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There's a storm in my teacup!

Well, in my dollar store mug.

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Work irks.
The one thing that annoys me most when I work is when I have to deal with people who come in solely to waste my time. There are a fair few of them - a few mobs of troublemakers.

On the grand scale of things, they're harmless, but they're persistently annoying and every time you see them you can expect mocking comments until they know they've annoyed you or until they get bored. Is it a bad thing that I just act grumpy and annoyed by them immediately, so they stop sooner?

Puyo Pop Fever to try tomorrow, if that's good I'll nab it. I've got £12.50 in Game vouchers to use up, or I can get it from my shop if it arrives tomorrow with £8 off, and the £12.50 safe for whenever something else appears. Though if I dislike the game, it is easier to return it to Game.

A trip has been arranged - I plan to visit Shaz next Sunday 7th, and arrive back on Tuesday 9th. She's not been feeling too grand recently, a visit will hopefully cheer her up.

Sleep now. And the Monty Hall Problem.

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Ah, yes, we had this problem in Statistics. I had heard of it before. What amused me was the girl who refused to accept the answer. Here we were, waist-deep in t-tests and bionomial probabilities, and she simply couldn't let it go. It'd be one thing if she didn't get it but resolved to work it out in her own time. No, she stopped the class and demanded to have her solution recognized as the correct one.

Here's another problem that may or may not surprise you -- I dub it the Bowser's Big Blast problem. There's a Mario Party mini-game in which the four players are lined up in a random order and set before 5 switches. One of the switches is fatal, the other four are harmless. Then, in order, the four players take turns hitting an unflipped switch. If they hit a harmless one, they return to the end of the line. If they hit the fatal one, they are removed from the mini-game. If all four players avoid the fatal switch, then all the switches are reset, and another random switch is chosen to be the fatal one.

The question is - which position in line stands the best chance of surviving? If you had a choice, where would you put yourself in the line?

I suppose they're actually not switches, they're detonators. Eh, close enough.

If you get that one, I've got a Pirate-based logic problem that I may or may not have shared in the past.

That's an interesting problem - the first player has an 80% chance of surviving the selection, the second 75%, the third 66.6%, and the fourth 50% (if one of the other players isn't killed first) - right? But then you've got to take into account the fact that there's a reasonably large probability that if you go later, someone else will die first, but I'm having difficulty working that out as a probability.

So far, so good. Although it can be calculated either way, you might find it easier to work out the odds of each particular player getting blasted, rather than each player surviving.

So, yes, you've got the odds of the first player getting blasted in the first round. Now you need to get the odds of the second player getting blasted in the first round. So he's got a 25% chance of hitting the wrong switch, but how often does this chance even come up?

Once you've got that, you should be able to get the rest from there.

20% each way?

So it's all the same odds. Very clever.

Here's the pirate problem.

You are the leader of a very democratic but also ruthelessly logical and blood-thirsty gang of pirates. There are 9 pirates beneath you, each with a very specific rank (such that you are the first-in-command, and there exists a second-in-command, a third-in-command, a fourth-in-command and so on all the way down). You've recently acquired a haul of 100 gold coins, and it's time to split the coins up amongst yourselves.

As leader, you can propose a possible way to split the coins, at which point everyone (including you) votes either "Keep" or "Kill." If you get at least 50% "Keep" votes, the coins are divided up as you suggested. If you fail to get 50% "Keep" votes, you are killed and the second-in-command takes over. He then proposes a split, and the gang votes between keeping it or offing him and letting the next-in-command take his place.

Now, as I said, these pirates are both fiercely logically and, well, fiercely fierce. If they think they can get at least as many gold coins by voting "Kill," they will. If they think they will get even one more coin by voting "Keep," they'll do that instead. They'll only consider what they'll make next round, though, which simplifies matters for you.

The question is -- what is the optimum distribution of wealth for you?

Once you've gotten that, you should be able to solve that question for any arbitrary number of pirates and coins. This is a tricky problem, but the answer is interesting. If need be, I can post questions that might help one who is trying to solve it.

Heh, I remember your posing this question in high school as well. Although, looking back I don't think I realized that "They'll only consider what they'll make next round" could be a reason for illogical reasoning. Especially if they thought farther ahead and also considered whether they would survive or not. Just saying. Asummably even pirates have a motive for self-preservation.

Also, about the whole BBB thing. . .is this assuming the person doesn't know the trick to reading the colors? In reality, the way the probabilities work out is really the only reason the mini-game exists. And because its so damn amusing.

Actually, Ian W. posted the pirate problem. I think I was the first to solve it, which is probably why I still remember it. As for "They'll only consider what they'll make next round," I think you have to include that, otherwise you enter into a "prisoners' dilemma" situation.

No, actually, I just worked it out, and I can definitively state that it doesn't matter whether the pirate compare their share this round to their share next round or whether they compare it to their predicted best share. Depending on which one you choose, the logic is ever-so-slightly different, but not much. In the end, the solution is the same. The "trying to get best possible amount" idea makes for a more realistic and more interesting "moral," I think.

So, just to make it clear -- the pirates have to be able to THINK as far ahead as need be. However, whether they consider the next round or their best round in comparison to the current round is irrelevant to the logical solution. If you can figure out why this is true, you might be a little closer to solving it for yourself.

As for the BBB problem:
A) Even if the BBB game wasn't weighted the way it is, it wouldn't really matter. Let's the first person has a 50% chance of winning -- as long as the line order is random each time, enough trials should make any particular player come out even.

B) Even knowing the solution is, it still doesn't seem that way intuitively. We've played a lot of pirate, and we never go up on that stage saying, "Well, here goes my (x-1)/x chance of survival." It's tense!

C) There is no trick to "reading the colors," you're just saying that to drive me INSANE LBAHRAGGL

Just to make it blindingly clear (which, really, doesn't sound very clear at all) -- each pirate will strive for an optimum solution. Even if a pirate foresees himself getting 100% if the first eight leaders are executed, if he also recognizes that at least one of the first eight leaders will get 50% of the vote, then that pirate won't just try to kill his way to top in the already-established-as-hopeless case that he might get 100%.

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