1
$\begingroup$

Question comes from Introduction to Probability, by J. Blitzstein.

A deck of cards is shuffled well. The cards are dealt one by one, until the first time an ace appears.

Find the probability that no kings, queens, or jacks appear before the first ace.

Approach 1

Lets simplify the problem by assuming all the cards are dealt.

There are $52!$ possible orderings of the cards, and all of them are equally likely.

Lets fix the first ace to be the ace of spades for now. To precede kings, queens and jacks, the ace can be any of the first $37$ cards. Suppose it is at position $1 \leq i \leq 37$. The $i-1$ cards to the left of the ace, selected from the $36$ valid ones (no kings, no jacks, no queens, and none of the remaining $3$ aces) can be permuted in $(i-1)!$ ways. The remaining $(52-i)$ cards to the right of the ace can be permuted in $(52-i)!$ ways.

Thus, the probability of an orderings where no king, queen, or jack precedes the first ace is $$\frac{4\sum_{i=1}^{37}\binom{36}{i-1}(i-1)!(52-i)!}{52!}.$$

Note the factor of $4$ that accounts for possible choices of the first ace.

Approach 2

This time, suppose the experiment terminates as soon as the first ace is dealt.

The first ace can be any of the $4$ aces. Once again, it can occur at either of the first $37$ positions. We select the cards to the left of the ace from the $36$ valid options and permute them.

Then, the probability of the desired even it $$4\sum_{i=1}^{37}\frac{\binom{36}{i-1}(i-1)!}{i!}.$$

Intuition

I really feel like stopping the experiment after an ace is dealt should give us the same probability as dealing all of the cards, since once an ace is observed, the remaining cards are essentially irrelevant in the sense that they won't make the ordering favorable if a king, queen, or jack was observed before the ace and vice versa. However, this intuition is really week. I wouldn't really be able to explain, let alone demonstrate it to someone.

Questions

  1. Are my calculations in the two approaches correct?
  2. Is my intuition stated above correct? If so, could you formulate your thoughts as to why that is and show how solutions in the two approaches equate?
$\endgroup$
2
  • $\begingroup$ Not an answer, but a simpler intuition (if I understand the question correctly): Discard all cards from $2$ through $10$, since they're irrelevant to the question. Now shuffle and deal the remaining $16$ cards. Is the first card an ace? With what probability is that first card an ace? $\endgroup$
    – Brian Tung
    Jun 19, 2020 at 20:15
  • $\begingroup$ Ahh, I see lulu has anticipated me. $\endgroup$
    – Brian Tung
    Jun 19, 2020 at 20:15

2 Answers 2

3
$\begingroup$

The calculation in your second approach is wrong. The numerator correctly calculates the number of ways of dealing $i-1$ cards out of the set of 36 acceptable cards, and then a particular ace as the $i$th card. But the denominator of $i!$ counts the number of ways of dealing a deck of $i$ cards, and that's not what you want. You want the total number of ways of dealing the first $i$ cards from a deck of 52 cards: $$4\sum_{i=1}^{37}\frac{\binom{36}{i-1}(i-1)!}{\binom{52}i i!}.$$

It's now very easy to see that this is equivalent to your first approach, using the identity: $$\binom{52}i = \frac{52!}{(52-i)!i!}$$

It's a little harder to see that they're both equal to the correct answer of 1/4.

$\endgroup$
3
$\begingroup$

Ignore all the cards except $J,Q,K,A$. There are $16$ of those, $4$ of which are aces. Each card has an equal chance of being first in the list, so the answer is $\frac 14$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.