Coin flipping probability game ; 7 flips vs 8 flips Your friend flips a coin 7 times and you flip a coin 8 times; the person who got the most tails wins. If you get an equal amount, your friend wins.

There is a 50% chance of you winning the game and a 50% chance of your friend winning.

How can I prove this? The way I see it, you get one more flip than your friend so you have a 50% chance of winning if there is a 50% chance of getting a tails.

I even wrote a little script to confirm this suspicion:

from random import choice

coin = ['H', 'T']

def flipCoin(count, side):
    num = 0
    for i in range(0, count):
        if choice(coin) == side:
            num += 1
    return num


games = 0
wins = 0
plays = 88888

for i in range(0, plays):
    you = flipCoin(8, 'T')
    friend = flipCoin(7, 'T')
    games += 1
    if you &gt; friend:
        wins += 1

print('Games: ' + str(games) + ' Wins: ' + str(wins))
probability = wins/games * 100.0
print('Probability: ' + str(probability) + ' from ' + str(plays) + ' games.')


and as expected,

Games: 88888 Wins: 44603
Probability: 50.17887678876789 from 88888 games.


But how can I prove this?
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 $0.5$.
Based on the answers so far you can get the following induction.
If 2 players A and B play a game with these rules: most number of tails wins. 
Each player tosses the coin N times. We have P(Awin-Ntosses)=P(Bwin-Ntosses) = 50%.
You tried to demonstrate that if player B gets an extra coin toss the odds of him winning will  remain 50%. 
that would mean that P(Awin-Ntosses)=P(Bwin-N+1 tosses)
If that is indeed true then using induction you can obtain P(Awin-Ntosses)=P(Bwin-N+x tosses). Induction is valid for any N or x. So therefore we can state that the following statement is true: Two players A and B play a game of toss. A tosses the coin 0 times and B tosses the coin 
5 million times. The winner is the player that tosses most tails. Both players have the same  50% chance to win the game.

I strongly advise you to evaluate this statement and if you still consider it to be true, then it means you would be willing to play such a toss game for a small stake.