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Bayes factor blues

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Bayes factor blues

michael corning
after several reads on different occasions i'm still stumped on page 318. i'm sure it's me and not an error, but there are two calculations that I cannot follow. first, why is the bayes factor equal to the ratio of the probability Reds win to the ninth power times the probability Blues win once for each model?

second, where did the 82% chance of Reds winning under the gambler's model come from? when I multiply .99^9 * .09 I get an answer that's off by a digit: 0.082.
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Re: Bayes factor blues

norman@eecs.qmul.ac.uk
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Michael

For a model M we have to calculate for each match P(Di | M) where Di is the result of match i. Let's start with model M1. For every match EXCEPT MATCH 5 we have

P(Di | M1)=0.9.

That is because each of those matches results in reds winning and model M1 asserts that the probability of reds winning is 0.9 in each of those matches.

However, for match 5

P(D5 | M1) = 0.1

because Blues win match 5 and model M1 says the probability of that is only 0.1.
Now we have to compute the product of each of the P(Di | M1) which is
0.9 to the power of 9 times 0.1 which is 0.08327 (the numerator of the equation).

Now we do the same thing for model M2 (the denominator). The error in what you wrote is that this is
0.99 to the power of 9 times 0.91 (NOT times 0.09 as you state).
That is because for match 5

P(D5 | M2) = 0.91  (the result is a blues win and for that match model M2 said the probability of a blues win was 0.91).

So the denominator is 0.831 as stated on page 316, not 0.082 as you state.

Norman

 and each of the 0 matches Mi we have to  P(Di | M1)
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Re: Bayes factor blues

michael corning
second question: how did you compute 82% Reds winning rate?

my original use of terms like .09 and .082 were about this question, not the Bayes factor question. in other words, I was trying to answer my question above by myself, and I failed.

so how did you do it?
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Re: Bayes factor blues

norman@eecs.qmul.ac.uk
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>how did you compute 82% Reds winning rate?

You mean for model M2.
It is just the expectation.
For 9 of the matches it expects Reds to have a 0.9 chance of winning, so in those 9 matchs it 'expects' 8.1 victories. In the other match it expects Reds to have 0.09 chance of winning so in that match it 'expects' 0.09 victories. Ther total number of Reds victories M2 therefore expects over the 10 matches is 8.1 + 0.09 whihc is 8.2, i.e. it expects an 82% victory rate overall for Reds.
M1 expects Reds to have a 0.9 chances of winning each of the 10 games, i.e. it expects them to win 9 out of 10 matches, i.e. a 90% winning rate.

Since reds do actuallu win 90% of the games,  M1 is perfectly calibrated (M2 is not) even though it is less accurate.

Norman
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