My point is that in some situations, “two successes” doesn’t make sense. I picked the dating example because it’s cute, but for something more clear cut imagine you’re playing Russian Roulette with 10 rounds each with a 10% chance of death. There’s no such thing as “two successes”; you stop playing once you’re dead. The “are you dead yet” random variable is a boolean, not an integer.
Sure. For simplicity, say you play two rounds of Russian Roulette, each with a 60% chance of death, and you stop playing if you die. What’s the expected value of YouAreDead at the end?
With probability 0.6, you die on the first round
With probability 0.4*0.6 = 0.24, you die on the second round
With probability 0.4*0.4=0.16, you live through both rounds
So the expected value of the boolean YouAreDead random variable is 0.84.
Now say you’re monogamous and go on two dates, each with a 60% chance to go well, and if they both go well then you pick one person and say “sorry” to the other. Then:
With probability 0.4*0.4=0.16, both dates go badly and you have no partner.
With probability 20.40.6=0.48, one date goes well and you have one partner.
With probability 0.6*0.6=0.36, both dates go well and you select one partner.
So the expected value of the HowManyPartnersDoYouHave random variable is 0.84, and the expected value of the HowManyDatesWentWell random variable is 0.48+2*0.36 = 1.2.
Now say you’re polyamorous and go on two dates with the same chance of success. Then:
With probability 0.4*0.4=0.16, both dates go badly and you have no partners.
With probability 20.40.6=0.48, one date goes well and you have one partner.
With probability 0.6*0.6=0.36, both dates go well and you have two partners.
So the expected value of both the HowManyPartnersDoYouHave random variable and the HowManyDatesWentWell random variable is 1.2.
Note that I’ve only ever made statements about expected value, never about utility.
My point is that in some situations, “two successes” doesn’t make sense. I picked the dating example because it’s cute, but for something more clear cut imagine you’re playing Russian Roulette with 10 rounds each with a 10% chance of death. There’s no such thing as “two successes”; you stop playing once you’re dead. The “are you dead yet” random variable is a boolean, not an integer.
Yes. But I think you have mixed up expected value and expected utility. Please show your calculations.
Sure. For simplicity, say you play two rounds of Russian Roulette, each with a 60% chance of death, and you stop playing if you die. What’s the expected value of YouAreDead at the end?
With probability 0.6, you die on the first round
With probability 0.4*0.6 = 0.24, you die on the second round
With probability 0.4*0.4=0.16, you live through both rounds
So the expected value of the boolean YouAreDead random variable is 0.84.
Now say you’re monogamous and go on two dates, each with a 60% chance to go well, and if they both go well then you pick one person and say “sorry” to the other. Then:
With probability 0.4*0.4=0.16, both dates go badly and you have no partner.
With probability 20.40.6=0.48, one date goes well and you have one partner.
With probability 0.6*0.6=0.36, both dates go well and you select one partner.
So the expected value of the HowManyPartnersDoYouHave random variable is 0.84, and the expected value of the HowManyDatesWentWell random variable is 0.48+2*0.36 = 1.2.
Now say you’re polyamorous and go on two dates with the same chance of success. Then:
With probability 0.4*0.4=0.16, both dates go badly and you have no partners.
With probability 20.40.6=0.48, one date goes well and you have one partner.
With probability 0.6*0.6=0.36, both dates go well and you have two partners.
So the expected value of both the HowManyPartnersDoYouHave random variable and the HowManyDatesWentWell random variable is 1.2.
Note that I’ve only ever made statements about expected value, never about utility.