# Highest card expectation puzzle

There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.

Winning grants you \$50, and losing costs you only the \$10 you paid to play.

Would you accept this challenge?

PROPOSED SOLUTION (TRICK SOLUTION):

Divide into two halves of 50 cards each, say, deck-1 & deck-2. Now total outcomes are as follows:

1) Highest card in deck1 and second highest card in deck1

2) Highest card in deck2 and second highest card in deck2

3) Highest card in deck2 and second highest card in deck1

4) Highest card in deck1 and second highest card in deck2

These are the four mutually exclusive, equally likely and exhaustive events. This means each has a probability of 1/4

Now we rely on event fourth and turn up all the cards of deck2 and remember the highest card in deck2. Now we start turning up card in deck1 until we see card higher than the highest card we saw in deck2. Voila!

This means that probability of win = 1/4 and thus after paying \$10 each for 4 games (total \$40) player is likely to make a profit of \$10 (\$50 - \$40 = \$10). Then he should play.

Can you suggest any non-trick solution where one can just directly calculate the probability of win (negative binomial distribution?) and calculate the expected dollar value and show it is profitable.

• One can find a reasonable answer without it, but for optimal strategy please see the secretary problem. – André Nicolas Sep 29 '15 at 10:46

Firstly, your four outcomes are not all equally probable: the highest two cards will be in the same half with probability $49/99$, and in different halves with probability $50/99$. So outcomes $1$ and $2$ have probability $49/198$, and outcomes $3$ and $4$ have probability $50/198$.
Secondly, you are not necessarily lost in outcome 1 $-$ you will still pick the highest card if it is the first card in deck1 that is higher than all cards in deck2. To calculate the probabilities in this case, you might want to look at the link that André Nicolas posted in his comment.
With the help of that link, you can obtain an expression for the probability of success when deck2 contains $n$ cards, and then you can find which value of $n$ maximises this probability. It will be close to $1/e \approx 0.368$.