### die throwing problem

Here’s a link to a subtle probability problem.

You throw a die until you get 6. What is the expected number of throws (including the throw giving 6) conditioned on the event that all throws gave even numbers?

The “obvious” answer is incorrect.

The correct answer can be brute-forced by computing probabilities of this sort:

If \(s\) is a sequence of length \(n\), then

\(P_n\triangleq\) = P(\(s\) ends in 6 and \(s\) all even) = \((2/6)^{n-1} (1/6)\)

(The total probability over all \(n\) is therefore \(Z=(1/6)/(4/6)=1/4\).)

The expected length of \(s\) is \(\sum_{n=1}^\infty n (P_n / Z)\)

\(= \sum_{n=1}^\infty n (2/6)^{n-1} (1/6) / (1/4) = \sum_{n=1}^\infty n (2/6)^{n-1} (2/3) = 3/2\)

This is interesting since premature conditioning on the 3 even sides of a die in every roll produces an (incorrect) expected length of 3.

There is already an elegant and convincing derivation of the correct answer attributed to Paul Cuff if you follow the links from the original article, but here’s an intuitive explanation for why the *incorrect* answer overestimates the expected length in such a way –

Notice that in \(P_n\) the probability \((1/6)\) of obtaining the final 6 has no effect on the expected length. The expected length depends entirely on the shape of \(P_n\) over \(n\). The calculation for the “incorrect” reasoning would have made the sum \(\sum_{n=1}^\infty n (2/3)^{n-1} (1/3) = 3\), and the difference is in the \((2/3)^{n-1}\) vs. \((2/6)^{n-1}\) term. So the existence of 1, 3, and 5 matter; they make longer prefixes of 2 and 4 even less likely, as more of the longer sequences that would have been valid ends up containing a 1, 3, or 5 instead and getting thrown away.

Note: There is also a good discussion on Quora.