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The con is on
One of the things my new flatmate (![[livejournal.com profile]](https://www.dreamwidth.org/img/external/lj-userinfo.gif)
whodo_voodoo, for those of you keeping track) brought with him is a television. I'm a bit dubious, as my ability to work is low enough as it is, but it did mean that I was finally able to catch an episode of Hustle, the "complicated confidence trick" series starring Robert Vaughn. I love crime capers even more than I love kung fu movies, so it was deeply frustrating to me that there was a new one going out on telly every week and I was missing it. Anyway, I struck particularly lucky with this episode: halfway through, an actual guy with a katana came in and got all kenjutsu on the gang's asses. Result, quite frankly.
The katana dude was seeking revenge for his father's death (by seppuku, natch), and so he forced the gang to eat pieces of possibly-contaminated fugu one by one. There was much debate about who was the leader (I'm Spartacus!), and so should go first, and where one should go in the order to maximise one's chances of survival. As much debate as you can have while a scary Japanese guy is threatening you with a sword, that is. The general consensus was that it was best to go early, while you still had a decent number of fugu pieces to choose from.
[The mathematically-inclined may like to look away now. Sorry about lack of cut, but I needed one for the spoilers, and you can't nest them AFAIK.]
Here's the thing: if the order is chosen in advance, then it doesn't matter where you go. Suppose the gang has m members, and there's one piece of fugu each. Only one is poisonous. If you go first, you have a one in m chance of choosing the poisoned piece. If you go second, you have a one in (m-1) chance of choosing the poisoned piece when it's your go, but you'll only get to your go if the first guy survives. So your chance of dying is (m-1)/m * 1/(m-1) = 1/m, the same as the first guy. If you go in the n'th place, you'll only die if you choose the bad piece and everyone in front of you chose good pieces, so your chance of dying is (m-1)/m * (m-2)/(m-1) * ... * (m-n+1)/(m-n+2) * 1/(m-n+1) = 1/m.
If the order isn't fixed, however, it gets a bit more confusing. When there are n pieces left, you've got a 1/n chance of picking the bad one, by the reasoning above: obviously, this is going to go up as n goes down. But at each stage, choosing to eat now or wait doesn't affect your odds of survival. Maybe you can't choose not to die, but only choose the manner of your dying: an appropriately Bushidoesque thought.
Whoever said probability was easy?
whodo_voodoo, for those of you keeping track) brought with him is a television. I'm a bit dubious, as my ability to work is low enough as it is, but it did mean that I was finally able to catch an episode of Hustle, the "complicated confidence trick" series starring Robert Vaughn. I love crime capers even more than I love kung fu movies, so it was deeply frustrating to me that there was a new one going out on telly every week and I was missing it. Anyway, I struck particularly lucky with this episode: halfway through, an actual guy with a katana came in and got all kenjutsu on the gang's asses. Result, quite frankly.The katana dude was seeking revenge for his father's death (by seppuku, natch), and so he forced the gang to eat pieces of possibly-contaminated fugu one by one. There was much debate about who was the leader (I'm Spartacus!), and so should go first, and where one should go in the order to maximise one's chances of survival. As much debate as you can have while a scary Japanese guy is threatening you with a sword, that is. The general consensus was that it was best to go early, while you still had a decent number of fugu pieces to choose from.
[The mathematically-inclined may like to look away now. Sorry about lack of cut, but I needed one for the spoilers, and you can't nest them AFAIK.]
Here's the thing: if the order is chosen in advance, then it doesn't matter where you go. Suppose the gang has m members, and there's one piece of fugu each. Only one is poisonous. If you go first, you have a one in m chance of choosing the poisoned piece. If you go second, you have a one in (m-1) chance of choosing the poisoned piece when it's your go, but you'll only get to your go if the first guy survives. So your chance of dying is (m-1)/m * 1/(m-1) = 1/m, the same as the first guy. If you go in the n'th place, you'll only die if you choose the bad piece and everyone in front of you chose good pieces, so your chance of dying is (m-1)/m * (m-2)/(m-1) * ... * (m-n+1)/(m-n+2) * 1/(m-n+1) = 1/m.
If the order isn't fixed, however, it gets a bit more confusing. When there are n pieces left, you've got a 1/n chance of picking the bad one, by the reasoning above: obviously, this is going to go up as n goes down. But at each stage, choosing to eat now or wait doesn't affect your odds of survival. Maybe you can't choose not to die, but only choose the manner of your dying: an appropriately Bushidoesque thought.
Whoever said probability was easy?