Probability of getting heads given that first flip was a head? What's the probability of getting heads on the second toss given that the first toss was a head.  (Trying to refresh my probability a bit).  I've seen this analyzed like this:


*

*HH 1/4

*HT 1/4

*TH 1/4

*TT 1/4


So since we are given information (Head on first flip), then TT goes away and were are left with:


*

*HH 1/3

*HT 1/3

*TH 1/3


So we could say that HH now has a 1/3 probability.  Should we not also get rid of TH, since we know that the first flip is a head?  So now we have:


*

*HH 1/2

*HT 1/2

 A: 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.
A: In general, for two events $A, B$ 
$$
P(A\mid B)=\frac{P(A\cap B)}{P(B)}.
$$
Let $A$ be the event of heads on the second toss, and $B$ the event of first toss being a head. Then
$$
P(A\mid B)=\frac{1/4}{1/2}=1/2.
$$
The numerator is explained by noting that of the four possible sequences of two tosses (all equiprobable), we want $HH$.
A: 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.
A: 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.
A: If the coin is a fair coin, the results of the first toss and the second are independent, so there are exactly two possibilities for the second toss: H and T. The probability of getting H is 1/2. Don't forget, the coin may have been tossed thousands of times before the one we care about. None of those affects the result; there's nothing special about the last of those pre-tosses.
A: What confuses our intuition in this case is the difference between probabilities of events in the future and events that already appeared (past). If all the coin tosses lay in the future, the probability of a straight sequence of H decreases the more coin tosses we plan to do. So for only one toss the probability of H is 1/2, for two tosses (HH) it is 1/4, and so on, so that e.g. for 10 tosses the sequence HHHHHHHHHH has a probability of around 0.001. (Mathematically, this is P(AB) = P(A) * P(B) for independent events A and B.)
The probability decreases with every toss we add, so we expect the likelihood for a T coming up next increases. But this is only true when all tosses lie in the future. In that case the space of possible events comprises all strings of H and T with a length given by the number of tosses we plan (2 ^ number_of_tosses members). The reason why the probability of a straight sequence of H's is so small with a large number of tosses is that it compares to all sequences where one or more T's appear anywhere in the sequence (and not just one at the end).
But this is not true when we already have a prefix string of H's as a given (past) (mathematically P(A|B)). This is because then the event space for the next toss is the one with only two members, H...HH and H...HT. And here the likelihood of the string of only H's is just 1/2. And it is so for any length of the prefix string (however unlikely it was to get such a prefix in the first place), because the event space for the outcome of the next toss always ever only has two members.
