what's the probability of 4 consecutive aces (not drawing, just appearing) in shuffled deck?

Answer 1

1 in 20,825

The probability of an ace appearing somewhere in the deck is 1.
The probability that the very next card is an ace is 3/51 (3 aces left out of 51 remaining cards)
The probability that the next card is also an ace is 2/50
And the fourth is 1/49.

Total probability is therefore (3x2x1)/(51x50x49)

Answer 2

For the first ace you have 1 chance out of 52 cards. Then you only have 51 cards left. So the chances of getting a second ace is 1 chance out of 51. Now you are down to 50 cards and have 1 chance out of 50 to get the next one, and finally 1 chance out of 49 to get the fourth ace. Thus:
1/52 x 1/51 x 1/50 x 1/49 = 1 chance out of
(52x51x50x49)=6,497,400 but since they are in the deck then you increase your chances by 4x3x2x1=24 or
The chance of getting all four aces in a shuffled deck is 1 chance out of (6497400/24) or 270725

Answer 3

Since you have 4 aces in every deck of 52, the chances of one ace appearing is (4/52) = 0.076923. Since one random ace has nothing to do with the other each probability will be the exact same. So (4X0.076923) = 0.307692

So the probability of shuffling a deck and getting 4 aces in a row is 0.307692

Answer 4

4 aces in a row, at the top of the deck is 4/52*3/52*2/52*1/52
considering the aces could be anyplace in the deck there are 49 different combinations, so the answer is.
4/52*3/52*2/52*1/52*49=1.608*10^-4=1/6217

Answer 5

1 in 124,950.

Here's my logic. If you're not drawing, then you know the probability of the first ace appearing in the deck is 1/1. What you're really calculating is the increasingly remote probability that the 2nd, 3rd and 4th ace are adjacent to the first one.

Chances of the second being adjacent is 1 in 51. Chances of the third being adjacent is 1 in 50, and the fourth 1 in 49.

Answer 6

Okay, I say (4*3*2*1*49)/(52*51*50*49)

Answer 7

48 * 3/51 * 2/50 * 1/49 = 0.0058

Answer 8

1 in 6,497,400

Answer 9

slim