HMM for solving given coin output

Question

I have got this assignment question on HMM and I have solved it. I would like to know if I am correct. The problem is:

Suppose a dishonest dealer has two coins, one fair and one biased; the biased coin has heads probability 1/4. Assume that the dealer never switches the coins. Which coin is more likely to have generated the sequence HTTTHHHTTTTHTHHTT? It may be useful to know that log₂(3) = 1.585

I calculated the P for fair coin and biased coin. The P for fair coin is 7.6*10^-6 where as P for biased coin is 3.43*10^-6. I didn't use log term, which can be used if I solve it the other way. So, I concluded that it is more likely that the given sequence is generated by a fair coin.

Am I right?

Any help is greatly appreciated.

Doesn't it depend on the prior probability that the dealer chooses the biased or unbiased coin? — user97370

Manuel Salvadores Manuel Salvadores · Accepted Answer · 2011-12-07T23:01:35

So you are given the following.

P(H|Fake) = 1/4 P(T|Fake) = 3/4
P(H|Fair) = 1/2 P(T|Fair) = 1/2
P(Fair) = 1/2 P(Fake) = 1/2

To answer the question you need to answer P(Fake/HTTTHHHTTTTHTHHTT) and P(Fair/HTTTHHHTTTTHTHHTT) for which you need to apply bayes:

Let X be HTTTHHHTTTTHTHHTT

P(Fake|X) = (P(X|Fake) * P(Fake)) / P(X)
P(Fair|X) = (P(X|Fair) * P(Fair)) / P(X)

Where

P(X) = P(X|Fake) * P(Fake) + P(X|Fair) * P(Fair)
P(X) = (3.43710e-6 * 0.5) + (7.629e-6 * 0.5) = 5.533e-6

And therefore

P(Fake|X) = (3.43710e-6 * 0.5) / 5.533e-6 = 0.3106
P(Fair|X) = (7.629e-6 * 0.5) / 5.533e-6 = 0.6894

So therefore, is more likely that the used coin is the FAIR one. Even though intuitively one might think that the selected coin is the Fake it seems that this is not the case. The given distribution is closer to 0.5 tail 0.5 heads than to 0.25 heads 0.75 tails. For example, in the case of tails 10/17 is 0.58 that is closer to P(T|Fair)=.5 than to P(T|Fake)=.75

HMM for solving given coin output

2 Answers