Probability problems

[The Sum of Awe]

The easiest way I've heard to explain the Monte Hall problem is that when you make your first choice, you have a 1/3 probability of being right and a 2/3 probability of being wrong. Once the other box is opened, if you switch, you are betting that your first choice was wrong.

I’m very confused about the Monte Hall problem.

Even with that explanation, even though you were 2/3 wrong on your initial choice, there was always going to be two duds of the three, why does it make a difference if one is exposed? It becomes 50/50 either way.

If Monte knows what is behind the curtains, he’ll reveal a dud regardless of whether you hold the big prize or not. Revealing a dud does not change the probability that you have either a dud or a prize, because your 1/3 choice was out of his control.

 

[ratiocinator]

I’m very confused about the Monte Hall problem.

Even with that explanation, even though you were 2/3 wrong on your initial choice, there was always going to be two duds of the three, why does it make a difference if one is exposed? It becomes 50/50 either way.

If Monte knows what is behind the curtains, he’ll reveal a dud regardless of whether you hold the big prize or not. Revealing a dud does not change the probability that you have either a dud or a prize, because your 1/3 choice was out of his control.

Been away from the forum a while, but Ill try again.

If your first choice is correct (1/3 probability), then Monte can open either door because they are both duds and if you switch, then you'll lose.

If your first choice is wrong (2/3 probability), then Monte has no choice about which other door to open, he has to open the other dud, thus telling you where the prize is. Now if you switch, you will undoubtedly win.

Yet another way to think about it is to imagine there are more than three options, say 100. You pick one and Monte opens 98 of the 99 remaining doors. Now what would you do?

 

[Subduction Zone]

Been away from the forum a while, but Ill try again.

If your first choice is correct (1/3 probability), then Monte can open either door because they are both duds and if you switch, then you'll lose.

If your first choice is wrong (2/3 probability), then Monte has no choice about which other door to open, he has to open the other dud, thus telling you where the prize is. Now if you switch, you will undoubtedly win.

Yet another way to think about it is to imagine there are more than three options, say 100. You pick one and Monte opens 98 of the 99 remaining doors. Now what would you do?

I don't know. I have a really good feeling about my door

 

[FredVB]

I don't see a reason to change the first choice. It does not become more likely to be the wrong one.

 

[ratiocinator]

I don't see a reason to change the first choice. It does not become more likely to be the wrong one.

That's not the point though. From the start it's more likely to be wrong (2/3) than right (1/3). The point is that if it was wrong (which is more likely: 2/3) and you change your choice, then you will definitely win.

 

[The Sum of Awe]

That's not the point though. From the start it's more likely to be wrong (2/3) than right (1/3). The point is that if it was wrong (which is more likely: 2/3) and you change your choice, then you will definitely win.

I understand it now.

 

[FredVB]

That's not the point though. From the start it's more likely to be wrong (2/3) than right (1/3). The point is that if it was wrong (which is more likely: 2/3) and you change your choice, then you will definitely win.

If ... then you will definitely win. There's the catch, that you are left thinking you get ahead with odds. There is no real getting ahead.

 

[Brickjectivity]

I am unable to understand the directions. For problem 1 are you looking for the expected value?

Spoiler: Expected value answer P1

I have 0.25 chance of hitting any square, so I use 0.25 as the frequency and take the average.
0.25 * (0 + 0.25 + 0.25 + 0.50) = 0.25