Odds of nonconsecutive number draw

Question

What are the odds that you will randomly draw 10 non consecutive numbers from a deck of 40 cards (i.e. numbered 1-40)? (answer should be in a X:1 format, with X representing the average # of drawings you would have to do for one "successful" drawing; "consecutive" means the new card is not within 1 of the previous card in either direction)

Answer 1

You can think of this problem as counting the number of binary strings with 40 bits and 10 non-consecutive ones. So you'll have 40-10=30 zeros, which gives you 31 "gaps" between zeros. So you just need to fill 10 gaps now.

$31 \choose 10$ ways to fill the gap.

$40 \choose 10$ ways total to choose 10 numbers from 40.

So your ratio would be $31 \choose 10$:$40\choose 10$.

In the "$X:1$" format, that would be $19.11:1$