Earlier I issued a challenge to my nerdy readership regarding game theory. The problem was given a prognostication question where one answer had a 1-in-3 chance of occurring and another answer had a 1-in-50 chance of occurring, construct a set of circumstances where it is proper winning strategy to pick the 1-in-50 chance. I was sad that none of the nerdy contingent of my readership was piqued enough to answer!
Assume a field of more than 50 people and some number of evenly-balanced questions in addition to the Miss America Question (with the 1-in-3 and the 1-in-50). If an entrant guesses that no one else will pick the 1-in-50 chance and that the other answers will be chosen roughly according to probability, it is a wise move to pick the 1-in-50. The long shot is a "buy" because it is undervalued by the rest of the entrants. Picking that long shot gives the one person a 2% shot of getting a leg up on the field and, everything else being even, of winning. Picking the popular answer gives the entrant a less than 2% chance of winning (because there are more than 50 entrants).
Another recent fun statistical moment for me. Katie had a big case she was working on over the weeks of Christmas and New Years, which kept us from taking off large blocks of time. I therefore voluteered in lab to do holiday duty for some maintenance tasks that are usually done by others (who did want to take large blocks of time). One of these tasks is weaning the mice. When the pups reach a certain age, they need to be separated by sex in order to prevent unwanted adult behaviors (primarily procreation and male-male aggression) that occur when sexes are mixed for too long. So a litter of mice needs to be split into sex-specific cages, with no more than five female mice or four male mice to a cage (male mice are a little bigger).
The litter that I had to wean on the day after Christmas had 10 pups. I saw no new cages nearby, so I had to walk a long way down the hall to get some. As I walked, I tried to calculate how many cages I would need. It seemed like three cages would be most likely, where my only danger litter would be 9 males and 1 female, which would need four cages, or 10 females, which would need two cages. So, I got three cages, and that turned out to be a good call: the litter held 4 females and 6 males.
When I calculated later, 10 pups is actually the easiest multi-pup litter to predict. 99% of litters require three cages:
My guess is that if genders need to be separated, females cages can have no more than f individuals, and male cages can have no more than m individuals, then the easiest litter to predict will be f + m + 1, which will most often use 3 cages. Can anyone prove that?
Assume a field of more than 50 people and some number of evenly-balanced questions in addition to the Miss America Question (with the 1-in-3 and the 1-in-50). If an entrant guesses that no one else will pick the 1-in-50 chance and that the other answers will be chosen roughly according to probability, it is a wise move to pick the 1-in-50. The long shot is a "buy" because it is undervalued by the rest of the entrants. Picking that long shot gives the one person a 2% shot of getting a leg up on the field and, everything else being even, of winning. Picking the popular answer gives the entrant a less than 2% chance of winning (because there are more than 50 entrants).
Another recent fun statistical moment for me. Katie had a big case she was working on over the weeks of Christmas and New Years, which kept us from taking off large blocks of time. I therefore voluteered in lab to do holiday duty for some maintenance tasks that are usually done by others (who did want to take large blocks of time). One of these tasks is weaning the mice. When the pups reach a certain age, they need to be separated by sex in order to prevent unwanted adult behaviors (primarily procreation and male-male aggression) that occur when sexes are mixed for too long. So a litter of mice needs to be split into sex-specific cages, with no more than five female mice or four male mice to a cage (male mice are a little bigger).
The litter that I had to wean on the day after Christmas had 10 pups. I saw no new cages nearby, so I had to walk a long way down the hall to get some. As I walked, I tried to calculate how many cages I would need. It seemed like three cages would be most likely, where my only danger litter would be 9 males and 1 female, which would need four cages, or 10 females, which would need two cages. So, I got three cages, and that turned out to be a good call: the litter held 4 females and 6 males.
When I calculated later, 10 pups is actually the easiest multi-pup litter to predict. 99% of litters require three cages:
| Pups | 1 cage | 2 cages | 3 cages | 4 cages | 5 cages |
| 2 | 50.0% | 50.0% | |||
| 3 | 25.0% | 75.0% | |||
| 4 | 12.5% | 87.5% | |||
| 5 | 3.1% | 96.9% | |||
| 6 | 90.6% | 9.4% | |||
| 7 | 72.7% | 27.3% | |||
| 8 | 50.0% | 50.0% | |||
| 9 | 24.8% | 75.2% | |||
| 10 | 0.1% | 98.9% | 1.1% | ||
| 11 | 74.2% | 25.8% | |||
| 12 | 50.5% | 49.5% | |||
| 13 | 27.9% | 72.1% | |||
| 14 | 6.1% | 93.8% | 0.1% | ||
| 15 | 0.0% | 84.4% | 15.6% |
My guess is that if genders need to be separated, females cages can have no more than f individuals, and male cages can have no more than m individuals, then the easiest litter to predict will be f + m + 1, which will most often use 3 cages. Can anyone prove that?
Actually, no. The easiest number to predict will always be 1, requiring 1 cage 100% of the time.
ReplyDeleteYou got me! I did say "multi-pup litter" above, but I did not include the qualifier in my last paragraph.
ReplyDeleteOK, more seriously, I suspect that the answer might not be f+m+1 if f >> m or m >> f.
ReplyDeleteI note that f=m=2 gives you equally good solutions for all odd numbers of mice.
ReplyDelete(also, in addition to "multi-pup litter" you need to specify that there is a 50% chance of female versus male)
you had me at "challenege."
ReplyDelete