Should You Stick or Switch? A Modified Monty Hall Problem

[gptejms]

TL;DR Summary

Monty Hall problem is modified to 2 cars and 2 goats

I modify the Monty Hall problem a bit. Suppose there are 4 doors and behind
each door either a goat or a car out of a pool of 2 goats and 2 cars. You choose a door at random and in comes Monty Hall who always opens a door with a goat behind it. He opens just one door for you and asks you if you would like to stick to your door or pick either of the two remaining doors. What should you do?

 

[PeroK]

Summary:: Monty Hall problem is modified to 2 cars and 2 goats

I modify the Monty Hall problem a bit. Suppose there are 4 doors and behind
each door either a goat or a car out of a pool of 2 goats and 2 cars. You choose a door at random and in comes Monty Hall who always opens a door with a goat behind it. He opens just one door for you and asks you if you would like to stick to your door or pick either of the two remaining doors. What should you do?

Switch.

 

[PeroK]

Suppose there are $g$ goats and $c$ cars. If we stick, then the probability of winning is$$p_0 = \frac{c}{c+g}$$If we switch, then the probability of winning is
$$p_1 = \bigg ( \frac{c}{c+g} \bigg ) \bigg ( \frac{c - 1}{c + g -2} \bigg ) + \bigg ( \frac{g}{c+g}\bigg ) \bigg ( \frac{c}{c + g -2} \bigg ) = \bigg (\frac{c}{c+g} \bigg ) \bigg (\frac{c+g - 1}{c + g -2}\bigg )$$Hence:
$$p_1 = p_0 \bigg (\frac{c+g - 1}{c + g -2}\bigg )$$And, in all cases, we have $p_1 > p_0$. E.g. for $c = 1, g = 2$:
$$p_0 = \frac 1 3, \ p_1 = \frac 2 3$$Or, in your example, with $c =2, g = 2$:
$$p_0 = \frac 1 2, \ p_1 = \frac 3 4$$

 

[FactChecker]

always opens a door with a goat behind it. He opens just one door for you and asks you if you would like to stick to your door or pick either of the two remaining doors.

Anything that [EDIT} intentionally filters out some goat doors from the unchosen doors, increases the probability of the remaining unchosen doors having a prize. That makes their probability of having the prize greater than for your door.

 

[member 428835]

Not what OP was asking, but a great intuitive way to think about it is increase to 100 doors, 99 goats, 1 car. You choose, and then the host opens 98 doors and asks you if you want to switch. It feels very intuitive that switching is in your best interest.

 

[FactChecker]

Not what OP was asking, but a great intuitive way to think about it is increase to 100 doors, 99 goats, 1 car. You choose, and then the host opens 98 doors and asks you if you want to switch. It feels very intuitive that switching is in your best interest.

Yes. In post #4, I should have said "intentionally filters out some goat doors". If you know that Monte is avoiding the prize door, then the 98 goat doors that he opens are a clear indicator that the prize is behind the door he avoided.
On the other hand, if you know that Monte is just blindly opening doors with no knowledge of where the prize is, then you must resign yourself to the fact that you have just witnessed a very strange run of luck and your door is just as likely as the remaining door to have the prize.

 

[member 428835]

Yes. In post #4, I should have said "intentionally filters out some goat doors". If you know that Monte is avoiding the prize door, then the 98 goat doors that he opens are a clear indicator that the prize is behind the door he avoided.
On the other hand, if you know that Monte is just blindly opening doors with no knowledge of where the prize is, then you must resign yourself to the fact that you have just witnessed a very strange run of luck and your door is just as likely as the remaining door to have the prize.

Deal or no deal comes to mind (not sure if you've ever heard of this show). At the end, with two boxes left, it's entirely 50-50 on what you have and what you don't despite 24 boxes being opened owing to the reason you said: it was random and not intentional.

 

[gptejms]

Suppose there are $g$ goats and $c$ cars. If we stick, then the probability of winning is$$p_0 = \frac{c}{c+g}$$If we switch, then the probability of winning is
$$p_1 = \bigg ( \frac{c}{c+g} \bigg ) \bigg ( \frac{c - 1}{c + g -2} \bigg ) + \bigg ( \frac{g}{c+g}\bigg ) \bigg ( \frac{c}{c + g -2} \bigg ) = \bigg (\frac{c}{c+g} \bigg ) \bigg (\frac{c+g - 1}{c + g -2}\bigg )$$Hence:
$$p_1 = p_0 \bigg (\frac{c+g - 1}{c + g -2}\bigg )$$And, in all cases, we have $p_1 > p_0$. E.g. for $c = 1, g = 2$:
$$p_0 = \frac 1 3, \ p_1 = \frac 2 3$$Or, in your example, with $c =2, g = 2$:
$$p_0 = \frac 1 2, \ p_1 = \frac 3 4$$

Nice work! I would have been happy with 1/2, 3/4, but you have done it for the general case! If I define bias as the difference between probability of a biased outcome and that of an unbiased outcome, then we can say that Monty Hall by opening a goat door introduces a bias of $$p_1-p_0=\frac{p_0}{c+g-2}$$ (in this case) into the system.