Two Plus Two Older Archives  

Go Back   Two Plus Two Older Archives > General Gambling > Probability

Reply
 
Thread Tools Display Modes
  #1  
Old 11-22-2002, 07:04 PM
Dynasty Dynasty is offline
Senior Member
 
Join Date: Sep 2002
Location: Las Vegas
Posts: 4,044
Default 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%.

Reply With Quote
  #2  
Old 11-22-2002, 07:44 PM
Jimbo Jimbo is offline
Senior Member
 
Join Date: Sep 2002
Location: Planet Earth but relocating
Posts: 2,193
Default 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!

Reply With Quote
  #3  
Old 11-23-2002, 11:41 AM
hutz hutz is offline
Senior Member
 
Join Date: Sep 2002
Location: Houston
Posts: 545
Default 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]
Reply With Quote
  #4  
Old 11-23-2002, 02:38 PM
Robk Robk is offline
Senior Member
 
Join Date: Oct 2002
Location: Chicago
Posts: 1,242
Default 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
Reply With Quote
  #5  
Old 11-23-2002, 03:30 PM
Jim Brier Jim Brier is offline
Senior Member
 
Join Date: Sep 2002
Location: Las Vegas, NV
Posts: 189
Default 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%.
Reply With Quote
  #6  
Old 11-23-2002, 06:18 PM
Guest
 
Posts: n/a
Default 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??
Reply With Quote
  #7  
Old 11-23-2002, 06:58 PM
Dynasty Dynasty is offline
Senior Member
 
Join Date: Sep 2002
Location: Las Vegas
Posts: 4,044
Default 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.

Reply With Quote
  #8  
Old 11-23-2002, 07:06 PM
BruceZ BruceZ is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,636
Default 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.
Reply With Quote
  #9  
Old 11-24-2002, 03:18 AM
BruceZ BruceZ is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,636
Default Re: Question??

2*1 actually equals 2 not 1.
Reply With Quote
  #10  
Old 11-24-2002, 05:28 AM
Bob T. Bob T. is offline
Senior Member
 
Join Date: Sep 2002
Location: Shakopee, MN
Posts: 3,657
Default 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.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 05:30 AM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, vBulletin Solutions Inc.