Odds of pairing AK by the river 62.55%

I saw a thread a couple days ago that I just can't seem to find now. One guy said that Cardplayer magazine says the odds of pairing an AK in your hole is 60% by the river. Another guy did some math and said no, it's 50%. I can't find the thread to see how the discussion progressed and what conclusions were drawn. I also can't find this particular scenario in any odds tables.

I come down on the 60% side. I do this with quick approximate odds calculations. In order to pair a hole card, there are six outs and five streets. Six outs per street = about 12%. Add five street possibilities to arrive at 60%.

Now I'll do what I believe to be the actual odds. If I'm doing this wrong, then I've been doing it wrong for ages and somebody needs to set me straight.

Since I only know my two hole cards, I'm looking for one of six outs from a universe of 50 cards on the first street. The probability is therefore 6/50. If I don't hit, I still have 6 outs from a universe of 49 cards so the odds of hitting on the second card of the flop (if I didn't hit the first) is 6/49, and so on.

Adding these probabilities I get:
(6/50) + (6/49) + (6/48) + (6/47) + (6/46) =

.1200 + .1224 + .1250 + .1277 + .1304 = .6255

So, the way I get it, there's a 62.55% chance that you'll pair at least one of your hole cards by the river. If you're all in pre-flop and your opponent shows no ace or king, your chances are even higher since the universe of available cards now starts with 48 rather than 50.

I'm not saying AK has a 62.55% chance of winning. There are odds calculators which show those odds against random hands and against given hands.

If you don't pair a hole card 62.55% of the time by the flop, tell me where I'm wrong, please.

Gary

to early for me and numbers but it sounds like you're playing alone or in some other universe . I think your odds are like 6 or 7 to 1 but I could be wrong too . I do know you'll never have 50 available cards unless you're playing solitaire , you got to account for the cards already out . we have plenty of good math people to help but I'm not one of them .
just busting your chops cause you're a clayhole

I think the odds one should concentrate on is the odds of winning, not pairing! If a lot of players stay in it is quite likely there are more aces and kings on the board.

I think it would have to be less than 60%. The odds of pairing AK aren't going to beany different than say pairing 62 or 95....

Thinking from experience, I dont believe that I would get a pair 60% of the time.

I googled the odds of getting a pair on the flop, and they are 32%. If you dont get a pair on the flop, then the odds are, like you said, 6 outs times 4% = 24%... Even if you had 32% and 24%, you get a number less than 60%, and you can't just add percentages like that, when you add them properly, they end up being less than if you flat out added them, so the odds must be less than 58%...

However, if you factor in the way your opponent plays preflop, you can predict whether the odds are better or worse of pairing AK. If someone bets preflop, they could very well have an A or a K thereby reducing your number of outs (though of course they could have a pocket pair). Also, if nobody bets preflop, they might not have an A or a K and so your odds may be increased.

Well you just did the (correct) calculations of five cards running, hence you would have that chance of hitting your A/K by the river. For the flop, you would simply take the probabilities of the first three cards: (6/50) + (6/49) + (6/48) = 0.37 and 0.37*%100 = %37. But ya, your math was right for the river calculation.

Really when you take a step back and look at what were doing here, the calculations don't help us at all except in keeping our math skills sharp. The chance of hitting your ace or king doesn't matter at all, since a decent portion of the time your hand still won't be best, plus there are some times where your hand will be best but will get outdrawn on later streets. Basically the math we're doing doesn't matter :/

If you want to start looking at the maths of poker, start using equity calculators like pokerstove.

Sorry, but you're doing the percents wrong. You can't simply add the probabilities. The easiest way to do it is to figure out the chance that you never pair, then subtract that from one.

There are 44 cards that won't pair you the first time, 43 the second, 42 the third, 41 the fourth, and 40 the fifth.
The actual probability is 1 - (44/50)*(43/49)*(42/48)*(41/47)*(40/46) = 48.74%.

To get the probability of pairing on the flop, just leave out the last 2 factors.
1 - (44/50)*(43/49)*(42/48) = 32.43%, confirming what razzle dazzle said.

I'll try to explain the problem with your reasoning. When you add the probability of the 2nd card, you do not take into account the probability that the first card paired. Thus, you are double-counting the times when the first card pairs and the second card pairs.

Hope that clears it up.

Thanks for the thoughts, but not the kinds of responses I'm looking for. If I'm wrong, I'd like somebody to show me logically and mathematically. I'm not seeing where I'm wrong, so if I am, I need to find out where my blind spot is.

You're right, the odds of pairing and A or a K are the same as pairing a 7 and a 2, or any combination of unpaired hole cards for that matter.

You are in fact drawing from a universe of 50 unknown cards pre-flop. No fantasy there. Fifty-two cards in the deck, you have two in your hole, that leaves fifty unknown to you. Yes, some cards have already been distributed to other players - all unknown to you. What are the odds that you'll pair up if given five of those fifty cards? That's the basis of the question.

As to adding pre-flop and post-flop percentages, it actually does work out pretty close. Pre-flop, you're looking for one of six cards on three streets. 6 X 2 X 3 = 36%. Continuing with rough odds, the chances of hitting on the turn or the river are roughly 6 X 2 X 2 = 24%. Add those up and it's 60%.

Yes, the chance of winning is the important thing. We all know it's race with AK versus a lower pair. AK has a pretty big advantage over lower unpaired cards.

I ask the question because there was a discussion on another thread regarding the strength of AK pre-flop in a possible all-in situation. One guy claimed he read 60% of pairing in Cardplayer, another guy said 50%. I lost the thread and wanted to know how the discussion was resolved. Thanks for your thoughts, but I'm looking for something more concrete. It's definitely not 6 or 7 to 1 against.

I base my calculations on Mathew Hilger's book on Hold 'em odds, which I read again last night.

I'm pretty sure it's over 62%. Prove me wrong.

Gary

Thanks, this is a good explanation. But aren't you calculating the probability of pairing EXACTLY one card? I think my approach shows the chances of pairing AT LEAST one hole card, doesn't it? That's really what I'm looking for, what are the chances of pairing or tripling at least one hole card. If that's the premise, are my calculations correct?

Your right, so it'll be hard to prove you wrong. You'll hit an A or a K ~62.5% of the time. Your math is spot on.

Nope, if you reread my paragraph, I calculated the possibility that NO cards pair, then subtracted it from 1. This means that it counts when both pair, or even trips. I'm pretty sure that adding the probabilities is essentially meaningless.

I'm 99% sure that my math is correct, which seems to be consistent with the fact that AK beats 22-QQ a bit less than half the time. The percentage of winning is slightly lower than the chance of pairing because a pp could hit trips.

If someone has statistics from some tracking site that could back this up, it would be nice to get some validation. However, you certainly don't pair 62.5% of the time just from thinking about it.

Doesn't this really only matter in all ins pf? AKs vs a pr is about 48%. If you miss the flop and just have 2 overs, then you are about 25% to pr by the river.

[quote=jamesdadeliverer;873651]

I'm 99% sure that my math is correct, which seems to be consistent with the fact that AK beats 22-QQ a bit less than half the time. The percentage of winning is slightly lower than the chance of pairing because a pp could hit trips.

quote]


The reason AK beats lower pairs a little less than half the time is because the lower pairs will sometimes turn into trips, straights, flushes, 2 pair, etc. Even if you hit an A or a K on the flop you still lose.

The OPs calculations are correct....you will hit at least a pair ~62% of the time.

Haha this is all very interesting, but really doesn't matter at all. Pretty much, you have a top 5 hand, and you should be playing it lotz

Sorry, but you're simply wrong. You don't make a pair 62% of the time.

Someone please post some statistics to back this up, if they are available.

This is the best explanation I have seen so far. I am working on it. I am very interested in the math aspect of the game and this is challanging my tiny brain hard. I will post if I can come up with any other alternative but so far I concur.

Am awsome resource for these type of problems is a book called "Killer Poker By the Numbers" by Tony gurrera. It is by far the best book on math that I have found.

So what does the book say on the chance of either hole card pairing by the river?

The more I read the responses, the more confident I am in my approach and math. I'm not being obstinate but nobody's really offered a compelling argument otherwise. Saying "I'm pretty sure you don't add the percentages" doesn't sway me, because that's exactly what you do. That's what Hilger does in his stats book, though surprisingly he doesn't address this specific scenario.

All the people who say it doesn't matter - what's important are the odds of winning: you're RIGHT! That is the bottom line, and I do understand the race situation. But as a mathematical exercise, we still lack a compelling and conclusive argument whether either hole card should pair by the river 62.5% of the time. I believe this figure is correct and I've seen some good arguments and approaches otherwise, but so far nothing compelling.

I'll continue looking.

Gary

I don't know if anything will convince anyone at this point, but here's where you make the error in your logic.
""
If I don't hit, I still have 6 outs from a universe of 49 cards so the odds of hitting on the second card of the flop (if I didn't hit the first) is 6/49, and so on.
""

But the chances of hitting on the second card of the flop (if you didn't hit the first) is NOT 6/49, it is 6/49 * 44/50. This is because you aren't taking into account "if you didn't hit the first".

I'll post again soon showing an alternate way that involves adding percentages, as Hilger does.

Please explain. If we miss on the 1st card the probability on the next card is EXACTLY 6/49.

You are working out the odds of hitting only 1 pair James.

Which is the answer to a different question.

So here's how to add percentages the right way.

First card:

6/50 cards will pair us. This is 0.12.

Second card:

If the first card didn't pair (44/50), 6/49 cards will pair us. This is (44/50)*(6/49) = 0.1077.

If the first card paired, we already counted it.

Third card:

If neither the first card nor the second card paired (1 - 0.12 - 0.1077) = 0.7723, then 6/48 cards will pair us. This is 0.7723*(6/48) = 0.0965.

If either of the first 2 cards paired, we already counted it.

Fourth card:

If the flop didn't pair us (1 - 0.12 - 0.1077 - 0.0965) = 0.6758, then 6/47 cards will pair us. This is 0.0863.

If the flop paired us, we already counted it.

Fifth card:

If the flop & turn didn't pair us (1 - 0.12 - 0.1077 - 0.0965 - 0.0863) = 0.5895, then 6/46 cards will pair us. This is 0.0769.

If flop/turn paired us, we already counted it.

So adding the chance of pairing on the 1st - 5th cards gives us:
0.12 + 0.1077 + 0.0965 + 0.0863 + 0.0769 = 0.4874

0.4874*100% = 48.74% chance of pairing your ace or king (or any 2 nonpaired hole cards) by the river.

Hope this convinces someone.

Yes, the probability is exactly 6/49.

But the probability of you being in that situation is only 44/50.
You have to multiply the probability of an event by the probability of being in the siutation where the event occurs.

Imagine you have a coin that you will only keep flipping if you get heads. The probability of getting tails the second flip is 50% as always, but the chance that you will even flip the coin a second time is 50%. So the chances that you actually get tails the second time is 25%. I don't know if that makes sense, but its the same concept.

If you already made one pair, you shouldn't be taking into account the chance that you make another. Otherwise, you will double-count the times when you made 2 pair or trips.

To sum up: The chance that you will hit the second card is 6/49 when you miss the first card.
The chance that you will miss the first card and and then hit the second is (44/50)*(6/49).

You already counted any situations where you hit the first card.

OK. It's starting to make a little bit of sense, but I still can't get my mind completely around it. Are you a mathematician, James?

This reminds me of the classic gameshow question: You're offered a choice of three doors. Behind one is a car, behind the other too are popsicles. Assuming you want the car, after choosing a door, the host - who knows what's behind each door, opens one with a popsicle, then offers you the choice of trading your door for the remaining door. Should you do it? When Marilyn vos Savant presented this question in her column, she asserted that you should always switch; your chances of success are 2/1. Many angry mathematicians blasted her, saying the odds are obviously 50/50: two doors, behind either of which could be the car. Though not easily apparent, Savant is correct. You should always switch.

James sounds like he knows what he's talking about, and I sense he's right. I'm trying to understand it. I did find this 50% figure in a table on the website pokercompatability. If it addresses the exact same question, which I'm getting confused about, it supports James's figure.

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You're not paying attention to what the other posters are saying, namely:

You're doing exactly what Rex said, and are wrong for our the OP's example.

Using your coin analogy you explained what you're doing wrong perfectly:

And you're carrying that over to the poker example. We don't 'stop flipping', or stop calculating the probabilities if we hit an ace or king. We run all five cards. OP's not asking the probability of hitting exactly a single ace or king (no more, no less), he's asking the probability of hitting a pair or better.

Merely an aspiring mathematician gl, but I consider applied statistics my specialty. Often the hardest part of the problem is getting your head around it. I never really understood what was so difficult about the Savant problem.

You had a 2/3 chance of being wrong when you picked a door. So when you are given the option to switch, do you really want to stick with the 1/3 chance that you got it right?

Sorry if I was a bit vehement in my responses sometimes. I hope my calculations cleared it up for you though.

Whenever I play AK, it certainly feels like I hit it about 25% :P.

Luck evens out over time, but humans don't live forever. I got the crappy end.

Chuck T is right.

Cheers Chuck,

James, read the posts above.

(the table uses the same logic, one King or one Ace)

I am paying attention to what they're saying. I am not calculating the probability of hitting exactly a single ace or king; rather, the others are double-counting the probability of hitting more than one.

And for this argument, you should 'stop calculating the probabilites if we hit an ace or king' because you've already counted it. And if you already counted it, it doesn't matter whether it pairs or not. I don't think there's any more I can say to convince you though, just try to look at it again.

At least we agree you are answering the wrong question and the answer to the OP is nearer 60%.

NO!! We don't agree at all.

If you look at my original post, you can see that I calculated the probability of NEVER pairing anything. Then I subtracted that from 1, leaving us with the probability that ANYTHING paired, one pair or two pair or trips, a FH or even quads. I am taking into account all cases.

If you go by the original answer, you are counting the first pair you get, in addition to the second pair, and anything else. Thus you are counting two pair or trips twice and a full house or quads three times.

Ooo hang on you are saying you are measuring the same thing.

In which case your methodology is very flawed.

James, there is no animosity at all. I appreciate the discussion, even if I'm still not getting my head around it totally. It's an odd thing, isn't it, that we have all these poker players who can't agree on what is seemingly basic and common statistic. I am the OP; my logic makes sense to me and many posters are backing me up. But heck - maybe we're all wrong!

To add further to the confusion, here's an online calculator which came up with a COMPLETELY different answer. I think I entered the info correctly. We're drawing from 50 cards, drawing 5 cards with 6 outs, and only require one hit to fulfill the parameters. Now this thing is saying our probability of hitting is 38.4%.

Can you explain that one?

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Don't bother worrying about that calculator, it's probably only meant for postflop draws and adding the 5th card screwed something up. Just ignore that one.

Ok, I'm going to do an about turn here I think.

There is a 54.5% chance of not drawing an A or a K. And so therefore a 45.5% chance of drawing at least one.

The logic of your other posts makes sense too. So unless someone comes along with further evidence to the contrary, apologies James.

Zach, where are you??

I'm blaming this on not having eaten for two days due to stomach bug fwiw.