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ORIGINAL: cnjpratt

Interesting at first glance, but not exactly rocket science:

If x = first 3 #'s
y = last 4 #'s

Then the instructions become: ((((x*80)+1)*250)+2*y-250)/2

Simplify: x*40*250+y = x*10,000 +y

Therefore, whatever 3 digit number you put in for x ends up a 7-digit # ex. 756*10,000 = 7560000. Then you add the last 4 digits Ex 4156 = 7564156.

They just made it look complicated.

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After that, regardless of which door you pick the chance is 1:2

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ORIGINAL: Rainbow

This is a great example that I always use in discussing probability and Bayesian statistics with students. It stumped many professional statisticians at the time (and continues to, from my experience). The problem becomes clearer (only to some) if you start with 1000 doors instead of three. Would you stick with your first door (only 1/1000 chance) or would you switch to the other door remaining after Monty opens 998 doors?

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ORIGINAL: mavraam

I remember a puzzle about 3 people splitting a check that ended up shorting the waiter $1 even though it appeared that everyone had paid their fair portion but I can't remember the specifics. [:(]

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I'm not sure that makes sense, but one thing I've found that will convince some is to get 3 playing cards, 1 red, 2 black and simulate the situation 20 times. The first ten, never switch. The second ten always switch. The second 10 should almost always win over the first barring some statistical anomoly.

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ORIGINAL: Svennemir

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I'm not sure that makes sense, but one thing I've found that will convince some is to get 3 playing cards, 1 red, 2 black and simulate the situation 20 times. The first ten, never switch. The second ten always switch. The second 10 should almost always win over the first barring some statistical anomoly.

It's funny, I also like to use playing cards to convince people. When they've tried to not switch 10 times, it becomes clear that they could as well select the door/card right away without the host revealing anything, and then the 1/3 follows.

Lo and behold! The power of google enabled me to find the one about the waiter you mentioned. (and in the first try!)

Here it is: Three men go to stay at a motel and the clerk
charges them $30.00 for the room. They split the cost ten
dollars each. Later the manager tells the clerk that he over-
charged the men and that the actual cost should have been
$25.00. He gives the clerk $5.00 and tells him to give it to the
men. But he decides to cheat them and pockets $2.00. He then
gives each man a dollar. Now each man has paid $9.00 to stay in
the room and 3 X $9.00 = $27.00. The clerk pocketed $2.00.
$27.00 + $2.00 = $29.00. So where is the other $1.00?

Link

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ORIGINAL: Rainbow
@dinsdale: The probabilities don't get reset, because nothing has altered the fact that you only had a 1 in 3 chance with your first pick of a door (or 1 in 1000 chance with my 1000 door setup). What's telling is that Monty won't open your door and he won't open the correct door. So you could stick with your 1/1000 chance door. But not being a gambling man, I'd choose the remaining door at 999:1 odds. And this is quite different from the coin flip example because each coin flip is independent of all previous ones, whereas here you don't get a second choice after Monty opens a door (you're stuck with your first choice).

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@ mavraam: The brute force method isn't very intellectually satisying, and it doesn't really explain anything. That's my last resort.

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ORIGINAL: mavraam

I know. But some people who don't have a background in probability will never quite get it.

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Two out of 3 times, you WON'T pick the right door on the first try. Then, after the other wrong door is open, you'll pick the right door after you switch.
One out of 3 times, you WILL pick the right door on the first try. Then, after the other wrong door is open, you'll lose the right door after the switch.

So by switching, you'll win 2 out of 3 times.