4
$\begingroup$

101 people flip a fair coin. Everyone who tosses heads is on one team and everyone who tosses tails is on another other team. The team with more people on it wins. What are the odds that, given you are one of the 101 players, you will win? (101 players and coins eliminates ties but I am also interested the case where there are 100 players where you can win/lose/tie).

$\endgroup$
2
$\begingroup$

If the 100 other players divide into two teams of size 50, your chances of winning are 100%. Otherwise, the winning team will not change with your choice hence your chances of winning are 50%. Thus your overall probability of winning is 50%+50%P(50-50 divide), that is, $$\frac12\left(1+\frac1{2^{100}}{100\choose50}\right)\approx53.98\%.$$

$\endgroup$
1
$\begingroup$

The expected number of winners is:

$$2\sum\limits_{n=51}^{101}\frac{n\cdot\binom{101}{n}}{2^{101}}\approx54.5193$$

So the probability of being a winner is:

$$\frac{54.5193}{101}\approx0.5397$$

$\endgroup$
0
$\begingroup$

You and 100 other people flip coins. You always win if exactly 50 of the others got heads. In that case, you are the tie-breaker and always end up on the winning side. If the split was not even, then you win if your flip matched the larger half ($50\%$).

Thus, your odds of winning are: $${1\over 2} \times \left(1 + {{100 \choose 50}\over 2^{100}}\right) \approx 54\% $$

For a total of $100$ people, you win exactly $50\%$ of the time. The other side never splits evenly, and you win if your coin flip matches the larger half.

It's a tie if the split (including yours) is even: $${{100 \choose 50}\over 2^{100}} \approx 8\%$$

And you lose about $42\%$ of the time when there are $100$ people.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.