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  #11  
Old 11-24-2002, 02:00 PM
marbles marbles is offline
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Default your intuition has gotten the best of you!

Dynasty,

Another angle to consider:

Consider this as a pass/fail test with 18 trials (for the 18 cards dealt to everyone but you). A success is an ace or king, a failure is anything else. Here's how the probabilities would break out:

P(zero): .057
P(one): .228
P(two): .346
P(three): .255
P(four): .096
P(five): .017
P(six): .001

So yes, the most likely result is two ace/kings being dealt out when you hold AK against 9 opponents. But notice that when you take the weighted average of the 7 possible results, the expected value is actually 2.16 successes. If exactly two opponents hold ace and/or king, that's actually LESS than you should expect. There are actually an extra 0.16 ace/kings left in the deck for the flop.
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  #12  
Old 11-25-2002, 12:26 AM
JTG51 JTG51 is offline
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Default Dynasty, you are cheating and I think you know it

I'll admit, I didn't read the whole debate on the beginners forum. It started to make my head hurt. If my point has already been covered there, I appologize.

You are playing games with the numbers, and I'm sure you are smart enough to know it. Did you and bernie actually agree that there will be exactly one A and exactly one K out in the other players hands?

I'm assuming bernie's point was that there are probably some A's and K's already dead. Assuming that there's exactly one of each is just silly. The math is totally meaningless.

If you are really interested in this question, you shouldn't just randomly pick two of your cards as dead. If all 9 players are playing random cards (which I think is a good assumption since limping doesn't make someone more or less likely to have an A or K) then on average 2.16 A's and K's will be in their hands, I think. That's 6/50 (the chance that any random card is an A or K, given that you have AK) * 18 (the number of random cards). Using 2.16 dead cards instead of 2 in your equations below, and altering the number of unknows accordingly makes the chances of floping a pair 28.89%, or slightly lower than if all cards are unknowns.

To answer your question though, ifyou really are interested in that specific problem your math is correct.

Boy, did that sound like a Dynasty post or what? [img]/forums/images/icons/smile.gif[/img]
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  #13  
Old 11-25-2002, 06:10 PM
David Ottosen David Ottosen is offline
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Default My cheap way of thinking about it

There are 6 aces and kings that you don't have out of 50 cards. If you don't know your opponent's hands, aces and kings make up 0.12 of the remaining deck.

If you don know their hands, out of the next 18 cards (your opponents 9 hands), 2 aces and kings are gone. That means of the remaining 32 cards, there are 4 aces and kings left, or 0.125 of the deck.

Your chances of flopping a pair when 0.125 of the deck helps you should be better than when 0.12 of the deck helps you.

I like cheap math! No !Choose! for me.
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  #14  
Old 11-26-2002, 09:13 PM
Ed Miller Ed Miller is offline
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Default Re: Am I making this AK calculation correctly

In step two, you are counting flops that contain two or more A's or K's multiple times.
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  #15  
Old 11-27-2002, 12:36 AM
Dynasty Dynasty is offline
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Default Re: Am I making this AK calculation correctly

In step two, you are counting flops that contain two or more A's or K's multiple times.

I don't think so.

My calculation was: Step 2: The number of A,x,y or K,x,y flops are 4 (number of Aces and Kings left in the deck)*28*27/2 = 1,512

I'm not even counting the flops which contain two or more A's or King's- not even once. I'm only caculating the probabilty of floping a pair of Aces or a pair of Kings. x and y cannot be either an Ace or King. The 28 (and 27) used in the calculation is 52 -4 Aces -4 Kings -16 non Ace/King cards in other players' hands. 52-4-4-16 = 28.

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