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Originally Posted by fonzerelli_79 yeah that the way i would have worked it out as well
taking this further to the river we have
(2 / 50) + (2 / 49) + (2 / 48) + (2 / 47) + (2 / 46) = 0.208514
100/ 0.208514 = 4.796
which is 3.796 to one to make your set by the river
however i dont know if this is 100% right - ie. do you need to adjust the odds slightly in the off chance you make quads? |
You're close. However you've missed a concept which is important when calculating any poker odds. First i'll explain with words:
Imagine you've got a coin. Odds of flipping head? 1 in 2. Now say you've get TWO flips. Odds of catching that one heads (at least one)? Well, according to that math:
(1/2) + (1/2) = 100 %
Try it yourself, you won't guarentee yourself heads every time with two flips. So how can you figure in this seemingly statistical anomaly? Figure in the chances you'll even NEED a second flip to land heads. I'll show you:
(1/2) + (1/2)*
(.50) = 3/4 or 75%
The .50 comes from the first flip. There is a 50% chance you will need a second flip because you didn't get heads first flip. If it were a 1 in 10 chance of landing heads, this number would become .90 (90%).
So to accurately calculate the odds of catching trip from PP, you'd start like this:
(2/50) + (2/49)*(.96) + (2/48)*(.96)*(.9591) = 11.8%
This may seem a little bit irrelevant, but as you can see it does affect the odds some.
Maybe someone understant this information, or maybe nobody does. But it's ok, it makes sense to me. I think i'll start my own thread for calculating odds like this one.