what s the odds of villain getting trips on turn or river when i have top pair? i tend to not value bet my good hands when the board pairs because i get beat soooooooooooo many times by ppl turning trips...

typical hand :

i m in utg+1 i open AQ villain is a 40/2 fish calling station he calls. flop comes AKx i bet he calls turn x river is K. he shows KQ

It's 50/50

Assuming his pair is not the same rank as yours, 2/45 + 2/44 = 9% or 10:1.

Slight correction

2/45 + (1-2/45)*2/44 = 8.8%

Yes, I knew my number counts the runner runner quads, but it was simpler. My form would be:

2/45 + 2/44 - (2/45 * 2/44) = 8.8%

I actually think the 9% result is closer to the spirit of what OP wants to know, but of course your correction is right.

Wrong, your method 2/45 + 2/44 counted the quads 3 times. The equation above does count quads 1 time. It counts all times that we hit on the turn which includes quads. Notice that this method subtracts 2 times the probability of quads which is 2/45 * 2/44, since 1*P(quads) = 2/45 * 1/44, and since the whole expression counts quads once, 2/45 + 2/44 counts the quads 3 times.

If we take 1 minus the probability of missing on both cards, 1 - (43/45)*(42/44), this is counts quads 1 time, same as the formula above. If we add twice the probability of quads onto this, or 2/45 * 2/44, this counts quads 3 times, and you can verify that it is equal to 2/45 + 2/44.

2/45 + 2/45 double counts the quads, so another way to cmpute the probability of a set or quads is 2/45 + 2/45 - (2/45 * 1/44), where we subtract off the probability of quads by the inclusion-exclusion principle.

Wouldn't the answer be

P(getting it on the turn and not the river)+P(getting it on the river and not the turn)+P(making quads)

This should be the same as 1-P(not improving)

You can just leave off the quads if you don't want it. I suspect OP wants to include quads in his answer.

P1(getting it on the turn and not the river)=2/45*43/44
P2(getting it on the river and not the turn)=43/45*2/44
P3(making quads)=2/45*1/44

P1+P2=172/1980~.087
P3=2/1980~.001

P(not improving)=43/45*42/44=1806/1980~.912

1-P(not improving)=.088.

So 8.8% includes quads and 8.7% excludes quads.

Yes, that's all correct, so we have another method in addition to the 3 described above.

How/where do you see 2/45+2/44 triple counts the quads?

How/where do you see 2/45+2/45 double counts the quads?

I can't see that for the life of me. I can verify it with simple math, but how (the heck) do you see it?

All I can do is say "Nope, that's wrong" but I can't see how/why/where it's triple/double counting quads.

Before either card is turned over, they each have the same probability 2/45 of making our hand. 2/45 includes all cases where we hit on the turn, including those cases where we also hit on the river for quads. The second 2/45 includes all cases where we hit on the river, including those cases where we also hit on the turn for quads. So adding these double counts the case where we hit on both for quads. This is the whole basis of the inclusion-exclusion method, and why we can't add the probabilities of non-mutually exclusive events.

P(hit on turn OR hit on river) = P(hit on turn) + P(hit on river) - P(quads)

= 2/45 + 2/45 - (2/45 * 1/44).

Now the difference between 2/44 and 2/45 is 2/(45*44), so

2/45 + 2/44 = 2/45 + 2/45 + (2/44 - 2/45)

= 2/45 + 2/45 + 2/(45*44)

= 2/45 + 2/45 + (2/45 * 1/44)

Since 2/45 + 2/45 double counts the quads, and since 2/45 * 1/44 is the probability of quads, 2/45 + 2/44 triple counts the quads.

Also, as I wrote in my previous post, 2/45 + 2/44 - (2/44 * 2/45) is the probability of making our hand, including quads counted 1 time. This comes from multiplying out 2/45 + (1 - 2/45)*2/44. Now since this counts quads 1 time, and since 2/44 * 2/45 = 2*P(quads), this means that 2/45 + 2/44 must count the quads 3 times. This is how I first saw it.

I really appreciate you taking the time to explain this stuff to me.

But there's only 44 unseen cards on the river so using 45 in the denominator seems inappropriate. WTF?

How can we talk about the turn and river in isolation like that? I see P(hit on turn) as 2/45*43/44 and P(hit on river) as 43/45*2/44.

That's how I got:

P1(getting it on the turn and not the river)=2/45*43/44
P2(getting it on the river and not the turn)=43/45*2/44
P3(making quads)=2/45*1/44

P1+P2=172/1980~.087
P3=2/1980~.001.

I guess I'm not using the inclusion/exclusion principle in my calcs where you are? I'm just adding together the probabilities of all three possibilities and you're using the inclusion/exclusion principle to simplify the approach?

Thanks a lot for helping me better understand the subtle nuances here. It seems like this is all back of your hand obvious to you.

We are talking about the probabilities computed on the flop before either the turn or river cards are exposed. These are just 2 cards that have the same probability of hitting our hand. In fact, every unseen card in the deck has the same probability of being one of our 2 outs, and that probability is 2/45 since there are 2 outs and 45 unseen cards, each of which has the same probability of being an out.


Again, these are the probabilities of the turn card being an out, and of the river card being an out, BEFORE EITHER CARD IS DEALT.


No, these are P(hit on turn AND miss on river) and P(miss turn AND hit on river). You computed them as:

P(hit on turn AND miss on river) = P(hit on turn)*P(miss on river | hit on turn)
= 2/45 * 43/44

P(miss on turn AND hit on river) = P(miss on turn)*P(hit on river | miss on turn)

= 43/45 * 2/44

Where P(A | B) = P(A GIVEN B). You have created two mutually exclusive events (hit on turn AND miss on river) and (miss on turn AND hit on river) since only one of these can occur. The event (hit on both) is also mutually exclusive. Since these are all mutually exclusive, you can add their probabilities to get the probability that one or the other or both occurred. This is fine, and it is not the inclusion-exclusion principle. You don't need the inclusion-exclusion principle since you created mutually exclusive events.

The events (hit on turn) and (hit on river) are not mutually exclusive, so if I add their probabilities 2/45 + 2/45, I am double counting the cases where I hit on both, so I must subtract the probability that I hit on both 2/45 * 1/44 by the inclusion-exclusion principle.


That's correct.

Bruce,

Thanks. Your knowledge seems deep, rich, thorough, and complete. Not that I'm in a position to assess your level of knowledge.

Thank you for taking the time to explain these concepts to me. Your arguments are clear and compelling!

Did I meet you this last July at the 2+2 event in Vegas at the Hard Rock? I got there early and I met two guys, TylerCracker and one other guy who is a Phd teaching math at UNLV. Was that you? If it was I remember I talked with you briefly and said I got lost in math when math went abstract. Do you remember?

Anyways ...

Me: But there's only 44 unseen cards on the river so using 45 in the denominator seems inappropriate. WTF?

You: We are talking about the probabilities computed on the flop before either the turn or river cards are exposed. These are just 2 cards that have the same probability of hitting our hand. In fact, every unseen card in the deck has the same probability of being one of our 2 outs, and that probability is 2/45 since there are 2 outs and 45 unseen cards, each of which has the same probability of being an out.

OK. I still have a problem. May be I'm thinking about this inappropriately. I'm thinking time line - turn comes before river. Turn must be out in order to think about river. I think it's called time series or stochastic process. But may be I'm wrong. Before the turn or river there's 2 cards that help and 45 unseen cards. It's not necessary to think about the turn before the river. Either can come first. Turn before river is a restriction us humans put on it by design. Before the turn is dealt both the turn and river have a 2/45 chance of being the card we're focused on.

It's probably even more clear if you deal the turn and river face down and then ask "what's the probability of each being the card we need" before we turn them up. Then they're both obviously 2/45. Wow, just thought of that technique - that's really clear now.

Is that right? I'd hate to be thinking about this wrong and then apply this thought process in the analysis of some other situation only to make disastrous conclusions.

If that's true, then I can see it by the inclusion-exclusion principle. Even though I don't think I really know the inclusion-exclusion principle. I only know it in as far as it applies to a venn diagram with two NON mutually exclusive events. Or 3 NON mutually exclusive events. And I'm sure I could do n NON mutually exclusive events.

You: Also, as I wrote in my previous post, 2/45 + 2/44 - (2/44 * 2/45) is the probability of making our hand, including quads counted 1 time. This comes from multiplying out 2/45 + (1 - 2/45)*2/44. Now since this counts quads 1 time, and since 2/44 * 2/45 = 2*P(quads), this means that 2/45 + 2/44 must count the quads 3 times. This is how I first saw it.

So you saw that 2/45+2/44 counted quads 3 times just by your previous assertion that 2/45+2/45 double counts quads, the fact that P(quads)=2/45*1/44, and Master algabreic manipulation all in your head?

WTF? This blows me away!

I moved this response out to the top level since the exchange of posts between DiceyPlay and me has gotten so deep (in posts) that I can't even display the last post in threaded mode.

Nope, I wasn't there.


Yes, I had that very thing in mind.


I just saw that since 2/45 + 2/44 - (2/45 * 2/44) counts the quads 1 time, and since the thing it is subtracting is 2*P(quads), then 2/45 + 2/44 must count the quads 3 times, since 3 - 2 = 1. Then I just verified that with the algebraic manipulation as you said.