Probability of the observations, given the cause
Let the probability that out coin comes up "heads" be q.
If we knew the value of q, then we could work out the
probability of the outcomes above as follows:
Outcome 1: H    Probability: q 
Outcome 2: H    Probability: q 
Outcome 3: T    Probability: 1q 
Outcome 4: H    Probability: q 
Outcome 5: T    Probability: 1q 
Because each coin toss is independent, we can get the overall probability of the sequence
HHTHT by multiplying probabilities:
p(HHTHT) = q x q x (1q) x q x (1q)
So, for any given value of q, you should be able to tell me the probability of obtaining the sequence
HHTHT
 For example, say q=0.6, what is the probability of the sequence HHTHT?
?
p(HHTHT) = 0.6 x 0.6 x 0.4 x 0.6 x 0.4
p(HHTHT) = 0.0346
 And what about if q=0.4, what is the probability of the sequence HHTHT then?
?
p(HHTHT) = 0.4 x 0.4 x 0.6 x 0.4 x 0.6
p(HHTHT) = 0.0230

Was the sequence HHTHT more probable given that q, the probability the coin comes up heads,
is 0.4, or 0.6?
?
0.6

Why should this not surprise you?
?
In the observed sequence HHTHT there are more heads than tails.
So this sequence would be more probable given a coin which favours heads (q=0.6)
than a coin that favours tails (q=0.4).
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