This is a discussion on Odds of... within the online poker forums, in the General Poker section; Getting AA 4 times out of 7 hands? It just happened to me and i won 3 of them. :) 


#2




The same odds as getting AA preflop in a single hand ? Previous hands have no impact on the future hands.

#3




Consecutive hands do have different odds.
ie Probability of getting aces twice in a row =/= probability of getting aces getting aces twice in a row = 1/220*1/220 = 1/48400 I can't remember how to do it for weird combos like 4/7 hands, but let's just say it's pretty improbable 
#4




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So basically imagine getting 7 decks, shuffling them all and dealing a hand out of each deck. What are the odds that 4 of those hands would be AA? 
#5




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One time, in a SNG, I got AA five times  three of which were consecutive. 
#6




re: Poker & Odds of...
Yeah, I know it's improbable, but the odds of a single event happening at a single point in time never changes. It's very improbably for a coin to flip heads 30 times in a row (because in the long run = infinity, it will even out) but this doesn't mean that on the 29 flip, the odds of landing it heads were any worse than on the first try.
When you start to think in terms of I got AA 4 times in a row, you also start to think in terms of I never got AA in 2k hands, wich leads to Gambler Fallacy wich leads to believing running hot means an impending downswing, etc. I prefer not to think of "wow I had AA four times in a row" and more in terms of "I have AA now". 
#7




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Example is what are the odds of flipping 2 heads in a row? Well hte options are: H H, T T, H,T or T,H Therfore it is a 25% chance of hitting 2 consecutive heads although using your method jeff there is a 50% of hitting a head after flipping the first. Do you understand now jeffred? 
#14




I've had Aces back to back to back once.If I am right the odds of that are 1 in 10,648,000.
If this is correct,then maybe poker isn't my game.I should play the Michigan Lottery. Oh yeah,maybe I shouldn't.Come to think about it I lost 2 outa the 3. 
#15




I think basically what switch said above, the chance of getting aces at an any given time is what ever that is (chance of the first one =4/52 * chance of second one 3/52)
But the confusion lies when you step back and take a sample of hands and ask what the odds are of getting Aces X out of Y samples. Quote:
the above is going on a shady knowledge of odds from a post on cc if i'm wrong plz tell me 
#16




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Well it's a little more complicated than that. You explained how to get the probability of getting aces 4 times in a row, ie 1/220*1/220*1/220*1/220, but getting them 4 times out of 7 requires slightly more complicated stuff that I can't remember how to do unfortunately :/ 