Monty Hall problem in a different way: What if at the start there are 3 players instead of 1?

[ctwdream]

Hi!

I'm just a normal person who found a logic puzzle at facebook (Monty Hall problem) and wanted to ask somebody about it - so I found this site (FreeMathHelp).

I understand the point of it, but there is one idea I can't handle:

What if at the start there are 3 players instead of 1. Each door can be chosen by 1 person. After everyone has selected a door, the game host opens one of it where there is no car - and whoever chose it was eliminated. Then the host says the same as in the Monty Hall problem: the remained players can change the door if they want.

So, my question is: what should they do? If they think that they will have 2/3 chance to win if they change, then that's not a paradox? Because both doors can't be more chance than 100%. What kind of math works here?

 

[ctwdream]

Actually I sent this question more than 24 hours ago (the approve took some time it seems), and I realised that there is a 2 players situation on the internet which is almost same as my question that I sent here. So got my answer:

Not worth the change, the chances are 50-50%.

 

[Dr.Peterson]

I was locked out for hours (due to a bug I've seen before), so my answer never got sent.

Do you understand why the two-person case is different? Briefly, it's that Monty is no longer prevented from opening the door you chose, 1/3 of the time you will win by staying (because you chose the right door initially); 1/3 of the time you will win by changing (because you chose a wrong door and Monty opened the other); and 1/3 of the time you lose because he opens your door. In the one-person game, you would have won by changing in those last 1/3 of cases.

General lesson: probability problems are highly dependent on details, so you can't just expect that you can still do what would have worked in a different version.