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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

Roland
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  • @lulu - this is called a paradox for a reason. Your suggested short-cut produces a different answer to the full calculation – Henry May 31 '19 at 15:53
  • @Henry Absolutely right. I just worked that out (and deleted my misleading comment accordingly). – lulu May 31 '19 at 15:53

1 Answers1

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The answer to your question comes from the line

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 $3/4$).

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:

Bet B’ has the same mean payoff as $B+1/4$: B’ pays $3/4$ more when the last throw is 6 (probability $1/6$), $1/4$ less when it is odd (probability $3/6$).

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

Henry
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  • If you last throw is 6 then B' will pay 2. How is this $3/4$ more than what B pays? – Roland May 31 '19 at 18:52
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    If you last throw is $6$ then $B'$ will pay $2$; this is $1$ more than what $B$ pays and so $3/4$ more than what $B+1/4$ pays. If you last throw is odd then $B'$ will pay $x$; this is $0$ more than what $B$ pays and so $-1/4$ more than what $B+1/4$ pays. – Henry May 31 '19 at 21:41
  • Where do you pull the number $1/4$ from(in "Bet B’ has the same mean payoff as $B+1/4$")? I understand the verification step but where does the $1/4$ originally come from? – Roland Jun 01 '19 at 11:23
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    @Roland: I did not pull $1/4$ out of the air, Berend Roorda did. It might be justified as the conditional probability that you roll a $6$ given that you roll a $6$ or an odd number – Henry Jun 01 '19 at 13:43