Monty Hall, we have a PROBLEM

[LynahFan]

Re: Monty Hall, we have a PROBLEM

The following statement is true.

The previous statement is false.

Or you can just skip a step: This statement is false.

[FreshFish]

Re: Monty Hall, we have a PROBLEM

This statement is false.

So you say...but how reliable are you? did you discover that truth on your own, or is it from sometihing you read on the internet and re-posted without checking its validity first?

[FreshFish]

Re: Monty Hall, we have a PROBLEM

Something about this thread reminds me of the story of Carl Friedrich Gauss as an elementary school student. Even at that age, he knew more than his teacher.

His harried teacher finally thought of a way to quiet the precocious child. "Add the numbers from 1 to 100" he instructed, hoping to gain a few moments' respite from the child's unrelenting questions.

"Oh, that's easy," supposedly replied the young Gauss. "You take 1+99, 2+98, 3+97, ... and that get's you to 4,900. Add 50 to get 4,950. Then it depends upon whether you said '1 to 100' or '1 through 100' which would make it 5,050."

[The Rube]

Re: Monty Hall, we have a PROBLEM

If I were you and you were me and I shot you...who would die?

There is a verifiable answer, but it must be phrased correctly.

[Ralph Baer]

Re: Monty Hall, we have a PROBLEM

I have some reservations about Twitch Boy's auction, in white, in case anyone hasn't read his explanation:

In the abscence of spite, the key is that you only bid $99 if you're the *first* bidder. Even if the other guy has only bid $1 before your $99 bid, it would still be in his best interest to go ahead and bid $100 to improve his net position by $1 (form $1 lost to breaking even). In the real world, I wouldn't bid $99 even on the first bid with a 10-foot pole. Even if there's only a 0.0001% chance that there's a spiteful bidder in the room, throwing your hat into the ring at all exposes you to a huge potential loss, because as soon as the second bidder goes to $100, the infinite loop begins where bidding higher is not spiteful but is actually in both of your own self interests. Even if there's only that 0.0001% chance of a truly spiteful bidder, I guarantee there's at least a 10% chance of a moronic bidder who would bid $100 not out of spite but because he didn't know any better. Therefore, the optimal bid is not to bid at all.

http://www.heretical.com/pound/dollar.html

[owslachief]

You're arrested, thrown in jail, and given the death penalty.

The executioner, however, allows you to make one final statement in your last words. If it is a true statement, you'll be hanged. If it is false, you'll be beheaded.

Your statement should be "You're going to behead me." He would have to behead you to make it true, which means he should have hanged you instead. If he hangs you, then it was a false statement and he should have beheaded you instead.

Another paradoxical scenario was used (unsuccessfully) by the doomed right after sentencing. A creative judge handed him an unusual sentence: he was to be executed during a one-week window, and the day of his execution was to be a surprise.

The man reasoned his salvation as follows: if I survive until the last day, next Friday, then I know it's Friday i'm to die, and it's no longer a surprise. So I can't die on Friday. Thursday then becomes the last possible day, but then the same reasoning applies as for Friday! Safe in the knowledge that he can inductively apply that back through the entire week, he serenely waited out the week, until Friday came. In the morning, the executioner arrived, to the doomed man's complete surprise.

[owslachief]

Xml Something about this thread reminds me of the story of Carl Friedrich Gauss as an elementary school student. Even at that age, he knew more than his teacher.

His harried teacher finally thought of a way to quiet the precocious child. "Add the numbers from 1 to 100" he instructed, hoping to gain a few moments' respite from the child's unrelenting questions.

"Oh, that's easy," supposedly replied the young Gauss. "You take 1+99, 2+98, 3+97, ... and that get's you to 4,900. Add 50 to get 4,950. Then it depends upon whether you said '1 to 100' or '1 through 100' which would make it 5,050."[/QUOTE]

Yep the 100 x (avg. of #s between 1 and 99, which is 50)

[FlagDUDE08]

Re: Monty Hall, we have a PROBLEM

There are all kinds of series like that. Adding numbers = n(n+1)/2; the square of n is the summation (x = 1 to n) of 2x-1; so on and so forth.

Here's an interesting problem: On a site called moola.com, there is a game called Gold Rush. The idea is that you get "gold bars" weighed from 1 to 6, as does your opponent. Bars from 1 to 6 are up for grabs in each round. Whomever puts down the highest weight takes the round's pot. Ties put everything into the pot and carries over to the next round. How do you get yourself to win that?

[Twitch Boy]

Re: Monty Hall, we have a PROBLEM

You're at a party. I'll bet you $100 that at least two of the people in the room share a birthday.

How many people would need to be present for it to be an unfavorable bet for you?

Just 23. At 23 it's about a 50/50 shot that there will be a matching birthday. At 30 there's about a 70% chance of a match. At 50 it's a 97% chance.

Imagine it this way: the guests enter one at a time and state their birthday. If anyone repeats a birthday...THE ROOM EXPLODES!

First guest = can be any date. 365/365 odds of no match. (Assuming no leap years.)
Second guest = has to dodge the first date. 365/365 * 364/365 odds of no match.
Third guest = has to dodge the first and second dates. 365/365 * 364/365 * 363/365 odds of no match.
...
Tenth guest = has to dodge nine dates. 365/365 * 364/365 * ... * 356/365 odds of no match.
...
Twentieth guest = has to dodge NINETEEN dates. 365/365 * 364/365 * ... * 346/365 odds of no match.

It may seem like only a few days to dodge, but try dodging them 23 times in a row. Or 30. Or 50.

[UNH GH]

Re: Monty Hall, we have a PROBLEM

You're at a party. I'll bet you $100 that at least two of the people in the room share a birthday.

The Birthday Problem hurts my head. It's awesome though.

[jericho]

Re: Monty Hall, we have a PROBLEM

The Birthday Problem hurts my head. It's awesome though.

This whole thread hurts my head.

The previous statement is false.

[owslachief]

Re: Monty Hall, we have a PROBLEM

The Birthday Problem hurts my head. It's awesome though.

If you think of it as one person trying to dodge another person's birthday, you'd tend to to make that safe-bet number high.
If you think of it as N people trying to dodge N-1 persons' birthday's it's apparent that N would be pretty low (like 23) to make it a safe bet .

[bigblue_dl]

Re: Monty Hall, we have a PROBLEM

Lynah has the right answer. First time I heard that one was some years ago and it took me a little while to figure out, but the trick is to figure out a question where both the angel and devil will give the same answer.

Why not ask a question you know the answer to, and go in the door guarded by the one that gives the correct answer? There were no limits of what the question has to be, so ask: Who won Super Bowl I, or Who was the 1st president of the USA? or What is capital city of Ohio?

[owslachief]

Re: Monty Hall, we have a PROBLEM

Why not ask a question you know the answer to, and go in the door guarded by the one that gives the correct answer? There were no limits of what the question has to be, so ask: Who won Super Bowl I, or Who was the 1st president of the USA? or What is capital city of Ohio?

Thanks for bringing that up; I think we all missed that. There needs to be one more condition (thanks to my 13-year-old niece who said "Just ask one of the guardians if Elvis is dead."):
4. The question must be one whose answer is unknown to you beforehand (except regarding the facts given in the other conditions listed).

[LynahFan]

Re: Monty Hall, we have a PROBLEM

Thanks for bringing that up; I think we all missed that. There needs to be one more condition (thanks to my 13-year-old niece who said "Just ask one of the guardians if Elvis is dead."):
4. The question must be one whose answer is unknown to you beforehand (except regarding the facts given in the other conditions listed).

Or, again, just assign the guardians to random doors. The Elvis question then tells you which is the liar, but doesn't tell you which door to pick. It's just a much cleaner way to state the problem.