POKRDAV
Joined: Dec 21, 2017
• Posts: 1
December 21st, 2017 at 10:58:03 AM permalink
In a ten handed holdem game. If you are dealt AQ what are the odds of one of the 9 remaining people getting dealt AA, KK, QQ or AK?
beachbumbabs
Joined: May 21, 2013
• Posts: 14230
December 21st, 2017 at 1:08:01 PM permalink
Quote: POKRDAV

In a ten handed holdem game. If you are dealt AQ what are the odds of one of the 9 remaining people getting dealt AA, KK, QQ or AK?

1326 starting hands. You have 1 of them.
AA 3 different ways
KK 6 different ways
QQ 3 different ways
AK 12 different ways.

24/1325 = 1.8113% of hands, or 1 in 55, per opponent.

However, what each opponent gets affects the others, through effect of card removal, so their chances increase or decrease accordingly. Also, could be more than one of those combos out against you at the same time.

I'm not a math guy, but there are many here. Perhaps one or more of them will structure it correctly and give a valid answer.

I'm going to guess, for my own amusement (and probably theirs), it's approximately 1 time in 5 there will be AT LEAST one hand like that dealt against you, 10 handed, when you hold that.
If the House lost every hand, they wouldn't deal the game.
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
December 21st, 2017 at 4:44:43 PM permalink
Quote: POKRDAV

In a ten handed holdem game. If you are dealt AQ what are the odds of one of the 9 remaining people getting dealt AA, KK, QQ or AK?

Given your AQ starting hand
a quick 1 million simulation deals
shows about a 1 in 7 chance of one person (could be more)
getting 1 of those 4 other starting hands.

so the odds would be about 6 to 1 against it

*****
Here is a good site (shows math too)
I did not see if it could be answered there

https://www.pokerstrategy.com/strategy/various-poker/texas-holdem-probabilities/

interesting question
Sally
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