The problem:
You throw a dice 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 following is the solution from Berend Roorda using fixed point updating. Can someone break it down to me? I didn't understand why Bet B’ has the same mean payoff as $B+1/4$?
Consider a bet $B$ on the number of throws: you get $T$ dollars when the $T$-th throw is $6$ after only $2$s and $4$s; as soon as a throw gives an odd number, the bet stops. Upfront you pay the mean payoff, call it $x$, but you get it back in the latter case. Then $x$ is the answer to the puzzle.
It can be determined as follows. The payoff is
$1$ if $6$, $x$ if $1$,$3$ or $5$, $B’$ if $2$ or $4$
Here B’ is the induced bet as of the second throw, “2 if 6, x if 1,3,or 5, B’’ if 2 or 4”, with B’’ the bet as of the third throw, etc.
Bet B’ has the same mean payoff as $B+1/4$: B’ pays $3/4$ more when the last throw is 6 (probability $1/3$), $1/4$ less when it is odd (probability $3/4$). So $x$ must be the mean payoff of
1 if 6, $x$ if 1,3,or 5, $x+1/4$ if 2 or 4
The answer to your question comes from the line
I am not sure this is quite correct given $\frac34 \times \frac13 - \frac14 \times \frac34 = \frac1{16}$. Nor do I see where the $\frac13$ or $\frac34$ come from. I would find the following statement more convincing:
You now have $\frac34 \times \frac16 - \frac14 \times \frac36 = 0$ and the intuitively obvious probabilities. The $\frac34$ comes from $1- \frac14$ and the $-\frac14$ from $0-\frac14$. You can also make similar statements for $B''$ and $B'+\frac14$ and for $B'''$ and $B''+\frac14$ etc.
Knowing Bet $B'$ has the same mean payoff as $B+1/4$, so $x+\frac14$ here, gives you $x = 1 \times \frac16 + x \times \frac36 + \left(x+\frac14\right) \times \frac26$, easily solved