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


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!

  • $\begingroup$ Many thanks for your answer but you have assumed incorrectly. It makes no difference whether you chose heads or tails. At each toss of the coin, you can call heads or tails. If you have call the toss correctly, you win. If you call it incorrectly, you lose. $\endgroup$ May 25 '18 at 13:52
  • $\begingroup$ Hi Danny, in the case of this game, since the probability of heads or tails is the same (1/2), the above actually applies for your game. I have modified my answer slightly. Note, however, that if the coin was defective, and the probability of heads (or tails) was greater than 1/2, then the above would not be the case. $\endgroup$
    – quanty
    May 25 '18 at 13:56
  • $\begingroup$ Still confused- sorry. If I go to a casino and make a single bet on black or red the probability of me winning the bet 47.4%. The probability of me winning is the same whether I bet once or bet 1,000 times. If my odds improved the more times I played, the casino would surely limit the number of times I could play. I realise that the probability and the odds are different in this example from the coin toss, but surely the principle is the same? Playing more does not increase my chance of winning. $\endgroup$ May 25 '18 at 14:11
  • $\begingroup$ The probability of you winning any given bet on black or red is 47.4%, because each bet is independent of any of the others. Basic probability theory says that we multiply independent probabilities to get the probability of them all happening. So the probability of winning the casino bet 3 times is $0.474\times 0.474 \times 0.474 = 0.106$. It's a similar thing, but is different to asking how likely you are to win after losing loads of times. $\endgroup$
    – quanty
    May 25 '18 at 14:29
  • $\begingroup$ The casino don't limit the number of times you play because they take enough money from all the times you lose before winning $\endgroup$
    – quanty
    May 25 '18 at 14:33

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