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$10$ poker, $5$ of them are red, the rest of them are black. Randomly shuffled.

The gamer is asked to guess the color of each poker sequentially. After each guess, he will be told the correct color, i.e. the gamer will know the color of the $i$-th poker before he guesses the color of the $i+1$-th. If all of his guesses are right, he will win a prize of $\$1000$.

What is the optimal strategy? What is the fair price of this game?

Any thoughts? Thanks!

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    "Any thoughts?" - exactly. Do you have any thoughts on this problem?2017-01-09
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    What difference does it make telling the gamer about the colors as you go, when the only prize is for an all-correct answer? You just need to match whichever of the $\binom {10}{5}=252$ color patterns is correct.2017-01-09
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    @Joffan I've been trying to find a way there is an optimal strategy, but I just as it if phrased I can find none. Any idea of what OP might have thought?2017-01-09
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    @O.VonSeckendorff I think the optimal strategy, assuming this is a single game (or that there is no correlation between successive games), is simply to pick whatever ordering of $5$ red and $5$ black you want. If there is something more subtle going on over long runs of such games, that hasn't been hinted at so far in the specification. I answered the question in that spirit.2017-01-09

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It makes no difference to tell the gamer whether each color is correct as the game progresses, because the only payout is for an all-correct answer.

There are $\binom {10}{5}=252$ possible color mix orders from the specified $5$ red and $5$ black set given, so assuming that the gamer is sufficiently rational to pick a mix of $5$ red and $5$ black, they will have a $1$-in-$252$ chance of the payout.

The expected payout is therefore $\frac{\$1000}{252} \approx \$3.97$. As the setter, I would say that a $\$4$ game fee would probably give a "house advantage" that was slightly too low. Knowing human psychology I'm sure you could get away with a $\$5$ game fee because the $\$1000$ prize is big enough to quell serious calculation. :-)