Coin flipping probability game ; 7 flips vs 8 flips
Well, let there be two players $A$ and $B$. Let them flip $7$ coins each. Whoever gets more tails wins, ties are discounted. It's obvious that both players have an equal probability of winning $p=0.5$.
Now let's extend this. As both players have equal probability of winning the first seven tosses, I think we can discard them and view the 8th toss as a tiebreaker. So let's give player $A$ the 8th toss: if he gets a tail, he wins, otherwise, he loses. So with $p = 0.5$, he will either win or lose this 8th toss. Putting it like this, we can see that the 8th toss for player $A$ is equivalent to giving both players another toss and discarding ties, so both players have winning probabilities of $0.5$.
The probability distribution of the number of tails flipped by you is binomial with parameters $n = 8$, and $p$, where we will take $p$ to be the probability of obtaining tails in a single flip. Then the random number of tails you flipped $Y$ has the probability mass function $$\Pr[Y = k] = \binom{8}{k} p^k (1-p)^{8-k}.$$ Similarly, the number of tails $F$ flipped by your friend is $$\Pr[F = k] = \binom{7}{k} p^k (1-p)^{7-k},$$ assuming that the coin you flip and the coin your friend flips have the same probability of tails.
Now, suppose we are interested in calculating $$\Pr[Y > F],$$ the probability that you get strictly more tails than your friend (and therefore, you win). An exact calculation would then require the evaluation of the sum $$\begin{align*} \Pr[Y > F] &= \sum_{k=0}^7 \sum_{j=k+1}^8 \Pr[Y = j \cap F = k] \\ &= \sum_{k=0}^7 \sum_{j=k+1}^8 \binom{8}{j} p^j (1-p)^{8-j} \binom{7}{k} p^k (1-p)^{7-k}, \end{align*} $$ since your outcome $Y$ is independent of his outcome $F$. For such a small number of trials, this is not hard to compute: $$\begin{align*} \Pr[Y > F] &= p^{15}+7 p^{14} q+77 p^{13} q^2+203 p^{12} q^3+903 p^{11} q^4+1281 p^{10} q^5 \\ &+3115 p^9 q^6+2605 p^8 q^7+3830 p^7 q^8+1890 p^6 q^9+1722 p^5 q^{10} \\ &+462 p^4 q^{11}+252 p^3 q^{12}+28 p^2 q^{13}+8 p q^{14}, \end{align*}$$ where $q = 1-p$. For $p = 1/2$--a fair coin--this is exactly $1/2$.
That seems surprising! But there is an intuitive interpretation. Think of your final toss as a tiebreaker, in the event that both of you got the same number of tails after 7 trials each. If you win the final toss with a tail, you win because your tail count is now strictly higher. If not, your tail count is still the same as his, and under the rules, he wins. But the chances of either outcome for the tiebreaker is the same.
Another way to think of it: By swapping all heads and tails, the winner is always reversed. For example, 2T 5H would beat 2T 6H, but 2H 5T would lose to 2H 6T. Or, 7H would beat 8H, but 7T would lose to 8T.
This gives a unique one-to-one mapping between the set where you win and the set where your friend wins. The sets are of equal size, so they must each have 50% probability.
Isolate your eighth toss.
Consider the two cases:
The first seven tosses do not end in a tie. This decides the game because your eighth toss is irrelevant (your friend wins ties).
The first seven tosses end in a tie. Your eighth toss decides the game.
In each case, you have a $50\%$ chance of winning.
The probabilities have a binomial distribution, so we have $$\mathrm{Pr}[\text{friend gets } i \text{ tails}]=\binom{7}{i} 0.5^7$$ and $$\mathrm{Pr}[\text{I get } j \text{ tails}]=\binom{8}{j} 0.5^8.$$
Since these events are independent, we have \begin{align*} \mathrm{Pr}[\text{friend gets } i \text{ tails and I get } j \text{ tails}] &= \mathrm{Pr}[\text{friend gets } i \text{ tails}] \times \mathrm{Pr}[\text{I get } j \text{ tails}] \\ & = \binom{7}{i} \binom{8}{j} 0.5^{15}. \end{align*}
The possible $(i,j)$-values where I win are mutually exclusive, hence we find $$\mathrm{Pr}[\text{I get more heads than my friend}]=\sum_{i=0}^7 \sum_{j=i+1}^8 \binom{7}{i} \binom{8}{j} 0.5^{15}=0.5$$