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Probably a pretty trivial question that I used to know how to figure out, but here it is:

Since one player hits the flop about 1/3 of the time, how do I figure out the likelihood of at least 1 of 2 players hitting the flop? The likelihood of at least 1 of 3 players hitting the flop? The likelihood of 2 out of 2 players hitting the flop? etc. etc.

Likelihood of 1 out of 2 hitting: % of the time player A misses*% of the time player B misses = % of the time both miss. E.g. players both hit 34% of the time so .66*.66 = .435, one out of 2 hits 1-.435 = 56.5%.

Likelihood of 2 out of 2 hitting is % of the time player A hits*% of the time player B hits, so .34*.34 = .116

probability is pretty simple all you have to remember is AND: multiply OR: add

probability is pretty simple all you have to remember is AND: multiply OR: add

So if we are saying we want the probability that A *OR* B miss the flop, we add the probability of A missing to the probability of B missing? That doesn't seem right as .66 +.66 is more than 1. It is also different from what Deets posted above.

Basically I just want the probability that *at least 1* player hit the flop.

edit: just realized that *at least 1* hitting the flop is [1- (% of A *and* B missing)], which is in line with what you and Deets posted.

The probability that A or B misses is one minus the probability that they both hit. Deets calculated that already as .116 so the probability is 1 - .116 or 88.4% that one or more of them miss.

You can also do this by figuring out the chance that they both miss (.66 * .66 = .4356), and add that to the chance that A misses and B hits (.66 * .34 = .2244) and the chance that A hits and B misses (.34 * .66 = .2244). .4356 + .2244 + .2244 = 88.4%

matt9041 means that you add when there's an or in the sentence all the possible probabilities of at least one of them missing.

one out of 2 hits 1-.435 = 56.5%.

Just to be absolutely clear on my earlier post, the above should read "at least one out of 2 hits..."

Good way to check your answers is to use propokertools pql btw.

You also have to take into consideration that if they share any of their hole cards,your assumptions are wrong

the probability at least one of three hitting it would be 1 - the probability all three miss.
to figure out only one hitting you would have to take away the probablity of all 3 hitting and of 2 of them hitting

Probably a pretty trivial question that I used to know how to figure out, but here it is:

Since one player hits the flop about 1/3 of the time, how do I figure out the likelihood of at least 1 of 2 players hitting the flop? The likelihood of at least 1 of 3 players hitting the flop? The likelihood of 2 out of 2 players hitting the flop? etc. etc.

With probability it is sometimes a lot easier to rephrase the question.

at least 1 of 2 players hitting the flop

The 'at least' part means that player A hitting is enough, player B hitting is enough or both player A and B hitting. This is quite complicated to work out this way but by spotting the short cut that the only option left after 'A hitting is enough, player B hitting is enough or player A and B hitting' is none hitting so that
1.0 - chance of none hitting is what you want (it is due to the 'at least' part of the question)
This is basically what Deets says and you multiply the chances of A not hitting, by the chances of B not hitting etc.
You can work it out the other way but it is longer (say A+ is A hitting, and A- is A missing, B+ etc) and using 0.34 as the prob

A+B+ = 0.34 * 0.34 = 0.1156
A+B- = 0.34 * 0.66 = 0.2244
A-B+ = 0.66 * 0.34 = 0.2244

Add these together we get 0.1156 + 0.2244 + 0.2244 = 0.5644
So 56.44%

So you can see it is easier just to calc the one final case, (A-B-) and subtract this from 1.0
A-B- = 0.66 * 0.66 = 0.4356
1.0 - (0.4356) = 0.5644

It gets harder and harder to calc 'at least one of' without using the shortcut as the number of players increases but using the shortcut there is always only one calculation to do. Using this reverse approach you always only have to do the one calculation, eg for four players it is 1.0 - (A-B-C-D-)

There will also be a complication to get an exact figure as one player not hitting will alter the chance of the second not hitting but it is good way to estimate it.