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Author Topic: Does anyone know the odds?  (Read 1795 times)
Newmanseye
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« on: November 25, 2006, 07:48:40 PM »

What are the odds on being dealt aces consecutively one hand after another?

Just curious as it just happened to me the first 2 hands of a comp.
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roverthtaeh
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« Reply #1 on: November 25, 2006, 07:52:44 PM »

Just under 50,000 to 1.
Odds of having them cracked, about evens.  Cheesy
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« Reply #2 on: November 25, 2006, 07:53:45 PM »

Just under 50,000 to 1.
Odds of having them cracked, about evens.  Cheesy

Best part was I never got any action at all!!!!

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« Reply #3 on: November 25, 2006, 07:55:21 PM »

I do remember a tourney on Tribecca , the 7.5k ($15k as it is now) about 3yrs ago when I was dealt AA 5 times in 7 hands.

I did try and work out the odds, but my 39 digit calculator couldnt cope with it! lol

James will remember that one.
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thetank
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« Reply #4 on: November 25, 2006, 07:58:03 PM »

The probablility of specific consecutive events occuring will be the probality of the first one multiplied by the probability of the second one.

If you know the fractional odds (220 to 1 in this case) then you can mulitiply.....

220 to 1 x 220 to 1 thus

220 x 220  (48,400)
1 x 1  (1)

So 48,400 to 1
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« Reply #5 on: November 25, 2006, 08:01:00 PM »

Ahh Tanky, I knew you would be the man to give me the answers as well as the means to work it out for myself next time.

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« Reply #6 on: November 25, 2006, 08:08:24 PM »

That is only the odds of being dealt aces on on two pre-specified hands.  If the first occurance is allowed to happen anytime and is then followed by the two aces again it is only 1/220

 
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thetank
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« Reply #7 on: November 25, 2006, 08:10:52 PM »

and if both hands have already been dealt, the odds are nothing.
« Last Edit: November 25, 2006, 08:13:05 PM by thetank » Logged

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« Reply #8 on: November 25, 2006, 08:37:08 PM »

The probablility of specific consecutive events occuring will be the probality of the first one multiplied by the probability of the second one.

If you know the fractional odds (220 to 1 in this case) then you can mulitiply.....

220 to 1 x 220 to 1 thus

220 x 220  (48,400)
1 x 1  (1)

So 48,400 to 1


Ok Mr smart ar$e mathematician - what are the odds that the AA held up both times online?

 
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thetank
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« Reply #9 on: November 25, 2006, 08:44:41 PM »

2 x (365 - number of days since your last cash out) to 1.
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TightEnd
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« Reply #10 on: November 25, 2006, 08:47:24 PM »

In a self deal satellite on thursday I dealt AA to the player immediately to my left 4 times in 5 hands

he won one hand, split another with AA elsewhere and lost two. The four hands cost him two rebuys.

 
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« Reply #11 on: November 25, 2006, 08:47:40 PM »

blimey - your kinda screwed if you've never cashed out in the last year!! Cheesy
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« Reply #12 on: November 26, 2006, 11:28:04 AM »

I remember getting dealt back to back AA in the first two hands of a £10 rebuy in Derby about ten years ago.

They were both cracked and I got dealt 88 in the next hand which also cost a rebuy.

I guess the odds are 221x221=48,840/1
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« Reply #13 on: November 26, 2006, 11:39:00 AM »

I was once dealt J5 and then K6 on consecutive hands

seriously, it happened
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« Reply #14 on: November 26, 2006, 11:39:39 AM »

I was once dealt J5 and then K6 on consecutive hands

seriously, it happened

 Cheesy
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