I think Pokerstars is wrong on these odds.. Check it out!

[Toddy]

I sent pokerstars my hand history where I got AQ 3 hands in a row and this was their reply to what the odds were..

Hello Todd,

Thank you for your email.

The odds of getting AQ in any one hand is about 82 to 1. So the odds
of
being dealt AQ on your next three hands is about 248 to 1.

I would rather take my chances with the AQ than the lightning!

[BruceZ]

[ QUOTE ]
I sent pokerstars my hand history where I got AQ 3 hands in a row and this was their reply to what the odds were..

Hello Todd,

Thank you for your email.

The odds of getting AQ in any one hand is about 82 to 1. So the odds
of
being dealt AQ on your next three hands is about 248 to 1.

I would rather take my chances with the AQ than the lightning!

[/ QUOTE ]

The 82 to 1 is right. That is C(52,2)/16 - 1 = 1326/16 - 1 =~ 81.9 to 1. For AQ on your next 3 hands in a row it should be (1326/16)^3 - 1 or about 569,207 to 1.

This is great news. Offer them a side bet where you pay them a generous 1000 to 1 for 3 AQs in a row.

[Toddy]

[ QUOTE ]
[ QUOTE ]
I sent pokerstars my hand history where I got AQ 3 hands in a row and this was their reply to what the odds were..

Hello Todd,

Thank you for your email.

The odds of getting AQ in any one hand is about 82 to 1. So the odds
of
being dealt AQ on your next three hands is about 248 to 1.

I would rather take my chances with the AQ than the lightning!

[/ QUOTE ]

The 82 to 1 is right. That is C(52,2)/16 - 1 = 1326/16 - 1 =~ 81.9 to 1. For AQ 3 hands in a row it should be (1326/16)^3 - 1 or about 569,207 to 1.

This is great news. Offer them a sidebet where you pay them a generous 1000 to 1 for 3 AQs in a row.

[/ QUOTE ]

Ya I figured it would be 82 x 82 x82... I think he added them (and then added wrong!) IM gonna send him the sidebet I'll post his reply!

[BruceZ]

[ QUOTE ]
Ya I figured it would be 82 x 82 x82...

[/ QUOTE ]

83 x 83 x 83 since it is 1 in 83. The odds are 82 to 1, but you cube the probability, not the odds.

He correctly converted the odds to 1 in 83, but then he multiplied 83 by 3 to get 1 in 249, and then he correctly converted that to 248 to 1. The beginning and end were good, but the part in the middle was screwed up.

[AaronBrown]

If you pick one card from the deck, the odds are 1.6 to 1 against it being part of a royal flush. So the odds of flopping a royal flush are 12 to 1, 5*(1.6+1) - 1. Boy, am I going to get rich playing Poker. I'll just fold 12 hands and clean up with my royal flush on the 13th.

For a bit of perspective:

Party Poker deals over a million real money hands per day, PokerStars would be similar. That means that on average two people are going to get AQ three times in a row each day. Same for every other non pair combo (78 of them). So over 150 people EACH DAY are getting dealt the same hand three times in a row. Also, set over set over set is only about 100,000 to 1. There will be multiple each day. There will be thousands of victims of runner runner flushes. Tens of thousands of flush suckouts, enough for several people to fall victim to 5+ in a row. Hundreds of two outer sets hitting on the river, and one outer quads, with a few people experiencing it several times.

I wonder how many emails they get each day?

Also, note that Bruce's calculation is the chance of getting 3 AQ's in a row. For any 3 non-pair hands in row (AK AK AK; AJ,AJ,AJ; 72o,72o,72o etc) it is


(1326/16)^2 - 1 or about 6867 to 1.

Is there a closed form for getting three in a row given n hands, as opposed to just three hands? I've solved this problem using dynamic programming, I'm just wondering if you can get a closed formula based on p and n.

[BruceZ]

[ QUOTE ]
Is there a closed form for getting three in a row given n hands, as opposed to just three hands? I've solved this problem using dynamic programming, I'm just wondering if you can get a closed formula based on p and n.

[/ QUOTE ]

Here is a simple recursion.

This old thread describes the Markov chain solution which involves matrix multiplication, as well as several approximate closed form methods, including a very accurate one from Feller which depends on finding the dominant root of a polynomial.

If you can find all the roots of the polynomial that results from taking the z-transform of the difference equation, then in theory you could do a partial fraction expansion and write a closed-form solution in terms of exponentials.

[Phogster]

[ QUOTE ]
If you can find all the roots of the polynomial that results from taking the z-transform of the difference equation, then in theory you could do a partial fraction expansion and write a closed-form solution in terms of exponentials.

[/ QUOTE ]

Ahh, my head hurts!