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 > 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 $0.5$.