What are the odds of getting pocket aces twice in a row, and having someone else also have rockets both times

Question

What are the odds of getting pocket aces twice in a row, and having someone else also have rockets both times

Answer 1

How many players at the table?  The odds change depending on how many other hands have the chance of being rockets.

Edit-

For instance, at a 10 player table the odds of two players both having pocket aces is about 135-1.  So square that to get the odds of it happening twice, 18,225 to 1.  I'm including a link to a page that explains the math, it's a little lengthy to get into here.

Answer 2

The odds of both players holding aces in a heads up situation is 270,724 to 1 so I guess 2 times in a row would be 73,291,484,176 to 1

Visit my Poker Forum PokerGob.com this would be a good question to post there too.

Answer 3

Wow!  Great Answer ZMan!

But wrong. ..... Well, not wrong.  Just a little incomplete.  Well, not incomplete ..... ok, your answer's dead on, it just left out one small twist.

The odds of getting AA are 1 in 221.

The odds of getting AA and having your opponent get AA (at a 2 man table,) are 1 in 270,725.

As for the twice in a row bit, it depends on how many total hands you play.  For example, if you are playing all night, any time you get AA, the odds of getting AA on the next hand are ... 1 in 221.

And similarly, any time you and your opponent BOTH get AA, the odds of you and your opponent both getting AA the next hand is ... 1 in 270,725.

.....

Essentially, if you play continuously, the odds of getting AA or 'AA v AA' twice in a row are the same as getting AA or 'AA v AA' once in a row, because you can't do it a second time until the first time is already a past event, and a past event has a 100% chance of occurring.

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But if you're asking what the odds are of you getting AA on the first hand ... and your only opponent getting AA on the same hand... and then you getting AA on the second hand ... and your only opponent getting AA on THAT hand as well ... when you've only played TWO hands all night .... it's 1 in 73,292,025,625.

Answer 4

50-50.  Either it happens or it doesn't.

Answer 5

To get an answer two pieces of information are needed that you did not supply.

(1) How many pocket cards does each player get?

(2) How many players are there?

I will assume each player gets two pocket cards, as in hold 'em. If a player got more cards, such as four in Omaha, the odds would be much more likely.

Start by calculating the odds that you will get pocket aces. The odds are (4 / 52) x (3 / 51).

Now calculate the odds of the player to your left having pocket aces given that you have pocket aces.

The odds are (2 / 50) x (1 / 49).

That makes the odds of both you and the player to your left having pocket aces =

(4 / 52) x (3 / 51) x (2 / 50) x (1 / 49) = 1 in 270725.

If there are more than two players, the odds of you having pocket aces and any other specific player also having pocket aces are the same as the odds of the player on your left having pocket aces when you have pocket aces. So, with three players the odds are twice as high, with four players the odds are thrice as high, etc.

That makes the odds of you and another player each having packet aces

1 in 270725 if there are two players

2 in 270725 if there are three players

3 in 270725 if there are four players

4 in 270725 if there are five players

5 in 270725 if there are six players

6 in 270725 if there are seven players

7 in 270725 if there are eight players

8 in 270725 if there are nine players

9 in 270725 if there are ten players

10 in 270725 if there are eleven players.

When it comes to the second hand, I am not sure if you mean any other player also had pocket aces or if you mean the same player had pocket aces again.

If you mean any other player the odds are

(1 / 270725)^2 or 1 in 73292025625 if there are two players

(2 / 270725)^2 or 1 in 18323006406 if there are three players

(3 / 270725)^2 or 1 in 8143558403 if there are four players

(4 / 270725)^2  or 1 in 4580751602 if there are five players

(5 / 270725)^2 or 1 in 2931681025 if there are six players

(6 / 270725)^2 or 1 in 2035889601 if there are seven players

(7 / 270725)^2 or 1 in 1495755625 if there are eight players

(8 / 270725)^2 or 1 in 1145187900 if there are nine players

(9 / 270725)^2 or 1 in 904839822 if there are ten players

(10 / 270725)^2 or 1 in 732920256 if there are eleven players.

If you mean the same player the odds are

(1 / 270725) x (1 / 270725) or 1 in 73292025625 if there are two players

(2 / 270725) x (1 / 270725) or 1 in 36646012812 if there are three players

(3 / 270725) x (1 / 270725) or 1 in 24430675208 if there are four players

(4 / 270725) x (1 / 270725)  or 1 in 18323006406 if there are five players

(5 / 270725) x (1 / 270725) or 1 in 14658405125 if there are six players

(6 / 270725) x (1 / 270725) or 1 in 12215337604 if there are seven players

(7 / 270725) x (1 / 270725) or 1 in 10470289375 if there are eight players

(8 / 270725) x (1 / 270725) or 1 in 9161503203 if there are nine players

(9 / 270725) x (1 / 270725) or 1 in 8143558402 if there are ten players

(10 / 270725) x (1 / 270725) or 1 in 7329202562 if there are eleven players.