# Tossing Coin Expected Value Against Intuition

I have the following two coin toss games:

Game 1: A and B tosses a coin. At first the coin is unbiased. Through the game, if heads comes A wins and game stops. If tail comes, the coin is swapped with another coin which is biased such that the probability of heads showing up becomes 1/3 and game continues. For kth turn the probability of heads showing up is 1/(k+1), if tails comes the coin is swapped with another biased coin for which probability of heads showing up becomes 1/(k+2). What is the expected value of number of turns in which the game ends ie. heads shows up.

Game 2: The game is same, but the biases of coins are different. This time, for kth turn, if probability of heads is 1/(2^k), the next turn it is 1/(2^(k+1)). So it goes 1/2, 1/4, 1/8, 1/16, … .What is the expected value of number of turns in which the game ends.

After calculating I found the expected value in game 1 as divergent, and in game 2 as smaller than 2 which feels very counter-intuitive since A seems to be have a greater chance of winning in game 1.

So my question is what does a divergent expected value actually mean and do you think I made a mistake with the calculations or does the problem really act against intuition.

Some quick calculations, to showcase (presumably) what the OP has done.

For the first game, the probability that the game ends on the $$k$$th turn is $$\frac{(k-1)!}{(k+1)!}=\frac{1}{k(k+1)}$$. Thus when $$X$$ is the number of turns, $$E(X)=\sum_{i=1}^\infty\frac{1}{i+1}=\infty$$ which is known to diverge.

For game two, the probability that the game ends on the $$k$$th turn is strictly less than $$\frac{1}{2^k}$$ (well, equal for $$k=1$$, but strictly less otherwise). Thus when $$X$$ is the number of turns, $$E(X)<\sum_{i=1}^\infty \frac{i}{2^k}=2$$

as you said.

Now why is this?

Well, think about what we assumed. We assumed that the game would end. In 1, the probability that the game lasts at least $$k$$ moves is $$\frac{1}{k}$$ so the game has to end. But for game 2, the probability that the game lasts forever is...

$$\prod_{i=1}^\infty\frac{2^i-1}{2^i}\approx 0.288$$

(Oh and an OEIS for this product too, because why not?)

Thus, the second game actually has a chance that you'll never finish! What does this tell you about the calculated expected probability?

• So game 1 has to end with divergent expected value. At first, I thought, since expected value is divergent the game probably wouldn't end but that's obviously wrong as it has to end. What does the divergence of an expected value mean? – Stefan M. May 21 at 14:49
• Whoops, sorry for taking so long to reply. Divergent expected value simply means that the game will take, on average, forever to finish. That is to say, if you play the game over and over again, the average of the length of your games will increase without bound. Contrast this with rolling a die until you get 1 -- even if you play a trillion games, the average length of each game will still be around 6. Or against your second game -- it's possible that you only ever play three games, and your third game takes forever, thus instantly making the "average" infinity. – auscrypt May 21 at 15:26
• See this answer for a proof that the constant 0.288... is irrational. – Yuval Filmus May 21 at 15:36
• Wait a minute. Are you saying the second game has finite mean length (less than $2$) and a non-zero probability of lasting forever? That doesn't seem right, but maybe I misunderstood something. – Ned May 21 at 17:45
• Given that the second game terminates, the expected length is less than 2, otherwise the game goes on forever. Without the given, the expected length is “infinity” because as long as there is a nonzero chance of a game lasting forever, the expected length is infinite. Conditional expected probability is weird yeah. – auscrypt May 22 at 5:21