probability

you could only consider 14 cards out of 14 so just imagine there are only 14 cards, 13 hearts and 1 Ace of Spades
so it is easy to find that the probability that no Heart is drawn before the Ace of Spades is 1/14(the Ace of Spades is the first one out of 14)

any problem?

2007-02-07 23:44:29 補充：
correction:you could only consider 14 cards out of 52

2007-02-08 01:22:15 補充：
只要明白呢條係比較14隻牌o既先後次序而唔係比較52隻牌, 就好易搵到答案.

By means of permutation:
For 52 cards, there are 52! possibilities of arrangement of the cards where ! means the factorial operation.
Now, let A be the card of ace of spades and then take a look at the diagram below:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Probphlmf.jpg

【此圖乃本人自製圖片，未經本人同意勿擅自連結或使用】
In fact, card A can be from position 1 to 39 for any favourable outcome (since at least 13 positiosn should be left at the end for those hearts), i.e. 1 ≦ r ≦ 39
Now, for each value of r, if we want a favourable outcome, the preceding r-1 positions before A must be filled by non-heart and for the remaining 52 - r position behind A, the arrangement can be random.
So the general expression of number of permutations for any 1 ≦ r ≦ 39 is:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Probphlmf2.jpg

And hence the final probability will be summing up all these values for 1 ≦ r ≦ 39 divided by the total number of possible permutations, i.e.

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Probphlmf3.jpg

∴ The probability, if expressed in percentage, is about 7.14%.