In how many ways can 10 cards be drawn so that the 10th card is the repetition of a previous draw?

Question

Suppose that we draw a card from a deck of 52 cards and replace it before the next draw. In how many ways can 10 cards be drawn so that the 10th card is the repetition of a previous draw ?

Solution:

No of ways the 9 cards can be chosen $ = 52^9$

And for the 10th card that should be repetition of 9th card should have only $1$ possibility (i.e. Same as 9th Card)

So the answer should be $52^9$ but the answer is $52^{10} - (51)^9.52$

Can anyone explain how it is calculated ?

Does the previous draw means all the previous draws ? If that's the case then its correct.

Answer 1

Be careful about the distinction between the words the and a. If you see the word the, it means only one. The word a signifies that there may be more than one.

The problem says that the tenth card is a repetition of a previous draw, meaning that the tenth card can match any of the previous nine draws.

The justification for the answer is that we are subtracting those cases in which the tenth cards fails to match any of the previous draws from the $52^{10}$ ways of drawing ten cards from a standard deck of $52$ cards with replacement. There are $52$ possible selections for the tenth card. If none of the previous draws match that card, they must all have been selected from the other $51$ cards in the deck, which can occur in $51^9$ ways. Hence, there are $51^9 \cdot 52$ ways for the tenth card not to match any of the previous nine draws. Thus, there are $52^{10} - 51^9 \cdot 52$ ways for the tenth card to match at least one of the previous nine cards.

Answer 2

It implicitly says it is with replacement. But in that case the previous draws are irrelevant, the probability is just getting the same card in two draws, i.e. repeat whatever card you get the ninth draw in the tenth. If the last card is drawn at random (note that even how the ninth draw is done is irrelevant!), this is just $1/n$ for $n$ cards.