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I am analyzing solution to the below problem (variation of the problem)

4 cards are drawn from the 97 card deck. What is probability that they are drawn in ascending order?

The proposed solution says that the probability is $$1 * \frac{1}{2} * \frac{1}{3} * \frac{1}{4} = \frac{1}{4!} $$ regardless of number of cards in the stack.

I don't understand the solution.
How the cases where the same cards (AA, KK etc) are drawn are reflected in the calculation?
Why the number of cards in the deck doesn't play a role. Intuitively more cards in the deck would translate in more combinations.


Can anybody explain please?

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    $\begingroup$ This is not a deck of standard playing cards. It's a pile of sequentially numbered cards. Also, the four cards are drawn without replacement. $\endgroup$ – Arthur Nov 8 '16 at 23:07
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Each card has a unique value.

If you know the values of the four cards, there are $4!$ equally probably orderings of which $1$ is in ascending order

Overall there are $\displaystyle 97\times 96 \times 95 \times 94= 4!{97 \choose 4}$ equally probable ways you can choose the four cards, of which $\displaystyle {97 \choose 4}$ are in ascending order

So the probability the four cards are in ascending order is $\dfrac1{4!}$. The $97$ does not affect this

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