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Thanks for reading.
Earlier today, I asked this question on the Math StackExchange:
https://math.stackexchange.com/questions/3307837/do-i-break-even-in-a-fair-game
If someone were to run a computer simulation of the coin-tossing game, what would the score be for the player choosing heads every time? And why? What should we interpret as the "expected" value - the average, or the standard deviation?
Thanks!!!
The process you have described is a simple one-dimensional random walk. Let $X_n$ be the score of the player who always guesses heads (so $-X_n$ is the corresponding score of the player who always guesses tails). After $n$ tosses of the coin, the score has the distribution:
$$X_n \sim n - 2 \cdot\text{Bin}(n, \tfrac{1}{2}).$$
The expected score is $\mathbb{E}(X_n) = 0$ and the standard deviation of the score is $\mathbb{S}(X_n) = \sqrt{n/2}$. Thus, the score will be distributed with an "expected value" of zero, but with increasing variability. For large $n$ the distribution of the score can be approximated by the normal distribution:
$$X_n \overset{\text{Approx}}{\sim} \text{N}(0, \tfrac{n}{2}).$$
(I am being a bit rough here, since this is a continuous approximation to a discrete distribution with expanding variability. If you round the approximating value to an integer then it will be a reasonable approximation.) Now, you have asked what score the player should "expect", but that is a little ambiguous. The "expected value" here is zero, as is the mode. So, on average, the player's score will be zero, and this is also the most likely score to occur. If you are willing to say that this is what the player will "expect" then that is the answer.
