For a sequence of 10 coin flips, the probability of all heads is 1 in 1024 i.e. $$1/2^{10}.$$
And the probability for an individual coin flip to be heads is $1$ in $2$.
After $9$ coins have been flipped, is the probability of the last coin being heads still $\frac12$ for that last coin?
I ask with respect to the "Monty Hall Problem" -- https://en.wikipedia.org/wiki/Monty_Hall_problem -- in which the probability of an event changes in an unintuitive way.
Does anything like that apply in this coin flip example?
The Monty Hall problem is about dependency and changing knowledge.
You choose a door. There is a probability of 1/3 that there is a prize behind it. One of the remaining two doors is then opened revealing that there is no prize behind it. You are given the opportunity to change to the third door or stay with the initial choice, now that you know the identity of one of the doors without a prize.
There is still a probability of 1/3 that there was a prize behind the door you choose before that information was generated. But, what of the remaining door?
Because of dependency (since there is only one prize), if you were right first time there is certainly no prize behind the third door, but if you were wrong, there certainly is. So the probability of there being a prize behind the third door is $2/3$. ("Third door" being "neither the door you chose nor the door that was revealed".)
None of which has anything to do with your coin toss problem as the coins are independently tossed and you gain no knowledge of the final toss by knowing about any of the previous.
(It would be different if, say, you were told that the coin being tossed was either fair or had a specified bias, with equal likelihood, and a certain number of heads had shown in the first nine tosses. Then you are in the realm of Bayesian probability.)