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 Dynasty 11-22-2002 07:04 PM

Am I making this AK calculation correctly

bernie is telling me that I am incorrectly approaching/calculating a problem in a Beginners Questions thread. I'd like some feedback on whether I'm approaching the problem correctly.

Here's the problem: You've got AK in a pot being contested by yourself and 9 opponents. One Ace is in another opponent's hand. One King is in another opponent's hand. Have your chances of flopping a pair gone up or down?

My answer was that your chances of flopping a pair have gone up slightly and backed it up with these calculations.

First, without any extraordinary information, the chances of flopping a pair with AK is 28.96% or 2.45:1. These numbers are well known.

These are my calculation on how often you will flop a pair given the unusual information you have about one Ace and one King being in an opponent's hand.

Step 1: Calculate the the total # of possible flops given that only 32 cards remain in the deck- 32*31*30/6 = 4,960

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

1,512/4,960 = 30.48% or 2.28:1.

So, under normal circumstances when you don't make assumptions about what cards are in your opponents' hands, you will flop a pair 28.96% of the time. However, when you know exactly one Ace and one King are in your nine opponents' hands, the chances of flopping a pair slightly increase to 30.48%.

 Jimbo 11-22-2002 07:44 PM

Re: Am I making this AK calculation correctly

Dynasty I have followed that thread and I believe that where the misunderstanding arises is that you say exactly one ace and one king and Bernie is implying at least which is a more realistic assumption. In your scenario you must know all 18 cards in your opponnents hands where in Bernies presumption all you know are two being one ace and one king. I call what you are doing Vodoo math since although the calculations are correct the result is useless. That is my 2 cents worth!

 hutz 11-23-2002 11:41 AM

Re: Am I making this AK calculation correctly

Perhaps you two haven't fully defined the terms of your debate, so you're both right in a way. I'm sure I'm not going to articulate this very well, but it seems like there is a disconnect in your respective lines of thought. The odds of a king or ace flopping if one or more of each are out actually go down because there are fewer of them left in the dealer's hands. The odds calculation you went through assumes you know that only one of each are gone and that all of the other cards your opponents are holding are NOT aces or kings. The base 2.45:1 odds calculation you use for your comparison assumes only two cards are known (yours). I don't believe your analysis is comparing apples to apples. Perhaps that's the basis for your disagreement? [img]/forums/images/icons/confused.gif[/img]

 Robk 11-23-2002 02:38 PM

Re: Am I making this AK calculation correctly

Anyway, here's how I approach the problem. If you only know your two cards, the chances of flopping exactly one pair are what Dynasty said. But the chances of flopping one pair or better are
1-(No A or K) = 1- (44c3)/(50c3) =.324
If you know that one (additional) A and one K are out, and no other cards, the chances of flopping one pair or better are
1-(No A or K) =1- (44c3)/(48c3) =.234
If you also know one A one K and 16 other (nonaces/kings) are dead the chance of flopping one pair or better is
1-(28c3)/(32c3)= .340
So the chances have gone up slightly.
If you wanted the prob of flopping exactly one pair in this last case it is (28c2)(4c1)/(32c3)= what Dynasty said.
If you wanted the prob of exactly one pair in the second case it is (44c2)(4c1)/(48c3) = .219

 Jim Brier 11-23-2002 03:30 PM

My Calculation

Rather than focusing on flopping exactly one pair, I would focus on flopping one pair or better. This is simply one minus the probability of not having either an A or a K flop and it ignores the remote possibility of a flop containing a Q-J-T which is very small anyway.

If you have A-K, then there are 3 aces and 3 kings from 50 unseen cards. The probability of flopping a pair or better (ignoring flopping a straight) is 1 - (44/50)*(43/49)*(42/48) which is about 32%.

If you have nine opponents, one of whom has an ace and the other a king than you have 32 unknown cards of which 2 are aces and 2 are kings. The probability of flopping a pair or better (but ignoring flopping a straight) is 1-(28/32)*(27/31)*(26/30) which is 34%.

So, I believe you are correct. The probability of flopping one pair or better has increased from 32% to 34%.

 11-23-2002 06:18 PM

Question??

"Step 1: Calculate the the total # of possible flops
given that only 32 cards remain in the deck-
32*31*30/6 = 4,960"

Why divide by 6??

"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"

Why divide by 2??

 Dynasty 11-23-2002 06:58 PM

Re: Question??

Dividing by 6 prevents you from couting these flops as 6 flops rather than just 1 flop.

5c4c3c
5c3c4c
4c5c3c
4c3c5c
3c5c4c
3c4c5c

It does not matter what order the three flop cards come in.

 BruceZ 11-23-2002 07:06 PM

Re: Question??

Annie,

In step 1 we don't care about all the possible orderings of the 3 cards on the flop, so we divide by 6 which is the number of ways to order the 3 cards from each flop. 6 = 3 factorial = 3! = 3*2*1 = 6.

In step 2, 28*27 counts every possible xy pair twice, so we divide by 2. 2! = 2*1 = 1.

Not to confuse you, but another way to do this is to not do this division and count all possible orderings in both cases. In step 1 we would have 32*31*30. In step 2 we would have 4*28*27*3. We multiply by 3 because there are now 3 places the A or K an go. The final ratio is the same.

 BruceZ 11-24-2002 03:18 AM

Re: Question??

2*1 actually equals 2 not 1.

 Bob T. 11-24-2002 05:28 AM

Re: Am I making this AK calculation correctly

One Other way to look at it is, if you have an Ace and a King in your hand, there are 50 unknown cards of which 6 are aces and kings. If you have 9 opponents which jointly hold one ace and king, there are 32 unknown cards of which 4 are aces and kings. Since 4/32 is slightly more than 6/50, the chance of flopping a pair under those conditions is slightly more.

Good luck,
Play well,

Bob T.

 marbles 11-24-2002 02:00 PM

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.

 JTG51 11-25-2002 12:26 AM

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]

 David Ottosen 11-25-2002 06:10 PM

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.

 Ed Miller 11-26-2002 09:13 PM

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.

 Dynasty 11-27-2002 12:36 AM

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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