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Sorry if this question is too stupid... Why in general do we care about expected value (integrating outcomes over a probability distribution)? Here is a scenario that illustrates my confusion:

Suppose we are forced to participate in the following game:

We roll a 6-sided dice in a completely dark room, so we can't see the number. One number is showing face up and we can't touch the dice after rolling it. We are given two options and our objective is to live.

Option 1: If the number is 1 or 2, we die, otherwise we live.

Option 2: If the number is 3,4,5 or 6 we die, otherwise we live.

It seems obvious that option 1 is the better choice because there is a lower "probability" that we die compared to option 2. But I'm having some trouble articulating a rigorous/clear argument for why it's "better" ?

Argument 1: Assuming we have a uniform prior on what number the dice could be showing, option 1 gives us a higher probability of survival. Define U_i to be a universe in which the dice has rolled i. By choosing option 1, you survive in universes U_3,U_4,U_5,U_6. Since you are equally likely to be in each universe, you should go with option 1.

Response: What is the precise meaning of this "probability" and how does it connect to our goal (surviving)? What does it mean to be "equally likely" and why "should" I maximize the "likely"-weighted number of universes I survive in when I am in one particular universe?

Argument 2: If this game were repeated multiple times, you would die fewer times if you chose option 1 every time.

Response: Why do I care about this if I'm only playing the game once?

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    $\begingroup$ Choose utility function f('live') = 1, f('die') = 0. Expected utility = Ef(x) = P('live'). Maximizing Ef(x) is equivalent to maximizing P(''live'). You're free to not care. I'll choose option 1 (presuming I've evaluated that I don't think the die is sufficiently rigged against it), which maximizes my expected utility. $\endgroup$ – Mark L. Stone Aug 30 '16 at 20:25
  • $\begingroup$ Right I guess I should be more clear, why should we care about maximizing P('live")? $\endgroup$ – convolutedstatistic Aug 30 '16 at 20:34
  • $\begingroup$ Because you don't care, you might as well choose option 2 of the following. You will play Powerball once,(and never again) and there are only 2 possible outcomes, you win \$300 million or you win nothing. Option 1:"'Inverse Powerball'. If your submission does not match the Powerball combination, you win the $300 million, otherwise nothing. Option 2: "Powerball'. If your submission matches the Powerball combination, you win the \$300 million, otherwise nothing. As for me, I'll go with option 1. $\endgroup$ – Mark L. Stone Aug 30 '16 at 20:35
  • $\begingroup$ Sure I agree that Option 1 has a higher EV than Option 2 in the powerball example, and that intuitively I should choose Option 1. But is there a clear logical argument as to why higher EV -> better choice in a single case? $\endgroup$ – convolutedstatistic Aug 30 '16 at 20:38
  • $\begingroup$ All my risk preferences have been baked into the utility function I have chosen. And I am evaluating its expected value under various options per the probability measure which reflects my beliefs. $\endgroup$ – Mark L. Stone Aug 30 '16 at 20:42
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This is Mark Stone's answer from the comments, which is exactly what I was looking for!

https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem

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