High-School level probability and logic problem

Question

So the other day I took a math test, (not for class, its just an optional test, so this isn't and kind of cheating) which included all kinds of logical and problem solving exercises, among others this one:

-"Given a deck of cards with N cards in it and each card is numbered (so the cards are 1, 2, 3, 4, ... nth card). Two cards are drawn one after each other. What is the probability that the when a third card is drawn, that this third cards number is in between the first card's and the second card's number."

I had done most of this test without problems but I just couldn't get around the fact that N (the number of cards) is unknown, so I'm not sure if the result should be a function of N or whether its an independent number.

Any kind of help is greatly appreciated! I can post some more of the questions if anyone is willing to see, the exam is something a teacher does for those students who want to stay after school and test out their logical and problem solving skills

Answer 1

The answer is $\frac13$.

There are six different ways the top three cards can be arranged. Two of those arrangements have the middle valued card in the third spot from the top.

Answer 2

All three cards drawn have equal chance to be the card with the number "in the middle". So $\frac13$ is the answer.

Answer 3

Don't think either answer given so far is right. Let's work through it.

There are N cards, numbered 1 thru N. You draw two cards off the top, so that leaves N-2 cards. What are the odds that the next card is between the first two in numeric sequence? The answer is "The number of cards between those two cards" divided by "The total possible number of cards that can be drawn after the first two". The number of cards between is simply the smaller of the two cards subtracted from the larger, minus 1. Divide that number by N-2.

For example, say there are 22 cards. Draw 2 cards. That leaves 20 cards. Now, let's say that the cards drawn were 8 and 12. The possible cards that could be between them are 9, 10 and 11, a total of 3 cards. By the formula, 12-8-1=4-1=3. So the odds are 3 out of 20, or 3/20, 0r 15%.

So, the formula is: ( Y - X - 1 ) / N -2 where Y is the larger card drawn and X is the smaller card.