Assuming a significantly large distribution of athletes sent by other rational managers, where all athletes are bound to the same rules of random event selection, I would still send the best possible specialist in a single discipline in this case, because without certainty that all other rational managers know certainly that some generalists will be better in everything than other generalists and that each one is confident that theirs is best, I conclude that some of them attempt a gamble of probabilities and send a specialist, and thus I also send a specialist to maximize my chances of winning.
After all, there are higher chances of the event being my athlete’s specialty than there are chances of every single other athlete being less good at it if I pick a generalist, unless the number of possible events is large enough to outweigh the number of athletes. Throw in irrational managers and the possibility of other managers having information unavailable to you, and your father’s argument seems very weak.
Now, of course, I’m probably attacking something that wasn’t meant to be a strong defensible argument. However, I feel very strongly about the point that negative selection is wrong in many contexts it is currently used in (which I support), as well as the point that positive selection is so difficult and utterly impractical in so many cases (which I want to pound into tiny bits of forgotten wrongness).
I’m not sure where I’m going with this, however. I strongly agree with the article’s statements, but my attempts to formulate any further useful thought seem to come up short.
Well, the sports analogy was my own interpretation of what he said.
Game theory question time: you and N other players are playing a dice rolling game. Each player has the choice of rolling a single twenty-sided die, or rolling five four-sided dice. The player with the highest total wins. (Ties are broken by eliminating all non-tying players and then playing again.) Now, rolling 5d4 has an expected score of 12.5 and rolling 1d20 has an expected score of 10.5, so when N=2, it’s obviously better to roll 5d4. However, when N becomes sufficiently large, someone is going to roll a 20, so it’s better to pick the 20-sided die, which gives you a 1 in 20 chance of rolling a 20 instead of a 1 in 1024 chance of getting five 4s. For exactly what value of N does it become better?
Edit: Fixed stupid math mistakes. That’ll teach me to post after staying up all night!
Insightful question, if you ask me, though solving for N feels a lot more like a straight up actuary-level math problem than Game Theory to me. My maths above basic calculus is generally foggy, so I’d appreciate any corrections or nitpicks someone more fluent here might have.
Essentially, you have to solve when (odds of having highest result when rolling d20) >= (odds of having highest result when rolling 5d4). To simplify, let’s assume that all players are perfectly rational, and thus at N and higher will all roll 1d20. This still leaves you the problem of calculating N’s odds of rolling higher than you for both rolls, which is a simpler reformulation of the above parentheses.
For any roll result Y, there is (y/20)^N probability that you “win” here, assuming ties count as wins (or at least are preferable to losses). This means that with N=1 (you’re playing against one other person), you will win 52.5% of the time (and so will your opponent, because that 2.5% is for ties) when rolling 1d20.
Your odds of winning naturally decrease if you roll 1d20 such that for N=2 you have 35.875% chances of winning, and so on in a proportional manner since the odds are always even for everyone.
Where it gets more interesting is when you are playing an unfair game where you have to equate your total odds of winning when playing 1d20 vs d20s to those when playing 5d4 vs d20s. Since the math here is kind of foggy and hard to combine into one big formula, I’ve thrown the data at a spreadsheet (to calculate the sum of the odds of any N rolling higher than you for each roll Y multiplied by your odds of obtaining Y), and it turns out that at N=3 the 5d4 roll dips just below the odds of winning with 1d20 by about 0.2%.
However, if we want to compute for xDf die for N, with K possible ways to roll (which was 2 here), then the math yet eludes me. I’ve figured it out or been told what it was several times, but I just can’t seem to ever memorize this when I can only barely remember integration anyway when I don’t use it.
Edit: For those curious, here’s the spreadsheet mentioned above with all the raw data and brute-force formulas.
Yes, the reward system is very important in choosing the right strategy. If the first place gives you gold, and all other places give you nothing, use positive selection. If the last places gives you problem, and all other places give you nothing, use negative selected. Other point of view: if being average is good, play safe by using negative selection; if being average is bad, aim for greatness (and accept a certain risk of failure) by using positive selection.
So the question is what exactly do we want in elite colleges or academia (examples from the article)? I guess for elite colleges it is better to play safe. If your students are above average and everyone knows it, they don’t have to be exceptional—your diploma will help them get a decent job, which is why they pay you. A few bad apples could ruin your marketing. With academia, for an average university it is probably better to have “safe” professors who do their jobs, get grants, and don’t cause scandals; even if the price is having less Nobel-price winners.
Yes, that it does, or at least it assumes that the difference is trivial within this decision scheme and the expected utility returns of a specialist are higher than the expected utility of a generalist even when taking second place into account.
Assuming a significantly large distribution of athletes sent by other rational managers, where all athletes are bound to the same rules of random event selection, I would still send the best possible specialist in a single discipline in this case, because without certainty that all other rational managers know certainly that some generalists will be better in everything than other generalists and that each one is confident that theirs is best, I conclude that some of them attempt a gamble of probabilities and send a specialist, and thus I also send a specialist to maximize my chances of winning.
After all, there are higher chances of the event being my athlete’s specialty than there are chances of every single other athlete being less good at it if I pick a generalist, unless the number of possible events is large enough to outweigh the number of athletes. Throw in irrational managers and the possibility of other managers having information unavailable to you, and your father’s argument seems very weak.
Now, of course, I’m probably attacking something that wasn’t meant to be a strong defensible argument. However, I feel very strongly about the point that negative selection is wrong in many contexts it is currently used in (which I support), as well as the point that positive selection is so difficult and utterly impractical in so many cases (which I want to pound into tiny bits of forgotten wrongness).
I’m not sure where I’m going with this, however. I strongly agree with the article’s statements, but my attempts to formulate any further useful thought seem to come up short.
Well, the sports analogy was my own interpretation of what he said.
Game theory question time: you and N other players are playing a dice rolling game. Each player has the choice of rolling a single twenty-sided die, or rolling five four-sided dice. The player with the highest total wins. (Ties are broken by eliminating all non-tying players and then playing again.) Now, rolling 5d4 has an expected score of 12.5 and rolling 1d20 has an expected score of 10.5, so when N=2, it’s obviously better to roll 5d4. However, when N becomes sufficiently large, someone is going to roll a 20, so it’s better to pick the 20-sided die, which gives you a 1 in 20 chance of rolling a 20 instead of a 1 in 1024 chance of getting five 4s. For exactly what value of N does it become better?
Edit: Fixed stupid math mistakes. That’ll teach me to post after staying up all night!
Fixed, thanks.
4^5 = 2^10 = 1024
Fixed, thanks.
Insightful question, if you ask me, though solving for N feels a lot more like a straight up actuary-level math problem than Game Theory to me. My maths above basic calculus is generally foggy, so I’d appreciate any corrections or nitpicks someone more fluent here might have.
Essentially, you have to solve when (odds of having highest result when rolling d20) >= (odds of having highest result when rolling 5d4). To simplify, let’s assume that all players are perfectly rational, and thus at N and higher will all roll 1d20. This still leaves you the problem of calculating N’s odds of rolling higher than you for both rolls, which is a simpler reformulation of the above parentheses.
For any roll result Y, there is (y/20)^N probability that you “win” here, assuming ties count as wins (or at least are preferable to losses). This means that with N=1 (you’re playing against one other person), you will win 52.5% of the time (and so will your opponent, because that 2.5% is for ties) when rolling 1d20.
Your odds of winning naturally decrease if you roll 1d20 such that for N=2 you have 35.875% chances of winning, and so on in a proportional manner since the odds are always even for everyone.
Where it gets more interesting is when you are playing an unfair game where you have to equate your total odds of winning when playing 1d20 vs d20s to those when playing 5d4 vs d20s. Since the math here is kind of foggy and hard to combine into one big formula, I’ve thrown the data at a spreadsheet (to calculate the sum of the odds of any N rolling higher than you for each roll Y multiplied by your odds of obtaining Y), and it turns out that at N=3 the 5d4 roll dips just below the odds of winning with 1d20 by about 0.2%.
However, if we want to compute for xDf die for N, with K possible ways to roll (which was 2 here), then the math yet eludes me. I’ve figured it out or been told what it was several times, but I just can’t seem to ever memorize this when I can only barely remember integration anyway when I don’t use it.
Edit: For those curious, here’s the spreadsheet mentioned above with all the raw data and brute-force formulas.
Your analysis also assumes there’s no difference between second place and last place.
Yes, the reward system is very important in choosing the right strategy. If the first place gives you gold, and all other places give you nothing, use positive selection. If the last places gives you problem, and all other places give you nothing, use negative selected. Other point of view: if being average is good, play safe by using negative selection; if being average is bad, aim for greatness (and accept a certain risk of failure) by using positive selection.
So the question is what exactly do we want in elite colleges or academia (examples from the article)? I guess for elite colleges it is better to play safe. If your students are above average and everyone knows it, they don’t have to be exceptional—your diploma will help them get a decent job, which is why they pay you. A few bad apples could ruin your marketing. With academia, for an average university it is probably better to have “safe” professors who do their jobs, get grants, and don’t cause scandals; even if the price is having less Nobel-price winners.
Yes, that it does, or at least it assumes that the difference is trivial within this decision scheme and the expected utility returns of a specialist are higher than the expected utility of a generalist even when taking second place into account.