A and B toss a coin 10 times, what is the probability that A wins? A and B toss a fair coin 10 times. In each toss, if its a head A's score gets incremented by 1, if its a tail B's score gets incremented by 1.
After 10 tosses, the person with the greatest score wins the game.
What is the probability that A wins?
And if B alone gets an extra toss. What is the probability that A wins?
According to me, 
The cases where A can win are 
(Score of A, Score of B)
(6,4)
(7,3)
(8,2)
(9,1)
(10,0)
These are A's winning cases. Now I am confused on how to proceed.
One method I can think of is that in each of these 5 cases the probability them happening is (1/2)^10. So the probability of A winning is 5*(1/2)^10
But I think I am not taking into consideration the various occurrences of the winning tosses from the total tosses.
So should the probability of A winning be like 
(10C6 + 10C7 + 10C8 + 10C9 + 10C10 ) / 2^10
Which is the number of possible outcomes for A divided by the total number of outcomes. Where 10C6 is the number of ways of selecting 6 from 10 items
 A: It should indeed be (10C6 + 10C7 + 10C8 + 10C9 + 10C10 ) / 2^10
Here is why:
There is only one way for (10,0) to be the outcome:
HHHHHHHHHH ... which happens with a probability of $(\frac{1}{2})^{10}$
But there are 10 ways for (9,1) to be the outcome:
HHHHHHHHHT
HHHHHHHHTH
...
THHHHHHHHH
Each of these happens with a probability of $(\frac{1}{2})^{10}$, and you have 10 of them, since the 1 T can happen in one of 10 places. So, probability of getting 9 H and 1 T is $10*(\frac{1}{2})^{10}$
For (8,2) to be the outcome, you need 2T's among 8H's ... which can be done in ${10}\choose{2}$ ways .. and thus indeed you the general formula that you indicated at the end.
A: It should be obvious, from symmetry, that, in the first case, where the coin is flipped 10 times, that A and B have the same probability of winning.  But if there are 5 heads and 5 tails, neither wins.  The probability of that is $(1/2)^{10}\begin{pmatrix}10 \\ 5 \end{pmatrix}= \frac{252}{1024}= \frac{63}{251}$ so the probability that there is no tie, that either A or B wins is $1- \frac{63}{251}= \frac{181}{251}$.  Since A and B are equally likely to win, the probability A wins is half that, $\frac{181}{502}$.
I don't understand what you mean by "if B alone gets an extra toss."  In the original question, it didn't matter who flipped the coin- just whether the flip resulted in "heads" or "tails".  Are you just saying that B wins if "heads" and "tails" come up an equal number of times?  In that case, the probability that A wins is the [b]same[/b].  The only difference is that, now, the probability that B wins is the probability that A does NOT win.
