Re: Monty Hall's riddle

feeds@yuku.com (Screerider) — Mon, 11 Aug 2008 14:12:04 GMT

Another way to look at this as the chances of losing. The only way you lose by switching is if you picked the winning door in the first place. Since you had a 1/3 chance of picking the winning door, switching means you have a 1/3 chance of losing. And since 1 - 1/3 = 2/3, you have a 2/3 chance of winning by switching.

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Re: Monty Hall's riddle

feeds@yuku.com (glazerboy) — Mon, 11 Aug 2008 04:17:31 GMT

Damn you all for putting this in my head. I woke up in the middle of the night and couldn't fall asleep until I finally got the solution in my head.

The odds NEVER change. When there are 3 doors available, the odds are 1:3 that you will pick the right one. Once you pick a door, there are now 2 sets - Set A=your door=1:3 odds, and Set B=the other 2 doors=2:3 odds. Revealing a losing door doesn't change the odds that Set B has a 2:3 chance of holding the winning door.
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Re: Monty Hall's riddle

feeds@yuku.com (Roadfinder) — Sun, 10 Aug 2008 23:29:04 GMT

A bit out of focus, but amusing nonetheless. And oh yes, there were goats in the original show!

Gregoire, if you want a goat, don't switch. If you switch, most likely you'll win a brand new $9,959 Mazda...

in those days.

� Antithesys wrote:

The original show? Goats?

Yes!

And a brand new car is the prize.

As you see, switching wins 2/3 of the times.

Re: Monty Hall's riddle

feeds@yuku.com (Antithesys) — Sun, 10 Aug 2008 22:17:31 GMT

The original show? Goats?

Re: Monty Hall's riddle

feeds@yuku.com (nardinim) — Sun, 10 Aug 2008 22:15:48 GMT

But what if I want the goat?

Re: Monty Hall's riddle

feeds@yuku.com (Roadfinder) — Sun, 10 Aug 2008 22:13:32 GMT

Gregoire wrote:
What's the prize behind the door? I want to know what's at stake here.

In the original show, it was a brand new car in one door, and two goats in the other two.

You're right, glazerboy. The probability at the beginning is 1/3, and it's the host interfering that makes this problem (not a riddle) interesting.

The solution was given above.

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Re: Monty Hall's riddle

feeds@yuku.com (annie normas butt) — Sun, 10 Aug 2008 21:58:25 GMT

Antithesys wrote:
Well bassoon is right...the chances it's behind your original door are 1 in 3, while the chances it's behind the remaining door are 2 in 3.

That doesn't make any sense. Your chances that it's behind the original door were only 1 in 3 before the empty door was opened, then increased to 2 in 3. There would be equal chance the prize is between door #1 or #3 whether you hold onto your...

Re: Monty Hall's riddle

feeds@yuku.com (Antithesys) — Sun, 10 Aug 2008 21:52:43 GMT

A bust of Wendy's founder Dave Thomas.

Re: Monty Hall's riddle

feeds@yuku.com (Gregoire) — Sun, 10 Aug 2008 21:50:51 GMT

What's the prize behind the door? I want to know what's at stake here.

Re: Monty Hall's riddle

feeds@yuku.com (glazerboy) — Sun, 10 Aug 2008 21:31:33 GMT

Right - I get that. What I was pointing out was that with six possible scenarios, there are still three times you can win and three times you can lose. It's the added variable of being able to switch that tilts it off being a straight 50/50 proposition.

Re: Monty Hall's riddle

feeds@yuku.com (Roadfinder) — Sun, 10 Aug 2008 20:57:45 GMT

What is interesting is that out of the six possible scenarios, three result in winning and three result in losing, which supports the 50/50 solution.

Nope.

Say you picked door #1 and another door is opened with nada behind it.

For the first three games, you choose #1 and "switch" each time, for the second three games, you choose #1 and "stay" each time, and the host always opens a loser.

Here are the...

Re: Monty Hall's riddle

feeds@yuku.com (Cassidy666) — Sun, 10 Aug 2008 20:13:31 GMT

The key part of the situation is that the host can only remove one option which (he knows) has nothing behind it and basically doubling the chances that the "other door" has the prize.

Re: Monty Hall's riddle

feeds@yuku.com (glazerboy) — Sun, 10 Aug 2008 20:02:40 GMT

What is interesting is that out of the six possible scenarios, three result in winning and three result in losing, which supports the 50/50 solution. The monkey wrench comes with the 2 variables - the host revealing a losing door and the contestant given the option to switch. The math to calculate that kind of probability is beyond me.

Re: Monty Hall's riddle

feeds@yuku.com (ColbyRulesAll) — Sun, 10 Aug 2008 17:01:06 GMT

I got a splitting headache about 5 posts into this thread.

Re: Monty Hall's riddle

feeds@yuku.com (Antithesys) — Sun, 10 Aug 2008 16:54:38 GMT

I'd say the door would be 2/3 in favor of being a 1/2 shot. So 2/3 x 1/2 = 2/6 = 1/3 for switching? Which would be the same as not switching. That doesn't mean 50/50, because a third of the time the removed door would be the winner.

And in the spirit of the problem, the host couldn't show you what's behind the randomly removed door because if it was the winner you wouldn't need to switch, and switching's the whole point of the problem.

Re: Monty Hall's riddle

feeds@yuku.com (glazerboy) — Sun, 10 Aug 2008 16:21:04 GMT

I agree with the host's interference. I suspect it would be 50% if the host removed one of the two doors at random rather than always removing a losing door.

Monty Hall's riddle

feeds@yuku.com (Antithesys) — Sun, 10 Aug 2008 09:22:16 GMT

Let's see if you dopes know this one.

The game show host offers you a choice of three doors. Behind one is a prize, while the other two contain nothing. You pick a door. The host then opens one of the remaining doors to reveal that one was empty. He then offers you the chance to switch your choice to the third door. Should you do it?