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.