Five cards are drawn at random from a pack of cards that have been numbered consecutively from 1 to 97, and thoroughly shuffled.

Consider that  all the possible arrangements that can be made by choosing any 5 numbers from 97 = 97*96*95*94*93  = P(97,5)

Now ....consider all the possible sets that can be made by choosing any 5 cards from 97  =

C(97,5)........but .......since these sets can be arranged in any manner.....just order each one from the least number element to the greatest......and we will still have the same number of sets.....

So....the probability is given by :

[Number of ordered arrangements ]   /   [ Number of possible arrangements]  =

C(97,5)  / P(97/5)   =  (   97! /  [ (97 - 5)! *5! ] )  / [97! / (97 - 5)!]  =

[97! / 97!] * [ 97 - 5]!/ [97 - 5]! * [ 1 / 5! ]  = 1/120   =  about .833.....%

There are (5!) possible arrangements of the 5 cards; one arrangement is (completely) in increasing order of magnitude.

The probability is

\(\Large {\frac{1}{5!}} \small \text{ or } \Large {\frac {1}{120}}\\ \)

Consider a simpler problem first

Three answers, all different.

Actually though it looks like Alan has made a mistake in his arithmetic right at the end.

nCr(97,5)/nPr(97,5) = 1/5! = 1/120 = 0.00833... .

Interestingly, both answers show that the number of cards in the pack is irrelevant, (so long as it is five or more).

"Actually though it looks like Alan has made a mistake in his arithmetic right at the end."

I did indeed make a mistake in the arithmetic of the last line.  It should have been:

                                         nCr(97,5)/nPr(97,5) = 0.00833...   (= 1/5!)

In general:   nCr(n,k)/nPr(n,k) = n!/[(n-k)!k!]*(n-k)!/n!  = 1/k!   (as given directly by Nauseated).

I really like your explanation Chris :)