Question regarding Odds (Math Aspect)

[Maverick511]

Ok here is what I am trying to figure out. If I have Ad6d and the flop comes Kd7d2c I have a 19.0% chance to hit a diamond the the turn. Then a 19.5% chance on the river. However if I am all in on the flop I have read I have a 35% chance to hit a diamond on the turn or river. Can anyone show me how this 35% is calculated??


Thanks
Maverick511

[uuDevil]

One way is to calculate the odds of not getting a diamond on either the turn or river and subtract from 1:

1-(47-9)/47*(46-9)/46=0.349676226

[jdl22]

what UUDevil said works.

Another way is the following:
P(hit by river) = P(hit on turn) + P(hit on river given missed on turn)

= .19+.81*.195 = .34795

[Lost Wages]

I have a 35% chance to hit a diamond on the turn or river.

To be technical, 35% is the probability of hitting at least one diamond by the river. In other words, it includes those times that you get a diamond on both the turn and river. Of course since you are all-in and drawing to the nuts it doesn't really matter but if you are drawing to the non nuts you don't want two diamonds. Also, even if you are drawing to the nuts you don't want to turn the flush and then have the fourth suited card come on the river if there is still money left to bet.

P(Exactly One Out) = P(out on turn)*P(blank on river) + P(blank on turn)*P(out on river)
P(Exactly One Out) = (9/47)*(38/46)+(38/47)*(9/46) = 31.63%

P(Exactly Two Outs) = P(out on turn)*P(out on river)
P(Exactly Two Outs) = (9/47)*(8/46) = 3.33%

P(One or Two Outs) = P(one out) + P(two outs) = 34.96%

Lost Wages