Probability of getting heads given that first flip was a head?

Yeah, that is right. You can also use a concept called independence; if the two coin tosses are independent, then knowing that the first one is heads does not change at all the probability of heads for the second one.


If we are give the information that "the first coin was a head" then, from, HH, HT, TH, TT, would remove both TT and TH. That leaves only HH and HT so that the probability that the second flip is also a head is 1/2.


the two events are unrelated, the outcome of the second is (as mentioned) independent of the first. So, the odds of the second being heads is 1/2.

The odds of both being heads is 1/4.

If you did 49 flips - and they all came up heads - the odds of the next one being heads is still 1/2.

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Probability

Bayesian

Probability Theory

Bayes Theorem

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