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> big mooning opportunity If you invest $100 with the expectation that it will be lost with 99% probability, and that it will become $100,000 with 1% probabili

> big mooning opportunity If you invest $100 with the expectation that it will be lost with 99% probability, and that it will become $100,000 with 1% probability, you are expecting a profit of $901. That is an "expectation of profit" my friend.

dheera 5y ago
> you are expecting a profit of $901 No, it would be stupid to "expect" a profit of $901 because there is no scenario in which you would gain $901. You would either have -$100 or +$100000 and never +$901 in your stated scenario.* In your case since there is a 99% probability of -$100, my "expectation" would be -$100 and if "surprise" of $100000 happens that is "surprise", not part of the "expectation". * I realize there is a term called "expectation value" and I hate that term with a passion because in any distribution that doesn't roughly resemble a bell curve it would be stupid to "expect" the mean value. Statistics like the mean are not a good way to establish a strategy for a game that you only get to play once.
- josephcsible 5y ago
  "Expected value" means the mean of any distribution, not just ones shaped like bell curves. GP is correct.
  - dheera 5y ago
    Exactly, my point is it's stupid terminology. I use the word "mean", not "expectation". That brainwashes people actually "expecting" the mean to happen, which in this case, it never will. Treating the mean as an expectation of some sort is just not a good formulation to strategize about this game. GGP said "you are expecting a profit of $901" -- I'm not, personally. I'm "expecting" -$100. How about we take a bet? I'll expect -$100, you expect $901. If I win you give me $100, if you win I give you $100, and if neither of us wins we split the $100K. Deal? This is a game theory problem, NOT a statistics problem. In my case the utility of losing -$100 is zero, the utility of gaining $100K is very positive, the utility of missing out on the $100K is extremely negative.
    - josephcsible 5y ago
      Are you saying we should stop using the term "expected value" in cases where it isn't equal to the mode?
      - dheera 5y ago
        Yes. I don't know about you but if you have a coin labeled 0 and 1 and flip it it would be mind-bogglingly stupid to expect a 0.5 out of this coin. Just like if you have a coin labeled A and C it would be stupid to expect a B to come out of it. You'd be better off picking either A or C and expecting that instead.
        Daishiman 5y ago
        I guess you never took stat 101.
        dheera 5y ago
        Personal attacks not appreciated. No statistics class I took said that a coin with sides A and C can give you a B, or that you could "expect" 0.5 from a bimodal distribution with a delta function at 0 and a delta function at 1. The mean value may be 0.5 but it's not something you would expect to actually happen; if you wanted to maximize the probability that your prediction is correct you'd be better off picking 0 or 1.
        Daishiman 5y ago
        You're making an entire thread about defending your position against using a term of art the way it's used by professional and academics all over the world. Judging by every reply you're getting, everyone know what they mean when they talk about the expected value of a probability distribution. Just drop it.
        kortilla 5y ago
        It’s not a personal attack. “Expected value” is a very basic term of art introduced in stats 101. It’s like having a huge argument about a physics problem claiming “force” isn’t mass times acceleration because nobody is “forcing” you to do anything. Back to your 0 or 1 example. 0.5 is the meaningful value to know over any kind of large sampling because that is what matters for “expected value”. It tells me what I should expect overall if I play 1000 times. What you want is called the “mode”. Given a single instance of an event, what is the most frequent value.

dheera 5y ago

> you are expecting a profit of $901 No, it would be stupid to "expect" a profit of $901 because there is no scenario in which you would gain $901. You would either have -$100 or +$100000 and never +$901 in your stated scenario.* In your case since there is a 99% probability of -$100, my "expectation" would be -$100 and if "surprise" of $100000 happens that is "surprise", not part of the "expectation". * I realize there is a term called "expectation value" and I hate that term with a passion because in any distribution that doesn't roughly resemble a bell curve it would be stupid to "expect" the mean value. Statistics like the mean are not a good way to establish a strategy for a game that you only get to play once.

josephcsible 5y ago
"Expected value" means the mean of any distribution, not just ones shaped like bell curves. GP is correct.
- dheera 5y ago
  Exactly, my point is it's stupid terminology. I use the word "mean", not "expectation". That brainwashes people actually "expecting" the mean to happen, which in this case, it never will. Treating the mean as an expectation of some sort is just not a good formulation to strategize about this game. GGP said "you are expecting a profit of $901" -- I'm not, personally. I'm "expecting" -$100. How about we take a bet? I'll expect -$100, you expect $901. If I win you give me $100, if you win I give you $100, and if neither of us wins we split the $100K. Deal? This is a game theory problem, NOT a statistics problem. In my case the utility of losing -$100 is zero, the utility of gaining $100K is very positive, the utility of missing out on the $100K is extremely negative.
  - josephcsible 5y ago
    Are you saying we should stop using the term "expected value" in cases where it isn't equal to the mode?
    - dheera 5y ago
      Yes. I don't know about you but if you have a coin labeled 0 and 1 and flip it it would be mind-bogglingly stupid to expect a 0.5 out of this coin. Just like if you have a coin labeled A and C it would be stupid to expect a B to come out of it. You'd be better off picking either A or C and expecting that instead.
      - Daishiman 5y ago
        I guess you never took stat 101.
        dheera 5y ago
        Personal attacks not appreciated. No statistics class I took said that a coin with sides A and C can give you a B, or that you could "expect" 0.5 from a bimodal distribution with a delta function at 0 and a delta function at 1. The mean value may be 0.5 but it's not something you would expect to actually happen; if you wanted to maximize the probability that your prediction is correct you'd be better off picking 0 or 1.
        Daishiman 5y ago
        You're making an entire thread about defending your position against using a term of art the way it's used by professional and academics all over the world. Judging by every reply you're getting, everyone know what they mean when they talk about the expected value of a probability distribution. Just drop it.
        kortilla 5y ago
        It’s not a personal attack. “Expected value” is a very basic term of art introduced in stats 101. It’s like having a huge argument about a physics problem claiming “force” isn’t mass times acceleration because nobody is “forcing” you to do anything. Back to your 0 or 1 example. 0.5 is the meaningful value to know over any kind of large sampling because that is what matters for “expected value”. It tells me what I should expect overall if I play 1000 times. What you want is called the “mode”. Given a single instance of an event, what is the most frequent value.