Yet another wonderful puzzle from Martin Gardner’s book “The Colossal Book of Short Problems and Puzzles”. This one is a probability puzzle, and something you should try with your children.
A deck of 52 playing cards is shuffled and placed facedown on the table. Then, one at a time, the cards are dealt face up from the top. If you were asked to bet in advance on the distance from the top of the first black ace to be dealt, what position (first, second, third, …) would you pick so that if the game were repeated many times, you would maximize your chance in the long run of guessing correctly
As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.
Happy gambling!
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I think I would guess the first card
Probability that the nth card is first black ace is:
50/52*49/51*48/50*47/49……(52-n)/(54-n) * 2/(53-n)
= 2*(52-n)/(51*52)
To max prob across n – minimise n
So I will bet on n=1
1st card.
If we assume the black aces to be one card, the first card has 51 unique permutations whereas the second one would have only 50… Chances of finding an ace within 51 cards are higher than finding it within 50 cards.
I would guess it as the 27th card as the probability of getting a black ace in a random draw is 1/26.
Answer is not correct Ram, please do look at the solution link