A variant of the Monty Hall problem

[Pikkugnome]

There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host could just guess which door he opens and what is inside is random, maybe the host always opens an empty door or perhaps you get the choice only when you picked correctly initially.

 

[FactChecker]

maybe the host always opens an empty door or perhaps you get the choice only when you picked correctly initially.

This is fundamentally different. It is critical in the original puzzle that the host can not open a door with the prize and his offer does not depend on whether your door is the winner.

 

[PeroK]

This is fundamentally different. It is critical in the original puzzle that the host can not open a door with the prize and his offer does not depend on whether your door is the winner.

I think the point was the complication that Monty does not always give you the choice to switch. This is news to me! That adds a new parameter to the problem.

If the contestant picks the correct door, then the probability that Monty Hall offers a switch is $p_1$, say. And, if the contestant picks the wrong door, then the probability that Monty offers a switch is $p_2$.

Then the best strategy depends on these probabilities. For example, if $p_1 = 1$ and $p_2 = 0$, then you are only offered a switch when you have picked the correct door! If you know this, then you stick whenever you are given the chance.

It depends, therefore, whether $\frac 2 3 p_2 > \frac 1 3 p_1$. So, in fact, the problem is not very complicated, but there are some new variable parameters.

 

[PeroK]

PS the overall chance of winning is:

Stick strategy: $\frac 1 3 (1-p_1) + \frac 1 3 p_1 = \frac 1 3$

Switch strategy: $\frac 1 3 (1 - p_1) + \frac 2 3 p_2 = \frac 1 3 + \frac 1 3(2p_2 - p_1)$

So, dare I say, if $p_1 = 2p_2$, then the probability of winning with the switch strategy, in general, reduces to $\frac 1 3$!

And, of course, if $p_1 > 2p_2$, then it's better to stick.

 

[FactChecker]

I think the point was the complication that Monty does not always give you the choice to switch. This is news to me! That adds a new parameter to the problem.

Good point. This brings in aspects of game theory. If the host plays games to offer a switch more often when the contestant's door is a winner, the contestant might catch on and never switch. Then the host might fool the contestant by offering when the contestant's door is a loser. It is basic game theory. Von Neumann would love it.

 

[Office_Shredder]

If Monty is trying to be adverse to the contestant, they can always win 1/3 of the time by refusing to switch no matter what, and Monty can make sure they never win more than 1/3 of the time by refusing to offer them the other doors, so it's clear that any jointly optimal strategy has the contestant winning 1/3 of the time and we have also just described one. Doesn't make for good tv though.

 

[FactChecker]

If Monty is trying to be adverse to the contestant, they can always win 1/3 of the time by refusing to switch no matter what, and Monty can make sure they never win more than 1/3 of the time by refusing to offer them the other doors, so it's clear that any jointly optimal strategy has the contestant winning 1/3 of the time and we have also just described one. Doesn't make for good tv though.

I would not call that jointly optimal. It's only that simple if the strategies are pre-determined and they know each other's strategy. Otherwise, the contestant can try to do better than 1/3. He might succeed. Then the host can try to make it worse than 1/3, but might guess wrong. It can go either way. There are solutions where either one can do better than the 1/3, so "jointly optimal" is not optimal for either.

 

[Office_Shredder]

I would not call that jointly optimal. It's only that simple if the strategies are pre-determined and they know each other's strategy. Otherwise, the contestant can try to do better than 1/3. He might succeed. Then the host can try to make it worse than 1/3, but might guess wrong. It can go either way. There are solutions where either one can do better than the 1/3, so "jointly optimal" is not optimal for either.

By jointly optimal I meant a nash equilibrium.