Math question

I'm in the process of writing a c-betting article and I'm stuck at the moment...

Two unpaired hole cards miss a flop ~2/3 of the time, or hit it ~1/3 of the time (right?). That doesn't mean that two opponents will hit the flop ~2/3 of the time, and three opponents ~3/3 (%100 of the time ) does it?

My mind's still very jumbled from a bad mix of too much wine + beer last night, and I can't figure it out for the life of me...

No it would probably be around 75% to 5/6=83.3 repeating for 3 opponents it really depends on hole cards if they have you dominated then it would be even a higher percentage for your opponent.

How do we calculate it though?

agh...nevermind. I realize now that of course it depends on hole cards; if they have each other dominated it's much different than if they don't hold each other's cards etc...

I'm just gonna be really vague in the article

Is this a rock, paper, scissors joke?

You PM'd me but I'll post here.

It probably helps to understand where the 2/3 for one opponent actually comes from first.

P(you hit first flop card) = outs/cards left = 6/50 = 0.12 (call this P(1))
So P(you miss first flop card) = 44/50
P(you hit second flop card | you missed first flop card)*P(you missed first flop card) = P(you hit second flop card) = 6/49 * 44/50 = 0.11 (call this P(2))
P(you hit third flop card | you missed first two flop cards)*P(you missed first two flop cards) = P(you hit third flop card) = 6/48 * 44/50 * 44/49 = 0.1 (call this P(3))

So P(hitting any flop card) = P(1)+P(2)+P(3) = 0.12+0.11+0.1 = 0.33

Work through the above but with 12 outs for two players for a rough estimate, but bear in mind in assuming 12 outs we're not accounting for the possibility of 'shared' cards, which is somewhat significant, though not hugely so.









Oh ok, I'll do it.

P(you hit first flop card) = outs/cards left = 12/50 = 0.24 (call this P(1))
So P(you miss first flop card) = 38/50
P(you hit second flop card | you missed first flop card)*P(you missed first flop card) = P(you hit second flop card) = 12/49 * 38/50 = 0.19 (call this P(2))
P(you hit third flop card | you missed first two flop cards)*P(you missed first two flop cards) = P(you hit third flop card) = 12/48 * 38/50 * 38/49 = 0.15 (call this P(3))

So P(hitting any flop card) = P(1)+P(2)+P(3) = 0.24+0.19+0.15 = 0.58

I think, anyway. It's late. :/

Chris, you play poker?

If you sit 3 people at a table one of them is going to hit the flop, if it's 6 people, the odds are at least 2 of the 6 are going to hit substantially. Not every time are the same people going to hit.

P(hitting on flop) = 1 - P(not hitting) (we have 6 outs)


= 1 - (44/50*43/49*42/48) = 0.32428

Likewise for 2 opponents, there are 12 outs for them to hit collectively

= 1 - (38/50*37/49*36/48) = 0.5696

So with 2 opponents with 4 unique cards, they will hit about 57% of the time. Of course, if there is overlap of their cards that is different. Let's say they both have aces but a different kicker. If they have different cards than yours:

P (2 opponents with different aces -- at least 1 hitting flop) = (8 outs)

1 - (42/50*41/49*40/48) = 41.43%

Etc, etc.

i studied maths at uni and as such i like to overcomplicate things as much as possible

I have an engineering degree -- so me, too.

I don't have a degree yet, and I say you're all doing a very good job. Keep up the good work hurting my brain.

I think someone can hit the flop harder without pairing a card. Would you rather have bottom pair or and open ended straight with a flush draw?