The Mathematics of Poker / Bayes' Theorem
Join Date: Aug 2012
Posts: 99
Hi All,
In Mathematics of Poker pp. there is an example where a new player joins the game and raises their first hand.
Assuming:
1) there is a 10% chance he's a maniac who will raise 80% of hands, and
2) there is a 90% chance he's tight and will raise 10% of hands
...the authors calculate there is a 47% chance based on him raising his first hand that he is a maniac.
The authors say that if he raises the first two hands, this chance moves to 87%. Can anyone share how the 87% is calculated please?
Thanks,
Join Date: Jan 2017
Posts: 93
(.1*.8)/((.1*.8)+(.9*.1))
(.1*(.8*.8))/(((.1*(.8*.8))+((.9*(.1*.1)))
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Join Date: Sep 2012
Posts: 352
x e {tight, maniac}
y e {openraise, fold}
p(y = openraise | x = maniac} = 80% ==> p(y = fold | x = maniac} = 20%
p(y = openraise | x = tight} = 10% ==> p(y = fold | x = tight} = 90%
p(x = maniac) = 0.1 ==> p(x = tight) = 0.9
bayes rule:
p(x | y) = p(y | x) x p(x) / p(y)
p(maniac | openraise) = p(openraise | maniac) x p(maniac) / p(openraise)
with p(openraise) = p(openraise | maniac) x p(maniac) + p(openraise | tight) x p(tight)
=> p(maniac | openraise) = 0.8 x 0.1 / (0.8 x 0.1 + 0.1 x 0.9) = 0.47
Join Date: Sep 2012
Posts: 352
x e {tight, maniac}
y e {openraise_twice, other}
p(maniac | openraise_twice) = p(openraise_twice | maniac) x p(maniac)/p(openraise_twice)
p(openraise_twice | maniac) = 0.8 x 0.8 = 0.64
p(maniac) = 0.1
p(openraise_twice) = 0.8 x 0.8 x 0.1 + 0.1 x 0.1 x 0.9 = 0.064 + 0.009 = 0.073
p(maniac | openraise_twice) = 0.64 x 0.1 / 0.073 = 0.87
Join Date: Aug 2012
Posts: 99
TYVM both
