Odds of pairing AK by the river 62.55%

[jdeliverer]

Hmmm... that depends on how you've been calculating outs i suppose

If you are on the turn outs are really easy: #outs / 46 gives your probability of making it.

If you are on the flop, outs are a bit trickier: 1 - (47-#outs)/47 - (46-#outs)/46.

What you're doing is calculating the chance that you miss both times, then subtracting that from 1. This avoids double-counting if you hit 2 outs while still counting when you hit 2 outs.

 

[mdafka]

Lots of odds talk, but really its just a hugely over bet hand.
If anyone is floping pairs over 60% of the time, please contact me so I can stake you.. =)

 

[jaymfc]

lots of smart people but can't you just look it up easier , lol . thats what I thought op should have done , thats why I answered even with no clue myself .
since I don't understand all the math talk , I need to know if the answer I found means the same thing you guys are saying.

odds of hitting a pair or better by the river with AK (any two are same right ? ) are 3.2 to 1

 

[glworden]

OP is an idiot

Well, I guess I'm going to have to join the flip flop bandwagon myself. I kind of get it, although my brain is dragging a few yards behind.

Guess the OP is just an idiot.

Gary

 

[glworden]

PS -
Really glad I asked the question. The discussion surpassed my expectations as far as complexity and I really appreciate the fact that there are some smart people here. Plus, we've helped James well on his way towards fifty posts - and he didn't even know that was an issue.

James, you can play the CC money tourneys with only fifteen posts so you're all set for those. The best eV one is Tuesday's bodog tourney at 17:05 EDT.

Gary

 

[glworden]

Hmmmm,
"postScript" automatically comes out "pokerstars." What other little diversions are on here? Funny Talk. Ugly Betty.
Gary

 

[glworden]

Howw can that be?

lots of smart people but can't you just look it up easier , lol . thats what I thought op should have done , thats why I answered even with no clue myself .
since I don't understand all the math talk , I need to know if the answer I found means the same thing you guys are saying.

odds of hitting a pair or better by the river with AK (any two are same right ? ) are 3.2 to 1

Jay, I tried to look it up, but I found this information difficult to find. I look at the odds you quote here and they just don't seem right. If the probability is around fifty percent of pairing A or K by the river, then those are even odds. Where are you getting 3.2/1?

 

[Makwa]

K, I use the additive system many others do: Odds of pairing are 6 outs times 3 cards plus add one for flop (dont ask me why): 37%. By river add 6X2plus2= 20%, total 57% chance of hitting a pr w any 2 cards.
I have no idea why this system works or makes sense, but it seems to be in sync with other figures presented here.

 

[Dwilius]

48.74% chance of pairing your ace or king (or any 2 nonpaired hole cards) by the river.
Hope this convinces someone.

This includes making trip etc. This was the 20th response, we're nearing 60. I guess it didn't convince anyone. James is right, read that post.

For future reference, its about 1/3 to pair the flop and 1/2 to pair by the river. Are we done now?

Edit \/\/\/ btw Zach's way is faster, same answer.

 

[zachvac]

James is definitely right here. Think about it for a minute. You want the odds of pairing at least 1 of your cards. This is the same as the odds of not missing all 5 streets. So you calculate the odds of missing all 5 streets, do 1 - that and you've got it.

 

[zachvac]

btw if you don't like that logic, you'd have to do it this way:

6/46 + (40/46)*6/45 + (40/46)(39/45)*6/44 + (40/46)(39/45)(38/44)*6/43 + (40/46)(39/45)(38/44)(37/43)*6/42 = .5199 ~= 52%

On the other hand: 1 - (40/46)(39/45)(38/44)(37/43)(36/42) = 48%

ok wtf did I do wrong? These should be the same

 

[zachvac]

Oh I'm an idiot using 46 instead of 47, I'll test that and see if it works.

 

[dj11]

In SS I, Mike Caro states that if you have QQ, an A or a K will FLOP approximately 38% of the time (pg 567 in my version).

On page 566 he states that if you start with AK, and stay till the end, 49% of the time you will pair up. This should hold for ATC. But why you would want to pair up on a 35o holding is beyond me.....

The reference post in the op was the 60% figure which could easily be correct if you do not hold AK.

So, looks like everybody is right! And everybody is wrong!

The problem is that these related, but not equivalent stats are muddled into perceptional stats regarding AK, and no context is provided.

When dealing with stats specifics can be everything!

Example;
Take 25% away from 100. Increase the result by 33%.

We end up dealing with a set of numbers, which at first glance suggest that 33% =25%! WTF ??

It takes a while before we finally figure out that the real question should be;

% of what?

 

[jtberrym]

math

Math is a good thing to have in the back pocket but i rely mor eon situations and reading other players betting styles when i play. AK is way overplayed in my opinion

 

[Dwilius]

Point taken DJ, I had written a different response because I misunderstood your first sentence. I thought it was AK v QQ, anyway this was part of it and it goes along the lines of your post...

Knowing your opponent has an ace or king changes the odds significantly, and these are the hands AK is most powerful against.
If your opponent holds Ax or Kx you have a 2/7 chance of pairing the flop and a little more than 3/7 chance of pairing by the river.

 

[zachvac]

ok finally:

1 - (41/47)(40/46)(39/45)(38/44)(37/43) = (6/47) + (41/47)(6/46) + (41/47)(40/46)(6/45) + (41/47)(40/46)(39/45)(6/44) + (41/47)(40/46)(39/45)(38/44)(6/43) ~= 51.1%

Rationale for first one was explained, second one is because you want the odds that it flops on the first one, then to not count the times it hits 1st and 2nd you figure the odds it doesn't hit 1st and does 2nd.

So basically you're saying you want to combine the odds of it hitting first street, not hitting first street and hitting second street, not hitting the first two and hitting third, etc. If you don't add these conditionals you end up where if 2 aces or kings come you're counting it twice, and this isn't right.

Note that dj is right, if you don't hold AK then there are 8 outs for the A or K, and 1 - (39/47)(38/46)(37/45)(36/44)(35/43) ~= 62.5%

 

[Makwa]

K, I use the additive system many others do: Odds of pairing are 6 outs times 3 cards plus add one for flop (dont ask me why): 37%. By river add 6X2plus2= 20%, total 57% chance of hitting a pr w any 2 cards.
I have no idea why this system works or makes sense, but it seems to be in sync with other figures presented here.

Hey, I got the numbers closely right. What is wrong with my system that I got from Phil?

 

[jdeliverer]

Zach what is this that you are calculating?

Makwa I think you just got lucky. If you think about it there's a quite significant difference between the actual answers and your estimates.

 

[ChuckTs]

ok this thread is now causing head asplotion for me. Thought I knew what I was saying but now my mind is just going in circles.

Not saying you're right james, but I'm not saying you're wrong either. I'm just lost

Either way, what we're calculating here has little bearing or usefulness in poker...I mean aside from keeping our math skills in check. Equity is what we should be concerned with. For the beginners, don't go leaning on the (correct) math from this thread. Punch AK into an equity calculator like pokerstove and see how it does vs certain hands on all streets. That'll give you an idea of what you want to do with the hand.

 

[Toad]

Not to beat a dead horse but.....

I think I might know wher the confusion comes in. I remember reading in one of many poker books about using the additive sysytem for quick calcualations because it is relatively close. (It might have been Sklansky's Theory of Poker - not sure)

Anyway, I remember there being an asterick next to it saying the actaul calulation is done the way james says.

Adding the probabliities does get you somewhat close for the first few cards, but the more cards you take the higher the margin of error.

Example:

Odds of getting A or K on flop

with additive estimate method: 6/50 + 6/49 + 6/48 = 36.74%

with the correct method: 1 - (44/50)*(43/49)*(42/48) = 32.43%




The additive method falls apart with more cards though. Let's say you want to figure the odds of hitting a A or K with the first 10 cards:

(6/50)+(6/49)+(6/48)+(6/47)+(6/46)+(6/45)+(6/44)+(6/43)+(6/42)+(6/41)

.1200 + .1224 + .1250 + .1277 + .1304 + .1333 + .1364 + .1395 + .1429 + .1463 = 1.324 = 132%

probabilities can't be greater than one - so this is obviously not the right way to do it




Thank God for PokerStove

 

[hornellfred]

I always figure that the odds of catching a pair on two unmatched cards by the river is just less than 50%. Therefore I agree with James. Been awhile since I took statistics in college but I do know you don't add you multiply and have to account for the times you would hit before the river

 

[icepick007]

personally for me i feel happier to fold if i see a huge reraise,and wen i tend to think that the other player has a high pp,or top pair on da board...its not really worth sinking all ur chips in with nothin on da flop,hoping yu'll catch somehing on da turn or da river

 

[jdeliverer]

Well in that situation I guess it depends on pot odds, you need about 3:1 odds to call there.

I don't know what you're responding to with this post though, we've mainly been discussing poker odds rather than what to do with a reraise..

 

[Dashir]

'The chances of either an Ace or a King appearing on the board by the river are 60 percent.'

This does not say that those are the odds of pairing AK by the river. It is entirely possible tho that I extrapolated in another post (hopefully burried extra deep) that those odds were valid for pairing.

I think that 60% is from the full deck of 52, not a hand where you already have AK in your hand (so 6 instead of 8 left).

I just found this thread today and it's kind of funny. James posted the correct and clearly explained answer and then got three pages debating simple statistics.

You can never keep adding percentages together. If that were true, then if you flipped a coin twice you would have to have at least one heads and at least one tails.

 

[jdeliverer]

I think that 60% is from the full deck of 52, not a hand where you already have AK in your hand (so 6 instead of 8 left).

I just found this thread today and it's kind of funny. James posted the correct and clearly explained answer and then got three pages debating simple statistics.

You can never keep adding percentages together. If that were true, then if you flipped a coin twice you would have to have at least one heads and at least one tails.

Odds of an Ace or a King being dealt in 5 cards (all hands hidden):
1 - (44/52)*(43/51)*(42/50)*(41/49)*(40/48) = 58.12%

Odds of an Ace or a King being dealt in 5 cards (no A or K in your hand):
1 - (42/50)*(41/49)*(40/48)*(39/47)*(38/46) = 59.85%

Doesn't really matter I suppose, if you have QQ I guess this makes a difference. But the chances of an A or K coming are lower if someone else has an ace or king.