The Mechanics of the Game Show Puzzle

[alexkidd3d]

of course this isn't simply a thought experiment, it's a plausible scenario. I don't know why these people felt the need to argue over who was right, you can simply do the experiment and show who is right. I forced some college friends to play the game with me over and over and then done the stats, after 100 plays its works out in favor of marilyn. There is indeed a 2/3 chance of winning should you switch.

[MrCheeseJeans]

Step one: 3 doors

Step two: 2 doors

In step one we have all the information we need. 1/3 chance to be right, but more importantly 2/3 chance to be wrong. Now if they take a door away BEFORE we choose, it ups our odds to 1/2. But they don't, they make us choose a door while probability is against us.

Step 2 takes a door away after we pick, but doesn't make our choice any more likely to be right. The door we chose still has a 66% chance of being wrong, but we are just misdirected.

[bikeNdrum]

i never knew there was so much contraversy over this question. its definitely not obvious, but it doesn't take a genius to realize the genius, marilyn, is right.

[brn20]

I think the confusion just arises from the two stages of the problem, and not considering that the probability of the prize being behind one of the other doors remains the same even if one of those doors is eliminated.

The elimination phase gives the illusion of the probability changing, but quick reflection shows that cannot be the case; only your perception of the problem may change. Your explanation is fairly clear - there is a 2/3 chance of the prize being behind one of the other doors...

[jay4457]

I read this post....and every "male" math scholar hated her answer...called her flawed....but soon as they tested it, she was right.........why would you argue with someone with the highest IQ recorded in the world.....

[TriplePalindromous]

@jay4457 because everyone can be wrong.

[giovea]

silly discussion. your talking in familiar terms of pre-test and post test probabilities, and you still cant decide if the elimination of one door is actually a true event or not. so decide that first- is it a given fact that the friend is consider to be right?? that would change the parameters and you need a pos-event new calculation. otherwise you still in the same event. like if you trou a coin 2 times and chance of 2 equal is 0.25, but chance of a equal after 1st is 0.5.mental mastur..

[BriDirt]

If you pick the wrong door(2/3 chance),and then switch it to the only remaining door,then you will be right 2/3 of the time. Its called a changing variable. The only way you will lose by doing this logic is if your iinitial pick was the correct door (which would only be a 1/3 chance), and then you switch doors. Just think if you did this with 100 doors. You pick one door, and then 98 doors are removed, leaving just your door and the correct door. Which one would you choose?

[babolatpdtp]

That's exactly how I explain it to people...Don't think of it as having a 1/3 chance of picking it right the first time and not switching...think of it as having a 2/3 chance of not picking it and then switching.

[FredricF]

I think the 100-door argument (see edwincorp comment below) sound quite convincing.

But there is a problem. This argument appeal to "common sense", and not logic, much like the "there are only two doors left in the end so it must be 50-50"-arguement.

I would like to see the logical flaw in the "two-doors in the end"-argument.

[FriendlyCougar]

If the contestant chooses a door in the first place he has a 1/3 chance of being right. This leaves a 2/3 chance that the prize is behind one of the other doors. The host is inclined to open a door that the contestant did not choose, therefor the 2/3 chance of being wrong remains, and the remaining door has that 2/3 chance of being right. If this explanation is unclear I'll make a video with a new chart that should be sufficient.

[prashantpawar]

She is wrong.

Lemme explain it how, if your friend always eliminates one wrong door and you were always given a choice to change your selection, then its as good as you were always asked to choose between two doors, one has prize, one does not.

Take for example there are 5 doors. You chose 1st time, then ur friend eliminates a wrong one, then you can change the option, then your friend eliminates another one.

This goes on till only two doors are left, only 1 has prize, and that's 1/2 chance.

[prashantpawar]

Ok I take it back, I tried it using a computer simulation, and changing does help. 66 out of 100 cases you would win.

[FriendlyCougar]

Thank you very much, I appreciate the fact that you can admit that your first argument is inaccurate. If you need a different explanation of how it works, I won't mind giving it.

[edwincorp]

I dont think i understand the contreversy. To me its obvious that shes right... what am i missing?.. If scaled up (choose one door from 100 and eliminate 98 wrong ones) you have to change door...

[FriendlyCougar]

The controversy comes from the other scholars who responded to her answer. Even the person who came to her with the puzzle said that she was wrong. I think it is sad that it made it to her in the first place. What kind of world do we live in when something this simple is considered to be at academic standards?

[Doazic]

I think the problem was that their responses were instinctive. They didn't bother reading her arguments.

[prospero811]

If a person has a 2/3 chance of being wrong, and then then he has a 1/3 chance of being right. But we need to acknowledge in the math that every combination of 2 doors has a 2/3 chance of being wrong. If you do it that way, you can't take 2 doors in isolation. B4 u learn that no prize is bhind door 3, there is equally a 2/3 chance of ANY 2 doors being wrong - 1 and 2, 2 and 3 and 1 and 3. The character count here is low - but can you see how that effects the math?

[prospero811]

Isn't the woman's rationale dependent on the assumption that making the choice effects the probabilities? After all, nobody would argue that if there were only 2 doors the chances would be 50%. Therefore, if start with 2 doors, and then ADD a door stating at the same time that there is no prize behind it, is the probability not still 50% as between 1 and 2?