P(whosnext and heehaww being THE NUTS)
hope I did that right.
Huge thanks for helping out so much. Definitely appreciate you guys taking the time out to give such comprehensive explanations. The fog is lifting!
Quote:
Originally Posted by Bettafold
Maybe, "how often will I deal a straight flush to ONLY ONE player in ONLY five hands dealt between ten people?" Additionally, it was 2 different players in each instance, so I'm guessing those odds would differ from the SF's being dealt to the SAME player, making it 2 separate equations.
Yeah I believe that is true. There is less of a chance for the straight flushes being dealt to a specific player, rather than to any of the ten players. Take what I say with a grain of salt, but it seems like with probabilities the more specific you get the less likely the chances get. So, it would be two different equations. Although it looks like you can build on the same information to solve both. Same ingredients different recipes.
Quote:
Originally Posted by heehaww
Yes, which is why 1-q^n is only an estimate (albeit a rather good one in that case). Specifically, it's the lower bound, while the additive estimate is the upper bound. And I forgot to mention I was talking about hold'em, though I'm sure you gathered that.
Hmmm ahhhh this is what you were showing here?
Quote:
Originally Posted by heehaww
1-(1119/1225)^9 < P(at least one caller) < 9*6/1225
Does this concept apply only to dependent probabilities?
I'm a little confused about when its appropriate to add probabilities instead of raising their inverse to the nth then subtracting that from 1.
For example: we can find out exactly how likely we are to win a coin flip if we flip a coin twice by doing the ol 1-.5^2. This would be the same as two people flipping a coin once each right? Both examples are indys.
Adding the probability together wont work here.
I feel like adding would work, for instance, in calculating the combined chances of anyone being dealt a specific card:
If I deal in ten players then the chances of dealing the A
would be the chance of dealing A
to one player * 10?
I feel this is correct but I can't articulate why this is.
Quote:
Originally Posted by whosnext
If a member does not know what C(52,5) means, for example, they should certainly ask.
If my understanding of heehaww's explanation is correct then C(52,5) would be the number of different possible flops. wow!
I played around with this a little further for hand combos and ran into a little hiccup.
It works for pairs: Combos of AA= C(4,2)= 6
sets: combos of AA on board of Axx= (3,2) = 3
But doesnt work for unpaired hands.
Combos of AK = C(8,2)=28
There is only 16 combos of AK. Whats going on here?
Thanks again!
And wowww, whosnext. It looks like the chances are way more likely than I expected. Where do you learn to build these sims?