# Guessing cards: is 3 for 3 more improbable than 3 for 4?

1. Nov 12, 2015

### bahamagreen

3 for 3 means guessing the first, second, and third cards of a pack without trying a fourth time... and 3 for 4 means guessing the first, second, and third correctly, but trying and missing the fourth. I think both are conditional (dependent) without replacement.

I see a form for 2 guesses that reads p(a and b) = p(a) x p(b|a) where the "and" means "and then",

I suppose for 3 guess it would be:

p(a and b and c) = p(a) x p(b|a) x p(c|b|a)

I'm not sure how to interpret the "|" symbol arithmetically... is it just a reminder that the actual calculation must be without replacement?

The story... A long time ago someone grabbed a deck of cards and asked me to guess the first one. Amazingly I guessed right, and then guessed correctly the second and the third. She wanted to pull a fourth card, but intuitively I felt that guessing three out of three was more improbable than guessing three out of four if I missed the last one, so I declined and said let's stay with the improbability of just three tries...

Recalling this event, I decided to check if my intuition was right, but am a bit confused on the calculation... and it may be that the calculation is additionally complicated by the guesser being allowed to halt the series of card guesses whenever he wants - another aspect of conditional dependent?

Any help on untangling the process to get a calculation is appreciated.

2. Nov 12, 2015

### andrewkirk

Assume guesses are completely uninformed, with no hints or indicators available.

The probability of guessing 3 out of 3 is $\frac{1}{52}\cdot\frac{1}{51}\cdot\frac{1}{50}=7.5\times 10^{-6}$

The probability of guessing 4 out of 4 is $\frac{1}{52}\cdot\frac{1}{51}\cdot\frac{1}{50}\cdot\frac{1}{49}=1.5\times 10^{-7}$

The probability of guessing 3 out of 4, with the fourth guess being wrong, is $\frac{1}{52}\cdot\frac{1}{51}\cdot\frac{1}{50}\cdot\frac{48}{49}=7.4\times 10^{-6}$, which is slightly less than the probability of 3 out of 3. Naturally, the first probability is equal to the sum of the next two.

It means more than that in general. But in this case, where the only dependence between successive guesses is the absence of previously drawn cards, your interpretation works fine.

3. Nov 12, 2015

Thanks :)