# Should You Stick or Switch? A Modified Monty Hall Problem

• B
• gptejms
In summary: So, if ##p_0## were to be ##\frac{1}{2}##, Monty Hall's action introduces a bias of $$\frac{1}{2(c+g-2)}$$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. So, if ##p_0## were to
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
gptejms said:
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.

member 428835
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##:
$$p_0 = \frac 1 2, \ p_1 = \frac 3 4$$

gptejms
gptejms said:
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.

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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
joshmccraney said:
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
FactChecker said:
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.

FactChecker
PeroK said:
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.

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## 1. What is the Modified Monty Hall problem?

The Modified Monty Hall problem is a variation of the original Monty Hall problem, which is a famous probability puzzle named after the host of the game show "Let's Make a Deal". In the original problem, a contestant is presented with three doors, one of which contains a prize. After the contestant chooses a door, the host reveals one of the remaining doors that does not contain the prize and gives the contestant the option to switch their choice. The Modified Monty Hall problem adds an extra step where the host may reveal a second door before giving the contestant the option to switch.

## 2. How is the Modified Monty Hall problem different from the original?

The main difference between the Modified Monty Hall problem and the original is the addition of an extra step where the host may reveal a second door before giving the contestant the option to switch. This changes the probabilities and outcomes of the game, making it a more complex and challenging puzzle.

## 3. What is the solution to the Modified Monty Hall problem?

The solution to the Modified Monty Hall problem is different from the original Monty Hall problem. In the Modified version, the contestant should always stick with their original choice. This is because the extra step of the host revealing a second door eliminates the advantage of switching, making it a 50/50 chance between the two remaining doors.

## 4. Why is the Modified Monty Hall problem important?

The Modified Monty Hall problem is important because it challenges our intuition and understanding of probability. It also demonstrates the importance of carefully considering all possible outcomes and their probabilities before making a decision. This can be applied to real-life situations where we need to make decisions based on uncertain information.

## 5. How can the Modified Monty Hall problem be applied in real life?

The Modified Monty Hall problem can be applied in real life situations where we need to make decisions based on uncertain information. For example, it can be used in risk assessment or decision-making in finance, business, or even everyday life. It teaches us to carefully consider all possible outcomes and their probabilities before making a decision, rather than relying on our intuition or initial choice.

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