An easy one for you guys. (I think)

[cold_cash]

Last night I went down to my semi-local Native American casino and played a little low-limit Hold 'Em.

During the course of one hand at the end the session a question arose in my mind that I realized I didn't know the answer to, and I felt like even though I'm not a seasoned player, I really should have.

This is probably very simple, and something I may be able to figure out on my own if I applied myself. But I also figured I might be able to get from someone in here the proper way to do it, thus saving myself the trouble of doing it wrong over and over before I got it right. [img]/images/graemlins/grin.gif[/img]

So here's the sitch:

The game is kind of breaking up and we're waiting for some seats to fill up. At the time we're 5 handed.

I'm UTG w/ TT and open raise. A really loose, semi-agressive player to my left calls, and everyone else folds.

The flop comes down K83r or something like that. (A King and two non-straight undercards.) And I think to myself, "Self, what are the chances that this flop hit your opponent?", and realized I had no clue.

More specifically, what are the chances of catching a pair (or better) on the flop if you hold two random cards? (I want to guess you're like a 2:1 underdog, but I don't know where I heard this.)

What are the chances of going all the way to the river with two random cards and not improving at all?

Thanks for the replies, and I apologize in advance for not knowing this, and also for being too lazy/stupid to figure it out on my own. [img]/images/graemlins/grin.gif[/img]

[pudley4]

The easiest way is to figure out the probability that all 3 cards missed you.

You have AK: What is the probability there is no A and no K on the flop?

There are 3 A and 3 K left in the deck. 44 cards are not A or K. 50 total cards left.

The probability of the first card not being a A or K is 44/50. The probability of the second also not being A or K is 43/49; the third, 42/48. We need all three of these to be true, so we multiply them together. We get: 44/50 * 43/49 * 42/48 = .676 = 67.6%

So the probability of at least one A or K is 1 - .676 = .324 = 32.4%

The probability that none of the 5 cards hits your A or K is: 44/50 * 43/49 * 42/48 * 41/47 * 40/46 = .513

So the probability of hitting at least one A or K by the river is 1 - .513 = .487 = 48.7%

[scottjack]

I agree with your analysis, with exceptions. There are other ways to improve that don't involve pairing either of your cards. If they're suited, you could catch your flush; your AK could catch 10,J,Q, for the straight. Of course, the calculations to figure these probabilities are done in exactly the same fashion as you've indicated in your reply.

[cold_cash]

Excellent.

Thanks a lot Jeff.

[well]

[ QUOTE ]

The probability of the first card not being a A or K is 44/50. The probability of the second also not being A or K is 43/49; the third, 42/48. We need all three of these to be true, so we multiply them together. We get: 44/50 * 43/49 * 42/48 = .676 = 67.6%

[/ QUOTE ]

Actually, what you're saying is not true.

The probability for the first card indeed is 44/50.
But the probability 43/49 is actually the conditional probability of the second card
not to be an Ace or a King given the first card is no Ace or King.
For the third card I think it should be clear.

And now, to get the answer to the initial question, we cannot talk about "we need all these to
be true", but we will use

IP(all three no AK)=IP(last two no AK | first no AK) * IP(first no AK) =
IP(last no AK | first two no AK) * IP(second no AK | first no AK) * IP(first no AK)

So indeed we'll take the product of those three, but we don't have three events that need all be true (in this example).

Next Time.

[BruceZ]

[ QUOTE ]
[ QUOTE ]

The probability of the first card not being a A or K is 44/50. The probability of the second also not being A or K is 43/49; the third, 42/48. We need all three of these to be true, so we multiply them together. We get: 44/50 * 43/49 * 42/48 = .676 = 67.6%

[/ QUOTE ]

Actually, what you're saying is not true.

The probability for the first card indeed is 44/50.
But the probability 43/49 is actually the conditional probability of the second card
not to be an Ace or a King given the first card is no Ace or King.
For the third card I think it should be clear.

And now, to get the answer to the initial question, we cannot talk about "we need all these to
be true", but we will use

IP(all three no AK)=IP(last two no AK | first no AK) * IP(first no AK) =
IP(last no AK | first two no AK) * IP(second no AK | first no AK) * IP(first no AK)

So indeed we'll take the product of those three, but we don't have three events that need all be true (in this example).

Next Time.

[/ QUOTE ]

This is a subtle point that everyone would do very well to understand. To clarify, Jeff multiplied the correct 3 numbers and got the correct answer; the only issue is what he called these numbers *. The 43/49 and the 42/48 are not the probabilities of "the second and third cards not being an A or K". We are not using P(A and B and C) = P(A)*P(B)*P(C). That is only true when A, B, and C are independent, and we do not have independent events in this case because as each flop card is turned, it changes the probability that the other flop cards will be an A or a K. However, it is always true that P(A and B and C) = P(A)*P(B given A)*P(C given A and B) even if A, B and C are not independent. The 43/49 and the 42/48 are these conditional probabilities. They are not the same as the probabilities of "each card not being an A or K"; those probabilities would all be the same number, 44/50. That is, each of the 3 flop cards has a 44/50 chance of not being an A or K before any of them are turned. Those events are not independent, so we do not multiply those probabilities, only the conditional probabilities.

* I personally don't really have a problem with the way Jeff stated this since he inserted the word "also", which can be taken to mean that the first statement is given. I'm not interested in quibbling about the translation of math to English which is often tenuous; however, it is of critical importance that the distinction well is raising be thoroughly understood.