What are the chances of 2 people being dealt an ace?

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luckytokenz

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If I have lets say Ace small suited, and I flop an ace. What are the chances that my opponent has atleast 1 ace in there hand?
 
PattyR

PattyR

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you cant really deciphyer if they have an ace or not lol, not sure if thats possible, there are programs though that will tell you how likely you are to have the best hand

one example is a program called poker stove i think, i dunno search around there are alot of different programs out there, obviously they are illegal to use in real money cash games or real money tournaments but they will work wonders for your game in play money, helped me to become an elite cash player

best of luck man
 
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Qutsemnie

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Depends on how many people are at the table originally, I think one of the things that doesn't get communicated properly is the filtering effect.

In micro, alot of people play an A no matter what its paired with if you want to assume that is true in a 9 way then your really asking what is the probability that any of my other 8 opponents were dealt an ace preflop even if there is only one left post flop. This is easiest calculated as the probability that none of the 8 people hit an ace. To do this we just imagine dealing out 16 cards out of a 47 card deck with 2 aces left in it.

The probability of not dealing an ace on the first card is 45/47. The probability of not dealing an ace on the second card is 44/46. On down to 30/32.

Multiplying all these together gives a bunch of fractiosn most of the terms can be canceled and you get 31*30 / (47*46) = .43.

So the probability one of the other 16 cards is an ace given what you flopped is .57. And you have to figure they would limp in wiht it at 1bb or if they are loose play it at 3bb at a low stakes table. So even if you are down to one person you have to put him on it 57% of the time.

Now your not supposed to play an ace with a weak kicker on a big table, and incidentally this is exactly why. 57% of the time some one else has it, but if your kicker is weak more than not they got you dominated.
 
jdeliverer

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With 2 aces used up there are 2 remaining. There are 47 unknown cards and villain has 2 of them.

So (45/47)*(44/46) is the probability that he has NO aces.

Taking 1 - (45/47)*(44/46) gives the probability that he has at least 1 ace.
This is 8.42%.

Now, keep in mind that villain is not calling with any random hand, so the chances that he actually has an ace in his hand (depending on preflop action) may be radically hgher. The above calculations assume villain blindly called or checked preflop.

Edit: My math is correct, not Qutsemnie. I don't know where you got the 30/32, but even thinking about it logically your answer makes no sense. Does your opponent really have top pair 57% of the time when you have it? Far from it.
 
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Qutsemnie

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James rules for your table:
There are 9 people. Everyone always plays to flop if they get an ace (loosely). You are facing one person after the flop. You have an Ace, the flop has an ace, what is the probability that person across from you has an ace?

Thats the question I answered. It is not at all the same as dealing someone two cards from a 47 card deck.

edit: My answer is actually slightly off, but only a bit, because it is the probability that one or two villans get dealt one or two of the remaining aces preflop conditioned on your hand and the flop, and since in this outcome only one villan could possibly have gotten an ace that is overcounting by a bit.

As a test if you have a deck of cards you can put an ace and a duece in your hand, put an ace king queen in a flop. And then deal out 8 other hands over and over and count how many times one or more of those 8 hands has an ace (alternatively you can just thumb through the top 16 cards of a 47 card deck shuffled over and over). I am saying its bout half. Now you figure if your at a table where everyone is playing their aces to flop, those other aces always end up at the flop.
 
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luckytokenz

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With 2 aces used up there are 2 remaining. There are 47 unknown cards and villain has 2 of them.

So (45/47)*(44/46) is the probability that he has NO aces.

Taking 1 - (45/47)*(44/46) gives the probability that he has at least 1 ace.
This is 8.42%.

Now, keep in mind that villain is not calling with any random hand, so the chances that he actually has an ace in his hand (depending on preflop action) may be radically hgher. The above calculations assume villain blindly called or checked preflop.

Edit: My math is correct, not Qutsemnie. I don't know where you got the 30/32, but even thinking about it logically your answer makes no sense. Does your opponent really have top pair 57% of the time when you have it? Far from it.



so would you put them as high as 15% if you figure you opponents to be playin slightly better than average hands?
 
jdeliverer

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Sure, that's probably reasonable, anywhere from 10-20% at a rough estimate.
 
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SlipperyPete

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I personally don't play too aggressively with ace-low card suited unless I hit decently hard on the flop with something like two pair or a straight or flush draw. If the board is all garbage besides the ace you can pretty much get an idea of what the other person has by the way they play back at you.

If I feel like they have the ace i wont try to draw out on them ill throw the hand away easy and just look for a better spot. I don't think calculating the odds they have an ace as well will help you out too much. IMO reading you're opponent on the specific hand at play is the only thing you should be looking at.
 
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BluffYou123

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Whatever the chances are, if someone does have one, they will likely have played it and it is very likely to be better than yours if you have a bad kicker.

Unless you hit your 2 pair, you might be best advised to get out of the hand, depending on betting of course.
 
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Bobmurphy07

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I think it's less important to know the odds of them having a bigger A, than it is to know what range of hands they play. I mean I saw someone mentioning that an opponent has 2 random cards out of 47 but he's not going to a flop with such a wide range. I would say if you're playing an A with a small suited kicker than it should be a 5 or lower and if you don't pickup a flush/straight draw that you be prepared to fold even if an A hits.
 
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Qutsemnie

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so would you put them as high as 15% if you figure you opponents to be playin slightly better than average hands?

That is one way to go at it (by counting the hands that opponent plays) but look at this link:
HTML:
http://en.wikipedia.org/wiki/Poker_probability_(Texas_hold_%27em)

Then go down to "Probability of facing an ace with larger kicker when holding" which does not account for the ace in the flop btw. If your playing A9 with 9 people in the ring then there is a 29% chance someone got dealt a whole card that was an ace and something bigger than a 9. If your playing A9 with six people in the ring then there is a 19% chance that someone in the ring got dealt an ace with a bigger kicker (not taking the information from the flop into account which lowers it a bit).

So one way to look at it is to take the hands your opponent would play to flop that beat you and divide by the hands that they would play. And another way to keep your intuition about the probabilities sharp is just be aware of the sifting nature of these card games. Good hands get sifted through the bets and raises and bad hands get sifted out (basically a variant on the monte carlo problem).

Both ways are informative in their own way.
 
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luckytokenz

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thanks! I'll check it out and let you know what I think.
 
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