5 cards are drawn from a standard deck. What is the probability that the 5th card is an ace?

Question

5 cards are drawn from a standard deck without replacement. What is the probability that the 5th card is an ace?

Note that this question does not say anything about the 4 cards before the 5th one. They may or may not include aces.

Answer 1

Scythian, it's not weird at all. Chris, the problem is as simple as the first answerer thinks.

What order the cards are chosen in does not matter if you do not measure the later events based on the earlier events. Think of it this way: with the way you have defined the problem, the 5th card is the first card you actually look at. None of the others matter.

BTW, if you don't fully understand the question, please don't give thumbs down to the right answers. It's retarded.

Answer 2

Now, that's weird. The computation of odds is as follows:

(1/52*51*50*49*48) (1*48*47*46*45*4 + 4*48*47*46*4*3 + 6*48*47*4*3*2 + 4*48*4*3*2*1 + 1*4*3*2*1*0)

which turns out to be exactly 1/13.

Addendum: Horatio has a point. If we DON'T pick any cards, but merely inquire whether or not the 5th card in the deck is an ace, the odds that it is is 1/13. Du-oh!

Answer 3

Chances of getting a non Ace card on the 1st card is 48/52,
2nd 47/51, 3rd 46/50, 4th 45/49, last card 44/48.

So you have a 91.67 % chance of drawing a non Ace card.

Therefore the probability of drawing an Ace is 100-91.67 or 8.33%

Answer 4

4 in 47

Answer 5

4 cards are aces.
There are 52 cards in a deck.
So the probability of any card being an ace is 4/52, or 1/13.

Answer 6

a. (4C4 x 48C1 ) / 52C5 b. 48C5 / 52C5 c. (13C4 x 39C1) / 52C5 d. ( 4C3 x 4C1 x 44C1 ) / 52C5 i could punch that out for you, different than i don't have my calculator obtainable. sorry.

Answer 7

one in thirteen.