3 ms·
Still, why? Why open 38 doors? I don't think this emphasizes the difference in probability, it just creates artificial bias. Let's analyze the classical proble
by alco 13y ago
Still, why? Why open 38 doors? I don't think this emphasizes the difference in probability, it just creates artificial bias.
Let's analyze the classical problem. After you pick a door, one of the remaining two doors is opened.
With more doors (say, 11), it wouldn't make for an interesting show if you opened 9 more doors after the first one. A more reasonable approach would be to open one additional door and let the player choose to switch or stay with their choice each time. But that would only complicate the explanation, of course.
Let's dissect the 11-door case. We will do the following: the player picks a door, then the host a) opens one other door; b) opens 50% of the remaining doors; c) leaves one door closed.
IMO, the (a) case is the closest generalization of the original problem. You have a 1/11 chance to pick the car door on your first try. Then, one other door is opened, and if you switch to another random door, the best case probability of you getting the car is 10/11 * 1/9 = 10/99. This number is slightly bigger than 1/11.
In the (b) case you get a better result when switching, but to me that just looks like adding artificial bias. Obviously, the (c) case looks best because you've removed so much choice by leaving only one other door closed. Adding more doors does one thing: improves the probability of switching. That's good if this is what you're aiming for. But this turns it into a completely different problem.
So, you can see that the result depends on what you want to get. If you try to stay faithful to the original problem statement, adding more doors actually makes the benefit of switching less obvious.
In summary, I'm convinced that 3 is not an accidental number in the original problem statement. Whoever came up with it, they knew how to remove space for speculation by providing the only reasonable thing to do: open 1 door (or 50% of the doors) and leave 1 other door closed.
- Jtsummers 13y ago(c), I feel, matches the original problem best when taken to an increasing number of doors, but makes for an absurd game show. It also most closely resembles the odds of the original game show (win by switching (n-1) of n times, or 2 of 3 with 3 doors). Anything else becomes a different game (you don't have a single choice, you have a multitude of choices). And it doesn't turn it into a different problem, it exagerates the odds of the original problem to better illustrate the statistics to people whose intuition is wrong. The case of 100 doors where switching results in winning 99% of the time gets passed the questioners intuition that it should be even odds (switching or staying). (a) demonstrates that the questioners intuition about the probabilities is off, but (again, IMO) doesn't really offer the clarity of (c). Re: Summary - I agree, it's not an accident. It was chosen because it was a good number for a game show, and leaving the participant with 2 options (switch or stay) ensures a certain swiftness of decision. While (a) (say they used 4 or 5 doors) offers only a slight improvement on the odds (contestants may lose too often) and (c) (again with 4 or 5 doors) offers a perhaps too great improvement of the odds (contestants win too much if they know the trick).