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As if algebra or calc was common sense. I did learn about the Monty Hall Paradox but I'm still doubting it

novembermike 10y ago
The Monty Hall problem is about being given two sets of information. First they give you a 33% shot, then they give you a 50% shot. Take the 50% shot.
- posterboy 10y ago
  No, the second chance is supposed to be better than a 0.5 and I could outline the argument, but I don't trust it.
  - retsibsi 10y ago
    The key is that Monty Hall is guaranteed to show you a losing door. If that weren't the case -- if he just picked one of the remaining doors at random -- then the fact that he happened to pick a losing door would count as evidence that your door is a winner, leaving you with a 50/50 shot. I think that's one source of intuitive discomfort, and even of lasting confusion if the scenario isn't specified precisely.
- hilop 10y ago
  That's not true. It's a 66% shot. It has to be, because there only two options, and (we agree) one of them was 33%, so they other must 66%.
- hilop 10y ago
  That's not true. It's a 66% shot. It has to be, because there only two options, and (we agree) one of them was 33%, so they other must 66%.
montyhallstuff 10y ago
Consider 3 doors: A, B, C. If you pick the right one you win the prize. Say you pick door A. Monty Hall opens B or C and reveals that there's nothing behind. Now he asks if you want to stay with A or switch to the closed door. What should you do to maximize your chances of winning? Right before any of the doors is opened we have no idea what to expect, so there are 3 equaly likely possibilities: 1. prize is behind A 2. prize is behind B 3. prize is behind C. In case (1) you win by staying with your choice of A. In case (2) Monty Hall must open C (by the rules), so you'll win if you switch to B and in (3) the host will open B(by the rules) and you should switch to C. In 2 cases out of 3 (2/3) you win by switching and in only one case out of 3 (1/3) you win by staying with your first choice. Since 2/3 > 1/3, you should switch. This problem is easy if you know the defintions of sample space, event and the Equally Likely Probability Formula.
- russellspitzer 10y ago
  I prefer thinking about the 100 doors case. You are shown 100 doors 1 ... 100. Imagine you pick 1 door. Of the other 99 doors Monty tells you which 98 of them don't have the prize. He now asks if you want to switch to the 99th door that he didn't cancel out. Intuitively that seems to make much more sense to me.
  - posterboy 10y ago
    Yes, indeed it did for me, as well, but then I'm not trusting my intuition that failed me for the three door version, and I'm desperately trying to prove it wrong, because I'm biased by my first commitment to an answer. I think that commitment bias or what it's called is also an explanation for the misunderstanding in the first place. The argument against the increased chance, as I would use, works both ways and that is much more apparent the smaller the initial chance is. But I only intuitively used it to defend my position.
    - montyhallstuff 10y ago
      Just in case.... A sample space is the set of all possible outcomes of a random experiment. An event is a subset of a sample space. Let S be the finite sample space with equally likely outcomes in it. Let E be an event. Then the probability of E is the number of outcomes in E over the total number of outcomes in S. Now consider 52 cards in a deck. The sample space of outcomes here are the 52 cards. What is the event that the chosen card is a black face card? E = {J-club,Q-club,K-club,J-spade,Q-spade,K-spade}. What is the probability that the chosen card is a black face card? 6/52 by the formula above. Monty Hall problem can be solved by using the above as a guide. Sample space here : {switch, switch, stay-with-your-original-choice}. By the formula above, the probability of switching is 2/3 and the probability of staying put is 1/3. If you're doubting the formula, I can provide you with a proof.
      - retsibsi 10y ago
        As I mentioned in another comment, it's important to note that Monty Hall is guaranteed to open a door with a goat behind it (a losing door). If he chose randomly and just happened to reveal a goat, the answer would be different.
inimino 10y ago
I think there is an easy and intuitive explanation: Two thirds of the time, you pick the wrong door on your first choice. So far, so obvious. Now, you picked a wrong door, Monty opens the other wrong door, and there are only three doors. Obviously the remaining door has the prize. Every time you pick the wrong door, and then switch, you win. That means two thirds of time, if you switch, you win. That makes switching the right strategy.

The Monty Hall problem is about being given two sets of information. First they give you a 33% shot, then they give you a 50% shot. Take the 50% shot.

posterboy 10y ago
No, the second chance is supposed to be better than a 0.5 and I could outline the argument, but I don't trust it.
- retsibsi 10y ago
  The key is that Monty Hall is guaranteed to show you a losing door. If that weren't the case -- if he just picked one of the remaining doors at random -- then the fact that he happened to pick a losing door would count as evidence that your door is a winner, leaving you with a 50/50 shot. I think that's one source of intuitive discomfort, and even of lasting confusion if the scenario isn't specified precisely.
hilop 10y ago
That's not true. It's a 66% shot. It has to be, because there only two options, and (we agree) one of them was 33%, so they other must 66%.
hilop 10y ago
That's not true. It's a 66% shot. It has to be, because there only two options, and (we agree) one of them was 33%, so they other must 66%.

montyhallstuff 10y ago

Consider 3 doors: A, B, C. If you pick the right one you win the prize. Say you pick door A. Monty Hall opens B or C and reveals that there's nothing behind. Now he asks if you want to stay with A or switch to the closed door. What should you do to maximize your chances of winning? Right before any of the doors is opened we have no idea what to expect, so there are 3 equaly likely possibilities: 1. prize is behind A 2. prize is behind B 3. prize is behind C. In case (1) you win by staying with your choice of A. In case (2) Monty Hall must open C (by the rules), so you'll win if you switch to B and in (3) the host will open B(by the rules) and you should switch to C. In 2 cases out of 3 (2/3) you win by switching and in only one case out of 3 (1/3) you win by staying with your first choice. Since 2/3 > 1/3, you should switch. This problem is easy if you know the defintions of sample space, event and the Equally Likely Probability Formula.

russellspitzer 10y ago
I prefer thinking about the 100 doors case. You are shown 100 doors 1 ... 100. Imagine you pick 1 door. Of the other 99 doors Monty tells you which 98 of them don't have the prize. He now asks if you want to switch to the 99th door that he didn't cancel out. Intuitively that seems to make much more sense to me.
- posterboy 10y ago
  Yes, indeed it did for me, as well, but then I'm not trusting my intuition that failed me for the three door version, and I'm desperately trying to prove it wrong, because I'm biased by my first commitment to an answer. I think that commitment bias or what it's called is also an explanation for the misunderstanding in the first place. The argument against the increased chance, as I would use, works both ways and that is much more apparent the smaller the initial chance is. But I only intuitively used it to defend my position.
  - montyhallstuff 10y ago
    Just in case.... A sample space is the set of all possible outcomes of a random experiment. An event is a subset of a sample space. Let S be the finite sample space with equally likely outcomes in it. Let E be an event. Then the probability of E is the number of outcomes in E over the total number of outcomes in S. Now consider 52 cards in a deck. The sample space of outcomes here are the 52 cards. What is the event that the chosen card is a black face card? E = {J-club,Q-club,K-club,J-spade,Q-spade,K-spade}. What is the probability that the chosen card is a black face card? 6/52 by the formula above. Monty Hall problem can be solved by using the above as a guide. Sample space here : {switch, switch, stay-with-your-original-choice}. By the formula above, the probability of switching is 2/3 and the probability of staying put is 1/3. If you're doubting the formula, I can provide you with a proof.
    - retsibsi 10y ago
      As I mentioned in another comment, it's important to note that Monty Hall is guaranteed to open a door with a goat behind it (a losing door). If he chose randomly and just happened to reveal a goat, the answer would be different.

inimino 10y ago

I think there is an easy and intuitive explanation: Two thirds of the time, you pick the wrong door on your first choice. So far, so obvious. Now, you picked a wrong door, Monty opens the other wrong door, and there are only three doors. Obviously the remaining door has the prize. Every time you pick the wrong door, and then switch, you win. That means two thirds of time, if you switch, you win. That makes switching the right strategy.