Guessing pack of cards colours

You have a pack of n red and n black cards. You try and guess what colour the next card is one at a time until you reach the end of the pack. If there are an equal or more number of red cards left, you guess red. If there are more black, guess black. How many are you expected to guess right? I don't know a formula or anything but for small n the fraction is remarkably higher than half..

For each position in the deck, indexed by the number of cards m cards dealt so far, compute the probability of a correct guess. By linearity, the desired expectation is the sum of these probabilities.

Given that we have seen m cards so far, the probability that exactly k were red is

p[n, m, k] = Binomial[m, k]Binomial[2n - m, n - k]/Binomial[2n, n]

in which case we will guess correctly with probability

q[n, m, k] = If[k <= m/2, n - k, n - m + k]/(2n - m)

Summing over k (0 to m in the first half of the deck, and m - n to n in the second half) and m yields the expected number of correct guesses:

Sum[ Sum[ p[n, m, k]q[n, m, k], {k, Max[0, m - n], Min[m, n]} ], {m, 0, 2n - 1} ]

For a standard deck/(2n - m)

Summing over k (0 to m in the first half of the deck, and m - n to n in the second half) and m yields the expected number of correct guesses:

Sum[ Sum/(2n - m)

Summing over k (0 to m in the first half of the deck, and m - n to n in the second half) and m yields the expected number of correct guesses:

Sum[ Sum[ p[n, m, k]q[n, m, k], {k, Max[0, m - n], Min[m, n]} ], {m, 0, 2n - 1} ]

For a standard deck/(2n - m)

Summing over k (0 to m in the first half of the deck, and m - n to n in the second half) and m yields the expected number of correct guesses:

Sum[ Sum[ p[n, m, k]q[n, m, k], {k, Max[0, m - n], Min[m, n]} ], {m, 0, 2n - 1} ]

For a standard deck of 52 cards (n = 26), expect to guess 30.0407 correctly on average, or 57.7705%.

ObPuzzle: Is it possible to guess _less_ than half correctly?

Eric Farmer

R

e

d

B

l

a

c

k

R

e

d

B

l

a

c

k

No. The number 2 * 'Correct guesses' + Abs|Red - Blackis strictly increasing, thus at least 56 at the end.

Alternative proof: By complete induction, with R red and B black cards left, at least Max(R, will be guessed correctly.

What happens if you follow the original strategy, but say 'black' when you are supposed to say 'red' and vice versa ??

Rj pease

***** SPOILER

As you deal, call the state of the cards a 'tie' whenever there is an equal number of each color remaining in the deck.

You begin with a tie and have a 50% chance of guessing card #1. After that, you always guess whatever color has the majority remaining until you reach another tie. Say this occurs after card #10. You are certain to get 5 of these right plus the 50% probability of guessing card #1.

You then begin this process again until the next tie is reached. By so doing, you will be certain to guess 26 cards correctly plus an expectation of one half of the number of ties reached while dealing the whole deck.

The problem then becomes one of determining the expected number of ties that occur while dealing the whole deck.

Let's deal 6 cards. One way of getting a tie is to deal RRRBBB. This probability is 26/52 x 25/51 x 24/50 x 26/49 x 25/48 x 24/47. This can be expressed ass: (26!/23!)^2 x 46!/52! = .01660…

Now RRRBBB is only one of 20 sequences of reaching this tie: [6!/3!^2 = 20]. (All sequences are equally probable.)

Multiplying yields a probability of .3320 of getting a tie after 6 cards.

I worked this out on a spreadsheet for all even numbers from 0 to 50. (The results are symmetrical around 26.) The expected number of ties is 8.08. Adding one half of this to 26 yields a expectation of guessing 30.04 cards correctly. This is the same answer that Eric Farmer got via a different method.

Bill Ryan