p_n$ denote the probability that you see a $5$ before a $6$ in no more than $n$ tosses. Of course, by symmetry, $p_n$ is also the probability that you see $6$ before $5$ within $n$ tosses. These won't add to $1$, indeed, $1-2p_n$ is the probability you see neither $5$ nor $6$ in $n$ tosses. Thus $$1-2p_n=\left( \frac 46\right)^n\implies p_n=\frac 12-\frac 12\times \left( \frac 46\right)^n$$

Letting $n$ tend to $\infty$ then yields the desired result.

Answer 2

Different method:
$\displaystyle P(A_n)=\frac{1}{6}\left(\frac{4}{6}\right)^{n-1}$, where $A_n$ is the event that you get a 5 in the n-th roll and that you haven't gotten either 5 or 6 before the n-th roll. $\displaystyle\frac{1}{6}$ is the probability of getting a 5 in a roll, and $\displaystyle\left(\frac{4}{6}\right)^{n-1}$ is the probability of not getting 5 or 6 before the n-th roll. From here $\displaystyle P(A)=\sum_{i=1}^\infty P(A_n)=\frac{1}{6}\sum_{i=1}^\infty\left(\frac{4}{6}\right)^{n-1}=\frac{1}{6}\cdot\frac{1}{1-\frac{4}{6}}=\frac{1}{2}$.

Answer 3

Another method:

Let $x$ be the probability that 5 appears before 6 in an infinite number of tosses. Let's consider the first toss in this experiment of infinite tosses. For event $A$ to be true, you must either observe a 5 in the first toss, or observe neither 5 or 6 in the first toss and 5 must appear before 6 in the subsequent tosses. Note that the subsequent tosses is also an infinite sequence, and hence the probability for 5 to appear before 6 in those tosses is again $x$. Hence, we can write the following equation: $$\frac 1{6}+\frac 4{6}(x)=x$$ Solving the above gives the answer as $x=0.5$.

Answer 4

Ignore the scores that are not $5$ or $6$. So we have a two-sided die (i.e. a coin). So the probability of getting $5$ before $6$ is $1/2$.

Answer 5

The proof you suggest is almost correct, there are only a few details to iron out. At least the intuition is correct.

The first detail is that the set of all issues can actually be partitioned into these three events, not just the first two:

• Event A "5 appears before 6";
• Event B "6 appears before 5";
• Event C "There are no 5 and 6 at all".

You didn't mention C, presumably because you had the correct intuition that its probability is $P(C) = 0$. But it needs to be mentioned if you want to partition the universe into disjoint events.

Because it's a partition, $P(A) + P(B) + P(C) = 1$. And because $P(C) = 0$, it follows that $P(A) + P(B) = 1$, which you correctly stated. Note that $P(B) = 1 - P(A)$, and yet we cannot write $B = A ^c$ since there are three events in the partition.

Your last step, the symmetry, is correct. In fact I would argue that it is much better to use symmetry rather than to carry out any complicated calculations as was done in some other answers.

However, you used the words "cardinality" and "number". This is incorrect. You should use the word "probability" instead. By symmetry, events $A$ and $B$ have the same probability, $P(A) = P(B)$.

Cardinality arguments are not helpful here. It is true that $A$ and $B$ have the same cardinality, but that doesn't help at all. In fact, all three events $A$, $B$ and $C$ have the same cardinality $|A| = |B| = |C| = 2^{\aleph_0}$, and yet $C$ has a different probability: $P(A) = P(B) = \frac 1 2$ and $P(C) = 0$.