Making inferences from Bayes networks and CPTs

Question

So I'm practicing working with Bayes Networks and conditional probability tables and I feel like some of my numbers simply don't make sense.

Here's the situation: I have a bag of three different biased coins. Coin A has P(heads) of 0.2, coin B has P(heads) of 0.6, and coin C has P(heads) of 0.8.

I pick a coin at random and then flip it three times. I'm trying to work out the probability of each coin being the chosen coin, given a result of two heads and a tails.

So, based on this, I drew up a Bayesian network with the following CPTs:

Coin chosen C: P(C): a .33 b .33 c .33

and

coin flip F: P(F|C=a): P(F|C=b): P(F|C=c): heads 0.2 0.6 0.8 tails 0.8 0.4 0.2

From Bayes' Rule, I know that P(C|f1,f2,f3) = P(C,f1,f2,f3) / P(f1,f2,f3), and that P(C,f1,f2,f3) = P(C)*P(f1|C)*P(f2|C)*P(f3|C) because the three flips are dependent upon C, but independent of one another. Also, I know that P(f1,f2,f3) = sum(P(C,f1,f2,f3)) over all values of C

Plugging in the numbers, I get the following:

P(C=a,f1=h,f2=h,f3=t) = P(C=a)*P(f1=h|C=a)*P(f2=h|C=a)*P(f3=t|C=a) = .33*.2*.2*.8 = .01056 P(C=b,f1=h,f2=h,f3=t) = P(C=b)*P(f1=h|C=b)*P(f2=h|C=b)*P(f3=t|C=b) = .33*.6*.6*.4 = .04752 P(C=c,f1=h,f2=h,f3=t) = P(C=c)*P(f1=h|C=c)*P(f2=h|C=c)*P(f3=t|C=c) = .33*.8*.8*.2 = .04224 P(f1=h,f2=h,f3=t) = P(f1=h,f2=h,f3=t|C=a) + P(f1=h,f2=h,f3=t|C=b) + P(f1=h,f2=h,f3=t|C=c) = (.2*.2*.8)+(.6*.6*.4)+(.8*.8*.2) = 0.304

With those values, I compute P(C|h,h,t):

P(C=a|h,h,t) = P(C=a,h,h,t) / P(h,h,t) = 0.01056/0.304 = 0.0347 P(C=b|h,h,t) = P(C=b,h,h,t) / P(h,h,t) = 0.04752/0.304 = 0.1558 P(C=c|h,h,t) = P(C=c,h,h,t) / P(h,h,t) = 0.04224/0.304 = 0.1384

I don't think these numbers make sense, however. Shouldn't the sum of P(C|h,h,t) for all values of C be equal to 1? I think I may have done something wrong in my calculation of P(h,h,t) above, as a value of 0.304 would imply that 90% of the time I would end up with two heads and a tails (as P(h,h,t) would be equal to P(h,t,h) and P(t,h,h))

Answer 1

I believe the issue is with this line:

$P(f1=h,f2=h,f3=t) = P(f1=h,f2=h,f3=t|C=a) + P(f1=h,f2=h,f3=t|C=b) + P(f1=h,f2=h,f3=t|C=c) = (.2*.2*.8)+(.6*.6*.4)+(.8*.8*.2) = 0.304$

Each of these three terms should be multiplied by the original probabilities of the assumed condition. i.e. $P(f1=h,f2=h,f3=t|C=a)*P(C=a)$

This scales the .304 (seemed high intuitively to me) of hht to about 0.10. Using this figure, you should get probabilities which sum to 1. I get 10.5%, 47.4%, and 42.1%.

Answer 2

Use,

$$ p(c|2H1T) = \frac{p(2H1T|c)}{\sum_{c} p(2H1T|c)} $$

which is a combination of Bayes' theorem and the law of total probability. With this expression you can obtain the probability of each coin given 2 heads and one tail.