This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

I show you a shuffled deck of standard playing cards, one card at a

time. At any point before I run out of cards, you must say "RED!".

If the next card I show is red (i.e. diamonds or hearts), you win. We

assume I the "dealer" don't have any control over what the order of

cards is.

The question is, what's the best strategy, and what is your

probability of winning ?

decision/red.s

If a deck has n cards, r red and b black, the best strategy wins

with a probability of r/n. Thus, you can say "red" on the first card,

the last card, or any other card you wish.

Proof by induction on n. The statement is clearly true for one-card decks.

Suppose it is true for n-card decks, and add a red card.

I will even allow a nondeterministic strategy, meaning you say "red"

on the first card with probability p. With probability 1-p,

you watch the first card go by, and then apply the "optimal" strategy

to the remaining n-card deck, since you now know its composition.

The odds of winning are therefore: p * (r+1)/(n+1) +

(1-p) * ((r+1)/(n+1) * r/n + b/(n+1) * (r+1)/n).

After some algebra, this becomes (r+1)/(n+1) as expected.

Adding a black card yields: p * r/(n+1) +

(1-p) * (r/(n+1) * (r-1)/n + (b+1)/(n+1) * r/n).

This becomes r/(n+1) as expected.

Continue to: