hey bit embarrassing really basic question but just curious if doing it right.

x happens 35% of time
y happens 35% of time

% of time either event occurs

is it just 57.7%?

Not enough information. If they are mutually exclusive, meaning that only one of them can occur at a time, then one of them will occur 70% of the time. If they are independent, meaning that the occurrence of one doesn't affect the probability of the other occurring, then at least one of them will occur 1 - 0.65*0.65 = 57.75% of the time. The answer could be 35% if they always occur together. Or it could be some other value between 35% and 70%.

well i was just trying to get an idea of how often someone has top pair with the the ranges i assigned

eg
2 players

flop Qxx
(random numbers)
70 combos of TP out of 200 so they have Qx 35% of time

2 villains have same 200 combo range

i know 1 has Qx 35% of time but what % does EITHER of them have Qx

just a rough sort of guess

wasnt sure

100%-65%^2 = 57.75% for either or both to occur, 57.75%-35%^2 = 45.5% for exactly one and not two to occur.

thank you all much appreciated

This is wrong. Come on. I just got through explaining that you can't assume independence. He then gave an example where they clearly are not independent. When one player has a Q, it changes the probability that the other player has a Q.

We still don't have enough info. Give the ranges, and the problem is very easy.

aware when player 1 has toppair it reduces the probability player 2 has toppair but isnt that irrelevant since im only trying to work out probability either has it not both?

NO, it isn't irrelevant. They aren't independent. You can only multiply 0.65*0.65 if they are independent. We need to multiply 0.65 times the probability that the second player doesn't have it GIVEN that the first player doesn't have it. And it's not even that simple because the probability for the second player depends on what cards the first player gets. We need the ranges for that.

yeah but lets say villains both call flop and i assign them 200 of the same combos

70 of those combos contain Qx

35% of their range is Qx

since their continuing range has both Qx and other non Qx random stuff i cant substract the Qx combos in player 2 range right?

what is the probability either have Qx when they both call

thnx for helping me out ><

You are obviously right, but I doubt op is looking for such a precise answer, he's just looking for an estimate to decide whether to cbet/2barrel or not.

I don't think it would affect the answer for at least one Q by more than 2%, I mean his assumptions about combos have greater uncertainty than this.

You count how many combinations player 2 can have that are not Qx when player 1 doesn't have Qx. See example below.


As I said, we need the ranges for that, and you won't tell us what they are. So I'll make up my own problem and solve that.

For simplicity, suppose both players will play any 2 cards T or higher, and nothing else. When the flop comes Qyy where y is < T, what is the probability that at least one of them has Qx, where x can be a Q or non-queen?

There are C(19,2) = 19*18/2 = 171 hands that they would see the flop with. Of these, 51 contain a Q. The probability that the first player has Qx is 51/171 or almost 30%. The probability that he doesn't have Qx then is 120/171. If player 1 doesn't have Qx, there are 3*14+3 = 45 ways that player 2 can have Qx out of C(17,2) = 17*16/2 = 136 possible hands. The probability that he has Qx when player 1 does not have Qx is 45/136,. The probability that he does have Qx when player 1 doesn't have Qx then is 91/136. So the probability that either one of them has Qx is 1 minus the probability that neither has Qx is

1 - (120/171)*(91/136)

=~ 53.0%

Now if you had simply computed 1 - (120/171)² as some people have so egregiously suggested, you would have gotten 50.75%.

The fact that there are unavoidable sources of error is all the more reason to not introduce any additional easily avoidable sources of error.

If the OP doesn't understand this concept, he will make other very big errors. If the flop comes Axx, and 2 players could only hold AA, KK, or AK, the probability that at least one of them has an ace is 100% 97.1%. Not 1 - (6/21)² = 91.8%. The probability that one of them has KK is 2*6/21 = 57.1% 54.3%. Not 1 - (15/21)² = 49%.

Omg, could this be an opportunity to correct Bruce for once? Not 100% because there's a chance they both have KK.

Oops. It's 100% that one of them has a king. It's 97.1% that one of them has an ace (not 91.8% which you would get by independence). Also the probability that someone has KK is only 54.3% since they can both have it. I made corrections in red. My point still stands, just slightly less dramatically.