Probability of winning multiple coin toss bets

Question

Please be gentle, I am no statistician.

Imagine that I offered you a coin toss bet in which you have to call heads or tails. If you call the toss correctly, I give you £200. If you call the toss incorrectly, you have to give me £100.

By my understanding you have a 50% chance of winning the bet at odds of 2-1. However, you refuse to take my bet.

I then say, "How about we toss the coin 100 times at the same odds?". You now take my bet.

Am I right to believe that you are no more likely to win money in the single toss game or the 100 toss games and that the probability of you winning or losing money is exactly the same in both scenarios?

I would really appreciate an authoritative answer – many thanks

Answer 1

I'm assuming that in the 100-toss game, you win as soon as you get a successful guess of heads or tails. The answer to your question comes down to the geometric distribution:

$$X \sim \text{Geometric}(p),$$

which can be interpreted as giving the probability of the number of flips required to get a successful toss. More technically and generally, this is the number of Bernoulli trials expected to get a single successful trial, where coin-tossing is a perfect example of a sequence of Bernoulli trials.

Under this distribution, if we have $x$ as the number of tosses required to win (i.e. correctly guess the coin toss), then with $p(x)$ being the probability of needing $x$ tosses, we have

$$p(x) = p(1 - p)^{x-1},$$

where $x$ is the number of tosses, and $p$ is the probability of our guess coming up, which is $1/2$. So we can see that the probability of needing just 1 toss to win is:

$$p(1)=0.5(1-0.5)^{0} = 0.5$$

This is to be expected. The probability of needing 2 tosses to win is

$$p(2) = 0.5(1-0.5)^1 = 0.25,$$

and the probability of needing $101$ tosses to win (i.e. you lose the game) is

$$p(101) = 0.5(1 - 0.5)^{101-1} = 3.94\times10^{-31}$$

So you are pretty much guaranteed to win the 100-toss game!