No, I think you are mixing the probability of at least one success in ten trails (with a 10% chance per trail), which is ~0.65=65%, with the expected value which is n=1 in both cases. You have the same chance of finding 1 partner in each case and you do the same number of trails. There is a 65% chance that you have at least 1 success in the 10 trails for each type of partner. The expected outcome in BOTH cases is 1 as in n=1 not 1 as in 100%
Probability of at least one success: ~65% Probability of at least two success: ~26%
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.
I think what Justin is saying is that finding a single monogamous partner is not significantly different from finding two, three, etc. For some things you only care about succeeding once. So a 63% chance of success (any number of times) means a .63 expected value (because all successes after the first have a value of 0).
Meanwhile for other things, such as polyamorous partners, 2 partners is meaningfully better than one, so the expected value truly is 1, because you will get one partner on average. (Though this assumes 2 partners is twice as good as one, we can complicate this even more if we assume that 2 partners is better, but not twice as good)
Sure: For a monogamous partner, finding a successful partner has a value of 1 Finding 2 successful partners also has a value of 1, because in a monogamous relationship, you only need one partner. The same holds for 3, 4, etc partners. All those outcomes also have a value of 1. So first, let’s find the probability of getting a value of 0. Then let’s calculate the probability of getting a value of 1. The probability of getting a value of 0 (not finding a partner):
(1−110)10≈0.349
There is one other mutually exclusive alternative: Finding at least one partner (which has a value of 1)
1−(1−110)10≈0.651
So we have a 34.9% chance of getting a value of 0 and a 65.1% chance of getting a value of 1. The expected value is:
0.349∗0+0.651∗1=0.651
If you did this experiment a million times and assigned a value of 1 to “getting at least one monogamous partner” and a value of 0 to “getting no monogamous partners,” you would get, on average, a reward of 0.651.
For the sake of brevity, I’ll skip the calculations for a polygamous partner because we both agree on what the answer should be for that.
I know I am a parrot here, but they are playing two different games. One wants to find One partner and the stop. The other one want to find as many partners as possible. You can not you compare utility across different goals. Yes. The poly person will have higher expected utility, but it is NOT comparable to the utility that the mono person derives.
The wording should have been: 10% chance of finding a monogamous partner 10 times yields 1 monogamous partners in expectation and 0.63 in expected utility. Not: 10% chance of finding a monogamous partner 10 times yields 0.63 monogamous partners in expectation.
and: 10% chance of finding a polyamorous partner 10 times yields 1 polyamorous partner in expectation and 1 in expected utility. instead of: 10% chance of finding a polyamorous partner 10 times yields 1.00 polyamorous partners in expectation.
So there was a mix up in expected number of successes and expected utility.
No, I think you are mixing the probability of at least one success in ten trails (with a 10% chance per trail), which is ~0.65=65%, with the expected value which is n=1 in both cases. You have the same chance of finding 1 partner in each case and you do the same number of trails. There is a 65% chance that you have at least 1 success in the 10 trails for each type of partner. The expected outcome in BOTH cases is 1 as in n=1 not 1 as in 100%
Probability of at least one success: ~65%
Probability of at least two success: ~26%
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.
I think what Justin is saying is that finding a single monogamous partner is not significantly different from finding two, three, etc. For some things you only care about succeeding once. So a 63% chance of success (any number of times) means a .63 expected value (because all successes after the first have a value of 0).
Meanwhile for other things, such as polyamorous partners, 2 partners is meaningfully better than one, so the expected value truly is 1, because you will get one partner on average. (Though this assumes 2 partners is twice as good as one, we can complicate this even more if we assume that 2 partners is better, but not twice as good)
I do not understand your reasoning. Please show your calculations.
Sure:
(1−110)10≈0.349For a monogamous partner, finding a successful partner has a value of 1
Finding 2 successful partners also has a value of 1, because in a monogamous relationship, you only need one partner.
The same holds for 3, 4, etc partners. All those outcomes also have a value of 1.
So first, let’s find the probability of getting a value of 0. Then let’s calculate the probability of getting a value of 1.
The probability of getting a value of 0 (not finding a partner):
There is one other mutually exclusive alternative: Finding at least one partner (which has a value of 1)
1−(1−110)10≈0.651So we have a 34.9% chance of getting a value of 0 and a 65.1% chance of getting a value of 1. The expected value is:
0.349∗0+0.651∗1=0.651If you did this experiment a million times and assigned a value of 1 to “getting at least one monogamous partner” and a value of 0 to “getting no monogamous partners,” you would get, on average, a reward of 0.651.
For the sake of brevity, I’ll skip the calculations for a polygamous partner because we both agree on what the answer should be for that.
I know I am a parrot here, but they are playing two different games. One wants to find One partner and the stop. The other one want to find as many partners as possible. You can not you compare utility across different goals. Yes. The poly person will have higher expected utility, but it is NOT comparable to the utility that the mono person derives.
The wording should have been:
10% chance of finding a monogamous partner 10 times yields 1 monogamous partners in expectation and 0.63 in expected utility.
Not:
10% chance of finding a monogamous partner 10 times yields 0.63 monogamous partners in expectation.
and:
10% chance of finding a polyamorous partner 10 times yields 1 polyamorous partner in expectation and 1 in expected utility.
instead of:
10% chance of finding a polyamorous partner 10 times yields 1.00 polyamorous partners in expectation.
So there was a mix up in expected number of successes and expected utility.
Yeah, I suppose we agree then.