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| Jul-04-05 | | farrooj: Is probability taught in high school? It's quite interesting |
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| Jul-04-05 | | InfiniteWombat: One interesting thing about probability is how it can thwart our intuition. A good example of this is the famous Monty Hall riddle. Even some mathematics professors got this one wrong. For those that are curious it goes something like this: There's a game show in which the contestant, Bob, has to select one of three closed doors. Behind one of the doors is a new car, but there's nothing behind the other two. First, Bob makes his selection. Then without opening the selected door, Monty Hall opens the unpicked door that has nothing behind it. Bob is then asked if he would like to stick with his current door, or switch to the other closed door. Should Bob switch ? |
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| Jul-04-05 | | OneBadDog: <InfiniteWombat> My guess is that the probabilities of the contestant winning the car are the same for both remaining doors, regardless of the contestant's choice. |
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| Jul-04-05 | | InfiniteWombat: OneBadDog, that sure seems reasonable, intuitively. It's not correct however. His chance of winning is 1/3 if he doesn't switch, and 2/3 if he does. One way to think about it is that if he switches, the only way he would lose is if he picked the car originally. Otherwise the unopened door must have the car behind it, since clearly it must be behind one of the two remaining doors after his selection, and Monty Hall always opens a door with nothing behind it. |
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| Jul-04-05 | | OneBadDog: <InfiniteWombat> Pardon my thick headedness, but why wouldn't both remaining doors have a 1 in 2 chance of being the correct door? |
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| Jul-04-05 | | InfiniteWombat: OneBadDog, I believe my previous post had an explanation. I can try to explain it another way if that doesn't work for you. |
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| Jul-04-05 | | OneBadDog: <InfiniteWombat> Yes, could you? I'm missing something. |
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| Jul-05-05 | | OneBadDog: <InfiniteWombat> I think I got it. Relative to the original selection, the probability for the door initially selected remains 1/3, but the remaining unselected door's probability goes up to 1/2. |
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| Jul-05-05 | | InfiniteWombat: It goes up to 2/3 not 1/2. The 2/3 represent the probability that either of the unselected doors have the car. |
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| Jul-05-05 | | euripides: I think the critical point is that it is part of the rules of the game that if there is a car behind one of the doors that Bob has not chosen, Monty opens the other one. If Monty made a random choice between the two doors, then the probability would indeed be 50%. But since Monty's choice is informed by the actual location of the car, his choice conveys information about the car's whereabouts. There are three possible sites for the car. In the 2/3s of cases where the car is not behind the door that Bob chose, it will always be behind the door that Monty does not open. So the probability that the car is behind the door Bob did not choose and Monty did not open is 2/3s. |
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| Jul-05-05 | | InfiniteWombat: Indeed that is the critical point, euripides. A very lucid description :) |
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| Jul-05-05 | | TheSlid: Surely the answer is that if the host is going to open a door with nothing behind it, the contestant's chances of winning are always 1/2. |
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| Jul-05-05 | | azaris: It's a completely ambiguous problem unless you specify exactly the method by which Monty Hall picks the door. We went through this for an hour in my stochastics class and ended up with six different answers, none of which anyone considered to be right. |
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| Jul-05-05 | | TheSlid: Good one <azaris>. The critical question is whether Monty Hall will behave differently depending on whether the contestant's first choice is correct. If he is just going to open a door with nothing behind it, irrespective of whether Bob is correct, then it really is trivial. |
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| Jul-05-05 | | TheSlid: If, on the other hand, the host will only open a door if the contestant has guessed wrongly, then the contestant will always be correct to switch. |
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| Jul-05-05 | | InfiniteWombat: Azaris, I believe the method was specified exactly. From the two remaining doors, Monty always selects the door that has nothing behind it. Anyway, I just mentioned this example to illustrate that intuition can be pretty bad at dealing with probability problems. |
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| Jul-05-05 | | azaris: <InfiniteWombat> In that case it's 1/3 if the contestant stays, 2/3 if he switches. This is the simple case. |
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Jul-05-05
 | | Sneaky: <Q> "In a gathering of 23 people, what is the probability that at least two people have the same birthday? (Assume a 365-day year.)" Try to guess the answer based on gut instinct. Two people out of 23 share their birthday how often... 1 in 15 time? 1 in 12? 1 in 5? 1 in 3? 1 in 2? <A> The last answer is the closest: The true odds are about 50.7% |
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| Jul-05-05 | | azaris: <Sneaky> More precisely 434508600510580 / 856516793531503. |
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| Jul-05-05 | | farrooj: where the hell did you get that answer <azaris>? |
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| Jul-05-05 | | Chesschatology: May I recommend Wikipedia’s entry on Hydra. It’s quite fascinating, and there are links to explain such concepts as “The minimax function” “Alpha-Beta Pruning” “Null move algorithms” etc.
Reading it has made me feel a little better about the mauling of Mickey, because some of the ideas are so ingenious. “Brute Force” is actually, it appears, a quite unfair name for a very clever little machine indeed. What struck me about it is that a player who REALLY studied how the algorithms worked could actually predict specific traps the computer would be likely to fall into in a far more sophisticated way then saying “Close up the position.” For example
- It appears that the more equally good continuations there are, the less effectively the computer can pair down its analysis tree, and therefore the fewer moves it can look ahead.
- Zugzwang type positions seem to confuse the critters.
- The fewer checks and captures available the less the tree can be pruned. Maybe we can still beat these things by putting as much work into understanding them as the programmers have put into making them so damn strong! |
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| Jul-05-05 | | csmath: <<Reading it has made me feel a little better about the mauling of Mickey, because some of the ideas are so ingenious. “Brute Force” is actually, it appears, a quite unfair name for a very clever little machine indeed.>> Yes, exactly. That is the point, these machines are not aliens, they were made by inteligent programmers/engineers and they are not a proof of machine dominating humans but of humans using their inteligence to develop better tools. |
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| Jul-05-05 | | azaris: <farrooj> That's in fact a bit rounded. I only had Excel to use at work. Mathematica 4 does it in integer precision and gives 38093904702297390785243708291056390518886454060947061 --- 75091883268515350125426207425223147563269805908203125 when you ask it to evaluate
1 - Product[k/365, k, 365 - 23 + 1, 365]. |
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| Jul-05-05 | | farrooj: I don't know how to calculate odds. Can you explain the formula a bit? |
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| Jul-05-05 | | Marvol: <OneBadDog, InfiniteWombat>
An interesting other way of looking at this problem is this: Suppose there are 100 doors. Bob picks one, and then Monty opens up 98 of the remaining 99. Should Bob switch?
He obviously should!
The same maths apply to this problem although the odds change - but they still favor switching. |
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