Probability that the next card is the ace of spades

Question

Say I have a standard $52$ count deck of cards in random order, and that I start flipping cards from the deck over until a king appears, which is card #$19$. What's the probability the next card is the ace of spades?

It's possible to brute force this using Bayes' theorem and slogging through a laborious calculation, but I was wondering if there were any tricks to arrive at the answer in a cleverer way.

Answer 1

There are $51!$ possible arrangements of the deck minus the ace of spades, in which there is only one "wanted" position for the ace of spades (immediately after the first king, at whatever serial number the king has come), against $52!$ unrestricted arrangements,

thus desired $Pr = \Large\frac{51!}{52!} =\Large\frac1{52}$

Interpretation issue

There is some confusion about the interpretation of the question. If it just happens that the first King is at #19, it could have as well been at #15, say, the answer is $\frac1{52}$

On the other hand, if it is given that the first King is at #19, the answer will become $\frac1{51}$

PS:

You haven't clarified whether by "flipping over..." means that you are seeing each of the first $19$ cards. If you are seeing, then. of course, the probability that the next card is tha ace of spades is simply $\frac1{33}$

Answer 2

If you cannot see the first $18$ cards

First we calculate all possibilities that satisfy the condition, namely that the first King appear as card number $19$. The remaining three Kings must appear later among the remaining $33$ cards. Then the Ace of spades must be one of the $48$ non-King cards. So the number of possibilities is given below:

$$ \binom{33}{3}\binom{48}{1} $$

Now we simply calculate number of possibilities that we want, namely that Ace of spades is card number $20$. Therefore, the three Kings must appear later among the remaining $32$ cards. The number of possibilities is given below:

$$ \binom{32}{3} $$

The probability is then simply the ratio between the two:

$$ \frac{\binom{32}{3}}{\binom{33}{3}\binom{48}{1}} $$

Notice that I intentionally ignore the King's symbols and permutation as it appear in both numerator and denominator and will cancel each other anyway.

If you can see the first $18$ cards

Simply $0$ if the Ace of spades is among the $18$ cards and simply $\frac{1}{33}$ if it is not.

Answer 3

You have 18 cards which are not kings, then one king, then 33 cards of which 3 are kings.

The probability that the next card is a king is 3/33 = 1/11, the probability that it is not a king is 10/11. If the next card is not a king, then we remove the three kings. There are now 18 cards before and 30 cards after the king. Ace of spades could be in any of these positions, so the probability that it is the card after the king is 1/48. Total probability is (10/11) * (1/48) = 10/528.

Note that each of the three remaining kings has a probability of 1/33 of being the next card. Accordingly all the other cards must have a lower probability. If we had just picked card #19 and it was whatever card, then any other card would come next with probability 1/51. And if the first king had come earlier, the probability would have been higher.