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 Post subject: Calculating E[HS2] ?
PostPosted: Wed Sep 04, 2013 11:58 am 
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Joined: Mon Aug 12, 2013 7:28 pm
Posts: 16
Hi, i'm looking at Michael Johanson thesis where he explains the E[HS2] metric.
poker.cs.ualberta.ca/publications/johanson.msc.pdf
Specifically on page 25 he writes:
E[HS2](b)=(0.2^2+0.8^2)/2=0.68 but that's wrong !
It's =0.34
So what's the correct formula ?
If the result is not bigger than 0.5 then it doesn't prove his point:
E[HS2] considers the hand with higher potential to be more similar to hands with high hand strength.


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 Post subject: Re: Calculating E[HS2] ?
PostPosted: Wed Sep 04, 2013 1:15 pm 
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Joined: Sun Feb 24, 2013 9:39 pm
Posts: 642
The formula is correct, and his point is made, even though his arithmetic is wrong.

a = (0.4,0.6)
b = (0.2,0.8)

ehs(a) = 0.5
ehs(b) = 0.5

ehs2(a) = 0.26
ehs2(b) = 0.34

ehs2 ranks high potential hands higher than low potential hands even when average strength (ehs) is the same.


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 Post subject: Re: Calculating E[HS2] ?
PostPosted: Wed Sep 04, 2013 4:09 pm 
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Joined: Mon Aug 12, 2013 7:28 pm
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Oh thanks a lot spears, it makes sense now!


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