getting 3 pocket Aces in a row

[valueplayer]

what's the prob. for that?
1/220 * 1/220 * 1/220 = 1/10,648,000 ?????
Thank you.

[BruceZ]

[Ray Zee]

its really only getting them twice in a row. see why? and actually its really only getting them the next hand. 220 to one. see that. if not think awhile as this concept will help you think about probability in the correct way.

[Cyrus]

(P/up)+(sLUG)+ [f(shuffle)]^m + Ins(mark) + |cut| + {Restore[Ch(pass)]} + D2(conf)

[DanS]

Oh mighty Z,
Yes, I see why this is true. It was obvious to me instantly, so I won't spoil it for those to who(m) it comes across as a trick question.

I have two questions for you: what is the name for this phenomenon (if there is one, and I get the feeling like I learned it in Psych 101), and how do you take advantage (poker wise, not Montana/Deliverance wise [img]/forums/images/icons/tongue.gif[/img] ) of people who think this way without even realizing it?

Dan

[Cyrus]

"What is the name for this phenomenon ?"

All those mistaken beliefs are variations of the Gambler's Fallacy (aka Monte Carlo Fallacy, etc). The gambler sees 5 Blacks in a row in roulette and either believes a Red is "overdue" or that the Black "streak" will continue. (In fact, each result has again the same probability of occuring.)

To the mathematicians the "Red overdue" is a mistaken belief in negative serial autocorrelation of a non-correlated process. The "Black streak" is a mistaken belief in positive serial autocorrelation. (Ask BruceZ for a translation to English.)

In the case of back-to-back, or three times in a row, of a "rare event" such as getting pocket Aces, the gambler confuses the probability before the fact of unrelated events happening in a sequence, with the probability of each event happening separately in the sequence.

"How do you take advantage (poker wise) of people who think this way without even realizing it?"

Here's one : If the other players are allowed to make or see my hand and then, in a round soon after, I have approximately or exactly the same hand in the hole, I can reasonably assume that the average opponent will estimate the probability of me having again something like I had previously, to be smaller than it really is.

A distant variation of that theme is a play immediately after a multiway pot in which a lot of blood was shed. The next round is often an opportunity for someone to steal quietly (but steal early) with relative impunity.

[David Ottosen]

Yes but according to Zeno's paradox, you will never get pocket aces. In fact, you will never even get one ace, as the dealer will never be able to deal it to you.

[Bama Boy]

I was trying to search some archives to see if I could figure it out, but I think I was doing it wrong.

Also, what would the odds be of not getting a pocket pair in 100 hands?

Thanks, you math guys are terrific!

[joeg]

I calculated the odds of not getting a pocket pair in 100 hands as 0.2% as follows

you have a 6% chance of getting a pocket pair so have a 94% chance of not getting a pocket pair, not getting a pocket pair two hands in a row is 94% * 94% or 88%, for n hands the answer is .94^n or for 100 hands 0.94^100 which roughly 0.002 or 0.2%.

this seems a little low to me, are my calculations correct?

[Mangatang]

It's like the old saying, "Lightning never stikes in the same place twice." Once lightning has stuck a spot, it is just as likely that it will stike that spot again as is any other previously non-stuck spot.