From Christopher Kyba comes a bippity boppity puzzle:
Each morning, your fairy godmother appears and gives you a chance to play a game. In this game, she deals 10 cards face down. Nine of the cards are winners, and one card is a loser. If you pick a winning card, you get a prize. You can then either take your prize and walk away or play again for the chance to win a second prize. But if you lose on that second play, you walk away with nothing and the game is over for the day. Each time you succeed, she invites you to play again under the same conditions (win yet another prize or lose everything).
What strategy maximizes the average number of prizes you win each day? And what is that average?
Extra credit: Suppose your fairy godmother deals N cards each day (instead of 10 cards), with N−1 winning cards and one losing card. Now what is your strategy, and how many prizes do you expect to win each day?
</div>The probability of winning exactly two prizes is:
$$\dfrac{N-1}{N} \cdot \dfrac{N-2}{N-1} = \dfrac{N-2}{N}$$
The Expected Value of prizes using this strategy is:
$$\dfrac{2\cdot(N-2)}{N}$$
The probability of winning exactly k prizes is:
$$\dfrac{N-1}{N} \cdot \dfrac{N-2}{N-1} ... \dfrac{N-k}{N-k+1}= \dfrac{N-k}{N}$$
The Expected Value of prizes using this strategy is:
$$\dfrac{k\cdot(N-k)}{N}$$
If $N$ is even, $k\cdot(N-k)$ is maximized when $k = \dfrac{N}{2}$, yielding an expected value of $\dfrac{N}{4}$.
If $N$ is odd, $k\cdot(N-k)$ is maximized when $k = \dfrac{N \pm 1}{2}$, yielding an expected value of $\dfrac{N^2-1}{4N} = \dfrac{N}{4} - \dfrac{1}{4N}$
If there are 10 cards, play 5 times, for an average win of 2.5 per day.