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 > 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 $0.5$.

Source : Link , Question Author : Jason , Answer Author : Newb

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