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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- 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 :
Is my solution correct?
Why does assuming indistinguishability still preserve the answer?I have seen this several times,but never really thought about it.
If we assumed distinguishability of the cards,we will get a huge expression. How can we compute that?