Question

# We play a game with a deck of 52 regular playing cards, of which 26 are...

1. We play a game with a deck of 52 regular playing cards, of which 26 are red and 26 are black. They’re randomly shuffled and placed face down on a table. You have the option of “taking” or “skipping” the top card. If you skip the top card, then that card is revealed and we continue playing with the remaining deck. If you take the top card, then the game ends; you win if the card you took was revealed to be black, and you lose if it was red. If we get to a point where there is only one card left in the deck, you must take it. Prove that you have no better strategy than to take the top card – which means your probability of winning is 1/2.

Hint: Prove by induction the more general claim that for a randomly shuffled deck of n cards that are red or black – not necessarily with the same number of red cards and black cards – there is no better strategy than taking the top card.

In case the top card is taken then the probability of winning is 1/2.

Foregoing the first card means that either a black or a red card is left out.

Then the probability that the next card is a winning card is:

P(first black card followed by a black card) + P( red card followed by a black card)

=1/2 * 25/51 + 1/2*(26/51) = 1/2 * (25+26)/51 = 1/2

Even when the second card was also foregone then the probability of winning in the third card is:

=p(black, black, black) + p(black, red, black) + p(red, black, black) + p( red, red, black)

=1/2 * 25/51 * 24/50 + 1/2 * 26/51 * 25/50 + 1/2*26/51*25/50 + 1/2*25/51*26/50

=1/2*25*(24+26+26+26)/(50*51) = 1/2

As we go on the probability of winning remains the same which is 1/2.

Hence the best strategy to follow is to pick the first card which has the same odds of winning as with picking the rest of the cards.

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