What are the odds that someone at a full table has a premium hand?

[bigredlemon]

Presuming ATo and 99 or better?
Ten people at the table.

I have no clue but would guess 75% chance at least one person has it? Feel free to ballpark it... I just want a rough idea.

[gaming_mouse]

[ QUOTE ]
Presuming ATo and 99 or better?


[/ QUOTE ]

Not specific enough. For example, would KJo qualify? KJs? What about QJs? KQo?

You need to actually list all the ranges. For example:

AA-99
ATo-AKo
KQ-KJ
etc...

gm

[bigredlemon]

ATo-AKs and 99-AA only. So something that can flop TPTK or a high pair. QKs would not qualify.

[gaming_mouse]

[ QUOTE ]
ATo-AKs and 99-AA only. So something that can flop TPTK or a high pair. QKs would not qualify.

[/ QUOTE ]

AT-AK: 16 hands each, 64 hands total
AA-99: 6 hands each, 36 hands total

So 100 hands are in the range you're concerned with. Because the inter-player hand dependence is loose, the following is an accurate approximation to the answer you want:

1 - (1226/1326)^10 = .54

So over half the time at least one person will have a hand in these ranges.

gm

[bigredlemon]

Awsome. I bow down to you!

[Position]

Neat question. I estimate about 60.6%, quite a bit higher than Gaming Mouse's quick 54%.

This method should handle the interdependences a little cleaner:

1) The biggish aces. For each ace, 20/52 that it's in a starting hand. Then, 4(5)/51 that it's a ATo+. So, 15% for each of the 4 aces. (1-(1-.15)^4) for none of the 4 aces being out there as biggish aces. Roughly 48% for at least one biggish ace being out there.

2) The highish pocket pairs: 6 of them. 10/(13*17) for each. The (1-(1-... method gives a 24.3% of at least one highish pocket pair being out there.

Combining the aces & the pocket pairs as rather independent events, meaning using the (1-(1-... method again, gives 60.6% for at least one qualifying starting hand among the ten players.

[gaming_mouse]

[ QUOTE ]
Neat question. I estimate about 60.6%, quite a bit higher than Gaming Mouse's quick 54%.


[/ QUOTE ]

That's not right either, but you are right that my assumption about the interhand dependence not being significant was wrong. The correct answer is about 71%:

ncr(52,2)=1326
ncr(50,2)=1225
ncr(48,2)=1128
ncr(46,2)=1035
ncr(44,2)=946
ncr(42,2)=861
ncr(40,2)=780
ncr(38,2)=703
ncr(36,2)=630
ncr(34,2)=561

1 - (1226/1326)*(1125/1225)*(1028/1128)*(935/1035)*(846/946)*(761/861)*(680/780)*(603/703)*(530/630)*(461/561)=.7148

Next time I shouldn't be so lazy /images/graemlins/smile.gif

gm

[Position]

Interesting method!

I'm thinking it can't possibly be sound, though.

1-(450/550) is a shockingly high chance of the tenth hand being *so* special -- even given that the first nine hands were not so special. A 20% chance of the tenth starting hand being a biggish ace or highish pair?!

?

[gaming_mouse]

[ QUOTE ]

Interesting method!

I'm thinking it can't possibly be sound, though.

1-(450/550) is a shockingly high chance of the tenth hand being *so* special -- even given that the first nine hands were not so special. A 20% chance of the tenth starting hand being a biggish ace or highish pair?!


[/ QUOTE ]

Sh*t! You are right. This method is incorrect. I am not accounting for the fact that one a premium-hand card (like an A) could have been used in a non-premium-hand, like A6.

I'll keep thinking....

gm

[Position]

BTW, our first estimates are a lot closer to each other than I realized: I just now caught a wrong number; there are 4 cards Ten-King, not 5 as I'd used in my counting the biggish aces.

I end up with 40.2% as my estimate for at least one player holding a biggish ace.

Also, refining a bit on the biggish pairs, I now get 24.06% for at least one player holding.

Putting those two results together as if they're independent gives 54.6% as the net estimate for at least one player holding something biggish.

I caught the error because I've now been writing little programs to estimate for practical situations: It's folded to you @ 2 off the button, so 4 players left to act. You figure they'll call/raise (fight back at your blind steal in this tournament situation) with any AT+, KQ+, or 77+. What's your risk?

[gaming_mouse]

Yeah,

I'm thining about writing a little program to calculate this answer. The more I think about, the more I realize doing this by hand to get an exact solution is just ugly and a big pain. The program would take about 15 minutes to write -- maybe I'll do it later, or maybe you can do it. In any case, the answer is within 10% of my original estimate, I'd say, if not closer.

gm

[Position]

Yeah, Monte Carlo would be easy here.

I now completely agree that the real answer is within a couple percentage of your first estimate (and my first estimate once I plugged in the right numbers!). We each got 54.something% for the original question.