Compare the following two games:

  1. You have a fair coin. After one toss, you will get 1 dollar if you get a head, and 0 dollars if you get a tail. How much will you be willing to pay to play this game 1000 times?

  2. You have a fair coin, after one toss, you will get 1,000 dollars if you get a head, and 0 dollar if you get a tail. How much will you be willing to pay to play this game once?

I know the distributions of the payoff are different. However, I cannot figure out how this affects (or does not affect) the amount of the money you wanted to pay to play either game. Could anyone help explain?

Other than distribution, are there any other differences between these two games?


  • $\begingroup$ How much do you value winning more money? And how much do you mind losing more money? That's the relevant question here. $\endgroup$ – celtschk Jul 29 '16 at 14:48

It's a little bit subjective.   Just comment on which game would you prefer to play, and why.

As you say, the distributions are different, but how are they different?

Compare their quantitative measures (expectation, variance, skewness, et cetera).   What do these mean?

Contrast their qualities (the results of some risk and reward scenarios, and such).   What does this suggest?

For instance: In a fair game, you would not want to invest more than the expected outcome, because that's what is paid back on average (by definition).   So what are the expected payouts?

Which game has the greatest risk of a low return?   Which has the greatest potential of a high reward?   What measure tells you about this?

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  • $\begingroup$ Just like @carmichael561 has explained, the expected payoffs are both 500 and I don't think I have a preference for one over the other and this is why I asking. Clearly, if we use variance to measure risk, game 1 is less risky and preferred. But I wanted to know if it is proper to ignore information provided by variance, skewness, etc and just base your decision on expectation? I don't see my expected payout will be greater after considering variance and picking game 1. If so, why do we consider it at all? $\endgroup$ – Map Jul 29 '16 at 15:23
  • $\begingroup$ @Map Which game is more "exciting"? Which is "safer"? What circumstances might you prefer playing one over the other? ...and such. $\endgroup$ – Graham Kemp Jul 30 '16 at 7:02
  • $\begingroup$ I see...it depends on circumstances or a person's risk tolerance - one does not absolutely dwarf another, right? Many thanks for the explanation. $\endgroup$ – Map Aug 1 '16 at 15:52

One basic way to determine how much you would be willing to pay to play a game is to calculate its expected payoff.

In the first scenario, the expected value of one game is $1\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{1}{2}$, hence the expected value of playing this game $1000$ times is $1000\cdot\frac{1}{2}=500$.

In the second scenario, the expected value of one game is $1000\cdot\frac{1}{2}=500$.

So in both scenarios, $500$ is a reasonable price to pay. Note however that the variance of the second scenario is much larger. Depending on your risk preferences, this might affect which game you regard as being more attractive.

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