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: 1-q|
|Outcome 4: H||Probability: q|
|Outcome 5: T||Probability: 1-q|
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 (1-q) x q x (1-q)
So, for any given value of q, you should be able to tell me the probability of obtaining the sequence
- For example, say q=0.6, what is the probability of the sequence HHTHT?
- And what about if q=0.4, what is the probability of the sequence HHTHT then?
Was the sequence HHTHT more probable given that q, the probability the coin comes up heads,
is 0.4, or 0.6?
Why should this not surprise you?