20080411, 14:56  #12 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 

20080411, 16:06  #13  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
I agree the confusion arose because I was previously using "you" to denote the quizmasters BTW I think we can safely dispense with the spoilers now David Last fiddled with by davieddy on 20080411 at 16:35 

20080420, 15:09  #14 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
I gave this to an old school friend of mine and he replied:
There are three scenarios: I pick the door with the car, hosts reveals tricycle 1 or 2, I swap and lose the car. I pick the door with tricycle 1, the hosts reveals tricycle 2, I swap and win the car. I pick the door with tricycle 2, the hosts reveals tricycle 1, I swap and win the car. So swapping is always worth it as it wins 2/3 of the time. I think that puts it as clearly as possible. 
20080421, 07:05  #15  
Jan 2006
JHB, South Africa
157 Posts 
Quote:
It looks like 2 scenarios are summarized into one: Should it not reflect: I pick the door with the car, hosts reveals tricycle 1, I swap and lose the car. I pick the door with the car, hosts reveals tricycle 2, I swap and lose the car. I pick the door with tricycle 1, the hosts reveals tricycle 2, I swap and win the car. I pick the door with tricycle 2, the hosts reveals tricycle 1, I swap and win the car. Bringing it back to a 50/50 chance? This is with the presumption that the hosts know what is behind each door and deliberately open a door that reveals a tricycle. Regards Patrick Last fiddled with by Patrick123 on 20080421 at 07:07 Reason: added the presumption. 

20080421, 10:30  #16 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Your 4 scenarios are not equally probable/frequent.
See my earlier posts. David 
20080421, 12:42  #17 
Jan 2006
JHB, South Africa
235_{8} Posts 
If it was a random selection that the hosts use to select the 'other door' to open then I would agree with you, but no matter which door you select, they will always ensure that they will open a door with a tricycle in it.
Thus what was initially a 1/3 probability, because they deliberately choose a door containing a tricycle no matter what door you choose, it becomes a standard 1/2 selection. 
20080421, 13:35  #18 
Jan 2006
JHB, South Africa
157 Posts 
Allow me to extrapolate:
You choose a door  66.7% chance it contains a tricycle. They choose a door(at random or a member of the audience chooses it) 50% or 100% chance it contains a tricycle. if it does contain one then it becomes logical to switch to the other door(66.7% chance you've eliminated the tricycles), but because they know the outcome and do choose a door with a tricycle, you are still left with a 50% chance. Last fiddled with by Patrick123 on 20080421 at 13:47 Reason: member of audience edit. 
20080421, 19:40  #19  
"Mark"
Apr 2003
Between here and the
3·1,973 Posts 
Quote:


20080421, 19:58  #20  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
the car (and you presumably lose it). But if he reveals a tricycle and offers you the swap, NOW it is a 50/50 proposition. (Why?) David 

20080422, 04:56  #21  
Mar 2008
100000_{2} Posts 
Quote:
(1/6) I pick the door with the car, hosts reveals tricycle 1, I swap and lose the car. (1/6) I pick the door with the car, hosts reveals tricycle 2, I swap and lose the car. (1/3) I pick the door with tricycle 1, the hosts reveals tricycle 2, I swap and win the car. (1/3) I pick the door with tricycle 2, the hosts reveals tricycle 1, I swap and win the car. If you don't understand why the first two scenarios must be 1/6, then consider that Monty's selection is only made after the contestant picks a door. If the contestant picks a winning door (with 1/3 prob), then the odds that Monty will pick each of the two wrong doors is 1/3 * 1/2 or 1/6. In this case it also really doesn't matter which one he picks, so it is simpler to roll them into one case where the contestant has picked the right door first. Last fiddled with by robo_mojo on 20080422 at 05:21 

20080422, 05:20  #22  
Mar 2008
20_{16} Posts 
Quote:
Because, suppose you have a 1/3 chance of picking the car at first. If the host COULD reveal any door, even a car, and have you lose it, then we cannot use his limits to increase our odds. So we have a 1/6 (1/3 * 1/2) chance that we had the car first, kept it and won; a 1/6 (1/3 * 1/2) chance we had the car first, but swapped and lost; 1/6 (2/3 * 1/2 * 1/2) chance we picked a wrong door and the host reveals a wrong door, and we kept the door and lost; a 1/6 (2/3 * 1/2 * 1/2) chance we had a wrong door, the host reveals a wrong door, we swap and win; and a 1/3 (2/3 * 1/2) chance that we picked a wrong door, and the host reveals the car, preventing us from having the opportunity to swap. Then, supposing we have an opportunity to swap, removing the last outcome (assuming the host hadn't revealed the car, though he COULD have), then there's a 1/2 chance that the swap will win and a 1/2 chance a swap would lose. Overall in that kind of game you'd still have only a total of 1/3 chance of winning the car. Again that's supposing that the host is NOT limited to picking a wrong door. Only when the host is limited to picking a wrong door can you increase your odds to 2/3 by swapping, as was already discussed. Where most people get confused about this kind of problem is that you must figure out how the outcomes are dependant of each other (what the restrictions are), and multiply the odds of each element together in the chain. Last fiddled with by robo_mojo on 20080422 at 05:23 

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