quick probability question
Join Date: Jun 2005
Posts: 503
Hello! We know villains range consists of 70% value hands and 30% trash.
Villain now bets, we assume that he bets his value hands all the time and bets his trash hands half the time: How is the ratio of valuebets/bluffs now?
Join Date: Mar 2009
Posts: 6,730
If I am understanding correctly, this is straightforward.
If you want, you can think of this as a direct application of Bayes Rule. Notation:
- V is a value hand
- T is a trash hand
- B is bet
- P is probability
Then at a certain point in the hand we know the following:
P(V)=.70
P(T)=.30
P(B|V)=1.00
P(B|T)=.50
We seek P(V|B) and P(T|B).
Bayes rule tells us:
P(V|B) = [P(B|V)*P(V)] / [P(B|V)*P(V) + P(B|T)*P(T)]
= [(1.00)*(.70)] / [(1.00)*(.70) + (.50)*(.30)]]
= .70/(.70+.15)
= .70/.85
= 70/85
Similarly, it is easy to see that P(T|B) = 15/85.
Thus, the ratio of value hands to trash hands given that he bets is 70/15.
Join Date: Jun 2005
Posts: 503
TY for the detailed explanation. I was actually looking for how the probability changed. I didnt know ratio was a different thing. So its 70/85, or in other words when he bets its a value hand in 82% and a bluff 18% of the time?
Join Date: Oct 2013
Posts: 3,664
Quote:
Originally Posted by Acryl2
TY for the detailed explanation. I was actually looking for how the probability changed. I didnt know ratio was a different thing. So its 70/85, or in other words when he bets its a value hand in 82% and a bluff 18% of the time?
ye
