Probability puzzle

[Zeta-Flux]

I thought a few of you would like this puzzle.

A card dealer has a deck with 13 black cards and 13 red cards. These are shuffled and placed before you. The card dealer is going to flip one card at a time until he finishes flipping through the entire deck. What you want to do is guess *one* red card before it is flipped. You have only *one* guess. If you haven't guessed when the dealer gets to the last card, then that is your guess.

So, for example, someone could ALWAYS guess the first card, and would win 50% of the time. Is there a strategy to be right more than 50% of the time?

Best,
Zeta-Flux

[Mystwalker]

I think so. Sloppy explanation:

Case 1: The last card is red. Probability: 50%
In this case, when you wait after 13 black and 12 red cards have been flipped, you just say "Red!" and win by 100%, thus a total probability of 50%.
If the last 2 (or 3 or 4) cards are red, you can say "Red!" even ealier, thus improving your total likeliness over 50% (and already meet the goal).

Case 2: The last card is black. Probability: 50%
In this case, after the 12th red card (which happens in 100% of all cases), you say "Red!".
It's probably not very likely to win, but better when nothing (which will be the case when you don't say anything and the next card is red...).

Right now, I don't have the time to calculate the overall percentage, but it's definitely greater than 50%. :banana:

[Wacky]

Originally Posted by Mystwalker

If the last 2 (or 3 or 4) cards are red, you can say "Red!" even earlier, thus improving your total likeliness

I don't think so.

Certainly you can say "Red!" earlier,
but it doesn't increase the likelyhood of a win
because that case has already been counted
in the 50% of the time that the last card is red .

[Mystwalker]

You're right.
Nevertheless, the probability stays above 50% due to the second case, does it?

[wblipp]

No.

At any point in the process, if there are "r" red cards and "b" black cards left,
your probability of success is r/(r+b), no matter what you do.

Prove it by induction on the number of cards left. If there is 1 card left, the
ratio is either 0/1 or 1/1, and the rule works.

For more than 1 card left, you can either "guess now" or "wait". If you guess
now, the probability of being right is r/(r+b).

If you wait, the next card is red with probabilty r/(r+b), and your probability
of success with (r-1) and b is known to be (r-1)/(r+b-1). The next card is
black with probability b/(r+b), and your probability of success with r and (b-1)
is known to be r/(r+b-1). Putting it together, your probability of success if
you wait is

(r/(r+b))*((r-1)/(r+b-1))+(b/(r+b))*(r/(r+b-1)) = r/r+b

I suspect there is a a trivial proof involving entropy, but I don't know enough
to make the statement and suspect that most of us wouldn't know enough
to understand it anyway.

[Zeta-Flux]

Sorry Mystwalker. Wacky is right. It all evens out in that solution, so you still have 50%. (That's what I tried at first too, so don't give up.)

[Zeta-Flux]

Wow! Good solution wblipp!