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Old 02-20-2017, 06:20 PM   #1
Acryl2
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quick probability question

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?
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Old 02-20-2017, 09:39 PM   #2
whosnext
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Re: quick probability question

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.
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Old 02-20-2017, 09:59 PM   #3
Acryl2
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Re: quick probability question

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?
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Old 02-24-2017, 10:42 PM   #4
aner0
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Re: quick probability question

Quote:
Originally Posted by Acryl2 View Post
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
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