Monty Hall's riddle

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Snuffy Smiff		#41	[-]
Posts: 1437 (08/10/08 2:58 PM) Registered user Reply Quote More Recent Posts Message Connection Block Invite	I had a really hard time with the logic of the riddle's answer until I thought of it like this: In your initial pick, your chances of picking an empty door are 2 in 3. So it makes sense to go with the odds and just assume you have an empty door. At this point Monty is left with only one choice: to open the only other empty door, thus guaranteeing that the the prize is behind the remaining closed door. So if you make the logical assumption that your original choice was wrong, you can go on to make the logical assumption that changing your choice will get you the prize. It's not for sure, but your odds definitely increase. Strange as it seems.
thriving sobi		#42	[-]
Posts: 274 (08/10/08 3:02 PM) Reply Quote More Recent Posts Message Connection Block Invite	Th TV show Numbers showed this a few seasons ago. Your first choice occured with an odd of 1/3, but by changing it, you went to 1/2 or fom a .33333 chance to a .5 chance. This is an improvement of .16666666666
Jennabridges18		#43	[-]
Posts: 171 (08/10/08 3:16 PM) Reply Quote More Recent Posts Message Connection Block Invite	Uh lets just accept that it is what it is and move on to a new problem we deal with in everyday life!
glazerboy		#44	[-]
Posts: 2677 (08/10/08 3:21 PM) Registered User Reply Quote More Recent Posts Message Connection Block Invite	The whole game is kind of a bait-and-switch probability problem. By starting with three doors, most people get into into the 1-in-3 chance mode. But once the host steps in and starts changing the conditions of the game, he's actually begun a whole new probability situation independent of the original 1-in-3 choice. But most people (myself included) miss that. The original 1-in-3 pick is a red herring to distract the player from the real probability situation of switch/no switch. Oh and Anti, you are right on the card thing.
Keitspid		#45	[-]
Posts: 5565 (08/10/08 3:27 PM) Registered user Reply Quote More Recent Posts Message Connection Block Invite	This question has always been hard for me to wrap my head around, because instinctively I want to say it's 50/50, obviously. But it only works because of the host's interference. Looking at the six possible games helps me understand it. It's still weird though.
PanamanianDevil		#46	[-]
Posts: 534 (08/10/08 4:09 PM) Reply Quote More Recent Posts Message Connection Block Invite	I knew the answer =D
glazerboy		#47	[-]
Posts: 2679 (08/10/08 4:21 PM) Registered User Reply Quote More Recent Posts Message Connection Block Invite	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.
Antithesys		#48	[-]
Posts: 22664 (08/10/08 4:54 PM) Gadabout Reply Quote More Recent Posts Message Connection Block Invite	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.
ColbyRulesAll		#49	[-]
Posts: 7686 (08/10/08 5:01 PM) Annoying sports freak Reply Quote More Recent Posts Message Connection Block Invite	I got a splitting headache about 5 posts into this thread.
glazerboy		#50	[-]
Posts: 2680 (08/10/08 8:02 PM) Registered User Reply Quote More Recent Posts Message Connection Block Invite	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.
Cassidy666		#51	[-]
Posts: 8103 (08/10/08 8:13 PM) Registered User Reply Quote More Recent Posts Message Connection Block Invite	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.
Roadfinder		#52	[-]
Posts: 4346 (08/10/08 8:57 PM) Registered user Reply Quote More Recent Posts Message Connection Block Invite	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 results. Switch each time: GAME 1: PRIZE NADA NADA ~ Switch and you lose. GAME 2: NADA PRIZE NADA ~ Switch and you win. GAME 3: NADA NADA PRIZE ~ Switch and you win. ............................................................ Stay each time: GAME 4: PRIZE NADA NADA ~ Stay and you win. GAME 5: NADA PRIZE NADA ~ Stay and you lose. GAME 6: NADA NADA PRIZE ~ Stay and you lose. When you switch, you win 2/3 of the time and lose 1/3, but when you don't switch, you only win 1/3 of the time and lose 2/3.
glazerboy		#53	[-]
Posts: 2681 (08/10/08 9:31 PM) Registered User Reply Quote More Recent Posts Message Connection Block Invite	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.
Gregoire		#54	[-]
Posts: 28370 (08/10/08 9:50 PM) Registered user Reply Quote More Recent Posts Message Connection Block Invite	What's the prize behind the door? I want to know what's at stake here.
Antithesys		#55	[-]
Posts: 22672 (08/10/08 9:52 PM) Gadabout Reply Quote More Recent Posts Message Connection Block Invite	A bust of Wendy's founder Dave Thomas.
annie normas butt		#56	[-]
Posts: 4382 (08/10/08 9:58 PM) Registered user Reply Quote More Recent Posts Message Connection Block Invite	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 original choice or switch doors. n/m...I see the formula above....odd Last Edited By: annie normas butt 08/10/08 10:01 PM. Edited 1 times.
Roadfinder		#57	[-]
Posts: 4347 (08/10/08 10:13 PM) Registered user Reply Quote More Recent Posts Message Connection Block Invite	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.
nardinim		#58	[-]
Posts: 154 (08/10/08 10:15 PM) Reply Quote More Recent Posts Message Connection Block Invite	But what if I want the goat?
Antithesys		#59	[-]
Posts: 22673 (08/10/08 10:17 PM) Gadabout Reply Quote More Recent Posts Message Connection Block Invite	The original show? Goats?
Roadfinder		#60	[-]
Posts: 4348 (08/10/08 10:30 PM) Registered user Reply Quote More Recent Posts Message Connection Block Invite	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.

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Monty Hall's riddle

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