Take a deck of cards labeled 1-60. Shuffle well and flip over the first 10 cards in a row from left to right.

Let x represent the number of cards in a row from the left that are in ascending order.

For example if the cards were 3 15 22 59 40 7 19 30 33 34 then x=4. The streak ends with the 59 because 40 is less.

Find a formula for p(x), the probability the first x cards are in ascending order and card x+1 is lower than card x, where x={1,...,10}

Generalize: For a deck of c cards labeled 1 to c and flipping over a row of n cards. Find a formula for p(x) where x={1,...,n}

(In reply to Wow! by Steve Herman)

Thanks.

In fact, they do not add to 1 (whoops) because I there is no x+1 card for the nth card.  Thus, when x=n, p(x)=1/x! (i.e. the probability that they are sorted with nothing about anything beyond).

If you look at this recursively the other way, S(i)=sum from x=i to n of p(x) = S(i+1)+i/(i+1)!
S(n)=1/n!
if S(i+1)=1/(i+1)! then S(i)=1/(i+1)! + 1/(i+1)! = (i+1)/(i+1)! = 1/i!
therefor by induction it is true.
S(1) = 1/1! = 1.

So, the correct answer is
p(x)={ x/(x+1)! if x<n, 1/x! if x=n }

Thanks

Posted by Joel on 2006-10-11 17:29:05