Probability of having more heads when flipping 17 coins compare to 16 coins. This is a classic brainteaser which I am not convinced by the popular answer.

You have 17 coins and I have 16 coins, we flip all coins at the same time. If you have more heads then you win, if we have the same number of heads or if you have less then I win. What's your probability of winning?


Notice that :


Probability that A obtains more heads than B is equal to Probability that A obtains more tails than B (by symmetry)


Given that A has n+1 coins and B has n coins, A either obtains more heads than B, or more tails than B but never both.


Combining (1) and (2), we get that the probability that A obtains more heads than B is 1/2.

But the question should only considering the number of heads. i.e. $P(\text{Your Heads}>\text{My Heads}) = P(\text{Your Heads}<\text{My Heads})$.
However, $P(P(\text{Your Heads}<\text{My Heads}))+P(\text{Your Heads}>\text{My Heads})\neq 1$. Because $P(\text{Your Heads}=\text{My Heads})\neq 0$ Since you can have 9 heads and 8 tails and I can have 9 heads and 7 tails.
Thus the popular answer would only be correct if it asks for the ratio between heads and tails, not whether one has more or equal heads than the other.
If the above deduction is correct, what would be the actual correct answer without resorting to solving convolution of two binomial random variable of $Bin(17,1/2)$ and $Bin(16,1/2)$?
 A: Your reasoning starts from a false statement, because you reject necessary facts of the situation.

$P(\text{Your Heads}>\text{My Heads}) = P(\text{Your Heads}<\text{My Heads})$.

is false because the players have a different number of coins; one player has a higher expected number of Heads results, so that player has a greater probability of having more Heads, so the probability of the other player having more Heads will not be the same.
Once you have assumed the wrong answer, you can get any answer you want, but it will still be wrong (or contradict itself, but either way, not correct).
This is because you didn't understand the problem correctly to begin with.
Your point about the possibility of having the same number of Heads is irrelevant; this is not necessary to consider because it is subsumed by the result of "Player A has more Tails" which directly equates to "Player A loses".
This in turn ties to the symmetry of Heads and Tails from before. If you switched every "Heads" in the problem statement to "Tails" and vice versa, you have the exact same probabilities of the game result, meaning that the probabilities of Player A getting more Heads or more Tails must be the same.
Since these results are the definitions of "Player A wins" and "Player A loses" (and the latter is equivalent to "Player B wins"), the conclusion reached by the popular answer is correct.
