Combinatorics – Expected Number of Cards Turned Before Finding an Ace

combinatoricsdiscrete mathematics

NOTE: I want to check my solution only

Same question here

The question is this:

Shuffle an ordinary deck of 52 playing cards containing four aces. Then turn up the cards
from the top until the first ace appears. On the average how many cards are required to be turned before
producing the first ace ?

I want to check my solution.It definitely matches with the answers in that other question but I don't understand their solutions(haven't done much study on probability yet). I want to check mine:

We consider all the cards except the aces indistinguishable,and we will consider the aces to be indistinguishable.Now a deck is just a binary string with $48$ $C$ and $4$ $A$ ($A$ stands for aces and $C$ stands for the other cards.Obviously,the number of such strings is $\dbinom{52}{4}$. Now we divide into cases:

1)Number of strings with the first A in the $1$st position:$\dbinom{51}{3}$

2)Number of strings with the first A in the $2$nd position:$\dbinom{50}{3}$

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  1. Number of strings with the first A in the $49$th position:$\dbinom{3}{3}$

Since there are $4$ A's,the first $A$ cannot be in the $50$th position.

Note that for case $1$ above,we need to turn $0$ cards before getting an ace.For case $2$,we need to turn $1$ card before getting an ace,…,in the $49$th case,we need to turn $48$ cards before getting an ace.Therefore,the average of it all is:

$$\dfrac{48\dbinom{3}{3}+47\dbinom{4}{3}+….+0\dbinom{51}{3}}{\dbinom{52}{4}}
=\dfrac{\dbinom{52}{5}}{\dbinom{52}{4}}
=\dfrac{48}{5}$$

which is indeed the answer given there.Note that the numerator was computed by repeated application of the hockey stick identity.My questions are :

  1. Is my solution correct?

  2. Why does assuming indistinguishability still preserve the answer?I have seen this several times,but never really thought about it.

  3. If we assumed distinguishability of the cards,we will get a huge expression. How can we compute that?

Best Answer

This is only an answer to your first question. I will show that your answer is correct by deriving the same answer in another way.

The probability that the seven of clubs turns up before the first ace is $\frac15.$ This is because each of the five relevant cards (the four aces and the seven of clubs) has the same one-in-five chance of being first.

Likewise, the probability that the jack of diamonds turns up before the first ace is $\frac15,$ and the same goes for the queen of hearts, the four of spades, and every other non-ace card in the deck. Since there are $48$ non-ace cards, the average number of (non-ace) cards preceding the first ace is $48\cdot\frac15.$

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