Probability problems

[It Aint Necessarily So]

[1] Imagine the 2x2 grid below to contain the following figures, one each in each square: 0%, 25%, 25%, and 50%. If you pick one at random by throwing a dart at the grid (assume that you are not good enough to direct the dart to any of the choices) and hit one of the squares, what is the chance you will be correct?

[2] This one is called the Monte Hall problem, and it vexes many. Perhaps you're familiar with the game show Let's Make A Deal. At the end, one contestant gets tot choose one of three prizes behind a curtain, door number 1, door number 2, or door number 3. One has a grand prize of great value, and the other two are worth much less. You pick a door, leaving two for Monte. He opens the curtain of one of them, and it is one of the two duds. He then asks you if you would like to trade your your previous choice for the unseen one he still has. What do you do, trade or not?

For those who like problems like these, it might behoove you to try to answer the questions before looking at the discussions of them below. @Polymath257 's input solicited.

 

[It Aint Necessarily So]

Spoiler: first problem discussed

I say that there is no correct answer. Whatever square you hit can't be right. If you hit the 50% square, that can't be right. You only had one chance in four, 25%, to hit that square. If you hit one of the two 25%s, you're wrong again. You had a 50% chance (2 in 4) of getting one of those. In the form of this problem I encountered on the Internet, the fourth choice was 60%. Some posters answered that the chance of being correct was 0%, so I made that the fourth choice. Now, 0% can't be correct, either, since you had a 25% chance of landing in that square.

Spoiler: second problem discussed

This one has a correct answer: you trade every time. At first blush, it may seem that the odds of having chosen the correct door changed from 1 in 3 to 1 in 2 when a dud was shown, but that's incorrect, it's still in 1 in 3 that you chose correctly, and still 2 chances in 3 that Monte began with it. That also hasn't changed just because he showed a dud. He always has one or two duds whether you guessed correctly or not, and showing one to you doesn't change the odds for your door (his doors have gone from 1 chance in 3 each before he revealed one to 0 in 3 for the revealed door and 2 in 3 fr the other).

If this is difficult to see, imagine that I have a bag of socks, 19 green and 1 red, and you get to pick one blindfolded to win a prize if you pick the red one. You pick one, and it is removed from view before you see it. I look into the bag and select 18 of my 19 socks to show you, and they're all green. Should you trade your unseen sock for mine? Yes, as quickly as possible. It was always true that I was 19 times more likely to have the red sock that you, and it was always true that I had 18 or 19 green socks and could show you 18 whether you picked correctly or not. Nothing changes when I show you 18 green socks. My sock is 19 times more likely to be red than yours, so you should trade.

 

[viole]

[1] Imagine the 2x2 grid below to contain the following figures, one each in each square: 0%, 25%, 25%, and 50%. If you pick one at random by throwing a dart at the grid (assume that you are not good enough to direct the dart to any of the choices) and hit one of the squares, what is the chance you will be correct?
View attachment 57608

[2] This one is called the Monte Hall problem, and it vexes many. Perhaps you're familiar with the game show Let's Make A Deal. At the end, one contestant gets tot choose one of three prizes behind a curtain, door number 1, door number 2, or door number 3. One has a grand prize of great value, and the other two are worth much less. You pick a door, leaving two for Monte. He opens the curtain of one of them, and it is one of the two duds. He then asks you if you would like to trade your your previous choice for the unseen one he still has. What do you do, trade or not?

For those who like problems like these, it might behoove you to try to answer the questions before looking at the discussions of them below. @Polymath257 's input solicited.

Increase the number of cells in your problem, to say 1 thousands, and you realise immediately why it is worthy to accept to change.

Ciao

- viole

 

[RestlessSoul]

With the first one, you have a 50% chance of landing in a square showing 25%, which is the probability of hitting any given square. So 50% chance of being right.

With the second puzzle I can’t see any reason to trade; you now have a 50% chance either way.

 

[RestlessSoul]

Regarding the 2nd puzzle, the odds have moved in your favour from 2/1 against you, to even money each of two. You are now facing a second contingency, and whichever box you choose, you have a 50/50 chance of being right.

 

[icehorse]

The first one is oddly worded, but I'd say that overall you have a 4/16 or 1/4 chance.

The Monty Hall problem is a classic !!

 

[It Aint Necessarily So]

Increase the number of cells in your problem, to say 1 thousands, and you realise immediately why it is worthy to accept to change.

Are you conflating the two problems? Cells sounds like the array, problem [1], and increasing to thousands is similar to what I did when I increased to 20 socks, but that was problem [2]. Agree, though that increasing the number of options in problem [2] helps reveal that the odds don't change when the person revealing doors or socks isn't choosing them at random.

With the first one, you have a 50% chance of landing in a square showing 25%, which is the probability of hitting any given square. So 50% chance of being right.

Remember, you are not being asked to calculate odds, just to hit a square with a number that represents your chance of getting that number. If the square had one 25% rather than two, and you hit it, you would have correctly (albeit randomly) hit an answer that was correct. You hit 25%, and you had a 25% chance of doing so. If it had two 50%s and you hit one, once again, your dart hit a number that tells you the chance of hitting that number.

Look at your comment again. If I have a fifty percent chance of hitting a 25% square, then if I do, I have not randomly selected the number that corresponds to my chances of hitting that number.

With the second puzzle I can’t see any reason to trade; you now have a 50% chance either way.

That would be correct if Monty had shown you one of his doors without knowing that it was a dud. At that point, the two remaining doors go from 1 chance of 3 of being the big prize to 1 in 2, and Monte's chance of having the big prize goes from 2 chances in 3 to 1 in 2.

But when Monte is known to have one dud if he has the grand prize, too, and he gets to look at his two doors and show you a dud, your chances of having been correct don't change - still 1 in 3. Nor does his chance of having been given the grand prize when you didn't select it change. It just goes from 1 in three for each door he has to 1 in 3 for both doors, one of them having no chance of being correct once revealed to be a dud, 0 in 3 for one and 2 in 3 for the other.

If you still don't see it, ask yourself whether you consider it more likely that I am wrong, that you just don't see that I am correct, or no way to tell? Is there any evidence to help make that assessment?

The first one is oddly worded, but I'd say that overall you have a 4/16 or 1/4 chance.

Then you disagree with the analysis provided? If so, what part? I say that there is no way to randomly hit a number that is the chance of having hit that number.

 

[RestlessSoul]

Okay, I concede on the first one; there’s no outcome that doesn’t contradict itself.

As for puzzle 2 though, if I’m playing blind and Monte isn’t, that certainly shifts the odds against me before I make my first pick. But once I make my pick, and he reveals his empty box, I am now faced with a second decision. And the chance of me getting that decision right is clearly 50/50. It makes no odds if he has information I don’t, because he has no part in my decision.

The odds at the beginning of the game are unrelated to the odds against my final decision. I no longer have a 1 in 3 chance of getting the box with the prize, and any calculation involving those historic odds bears no relation to the current situation.

 

[Subduction Zone]

Regarding the 2nd puzzle, the odds have moved in your favour from 2/1 against you, to even money each of two. You are now facing a second contingency, and whichever box you choose, you have a 50/50 chance of being right.

That is incorrect. You are forgetting that Monty had to know where the winner was from the start. This is a mistake that most people make when looking at this puzzle. Your initial odds do not change due to Monty showing you a dud. From the start your odds were one out of three of winning. Of the remaining two the odds were two out of three that the winner is in there. No matter what at least one of Monty's doors or boxes will be a dud. When he shows you that you know that the odds of it being a winner are two out of three.

You switch.

 

[It Aint Necessarily So]

if I’m playing blind and Monte isn’t, that certainly shifts the odds against me before I make my first pick.

I have to disagree. It doesn't matter if Monty is even in the studio when you choose, or whether he knows which door has the big prize until he has to decide which of his two doors to show you. At that time he needs to know to be certain that if he has the big prize, that he doesn't show it.

Imagine what that would be like - he's revealed the big prize, and only duds are left. You have one and so does Monty. Now it really doesn't matter whether you keep your door or trade.

But once I make my pick, and he reveals his empty box, I am now faced with a second decision. And the chance of me getting that decision right is clearly 50/50. It makes no odds if he has information I don’t, because he has no part in my decision.

Disagree, but you already know that, and you know why. You might try addressing that answer rather than just repeating what it was written to rebut. Until you do, we're at an impasse.

Dialectic is the cooperative interaction of two skilled critical thinkers trying to resolve some difference of opinion. It happens when each addresses all of the points the other makes with either a yes, I agree, as you did in your last post, a part I didn't recopy, because that issue had come to resolution through dialectic. You had made a comment, I rebutted it convincingly to you, and you changed your position. Had you been right, you would have changed my mind. That's exactly what dialectic is.

We can do that with this problem as well, but you'd need to state specifically why you disagree that the odds don't change after the reveal, and not just say, I still think it's 50/50. Did you see the red and green sock metaphor? If so, when you were down to your small sealed bag with the one sock out of twenty that you chose, and the other guy was down to one sock after looking in the bag and removing 18 green ones, would you think that your chances had become 50/50, or do you still think that he was 19 times more likely to have the red sock after you drew one, but now it's 50/50, knowing that showing you 18 green socks changes nothing?

 

[Subduction Zone]

Okay, I concede on the first one; there’s no outcome that doesn’t contradict itself.

As for puzzle 2 though, if I’m playing blind and Monte isn’t, that certainly shifts the odds against me before I make my first pick. But once I make my pick, and he reveals his empty box, I am now faced with a second decision. And the chance of me getting that decision right is clearly 50/50. It makes no odds if he has information I don’t, because he has no part in my decision.

The odds at the beginning of the game are unrelated to the odds against my final decision. I no longer have a 1 in 3 chance of getting the box with the prize, and any calculation involving those historic odds bears no relation to the current situation.

Once again, not so. When you make your initial choice the odds of that do not change. The odds were always one out of three that you would be right. The odds are that no matter what the odds were two out of three that the prize is in one of the remaining two boxes. Those odds do not change. Monty just exposes a box so that you have a clear choice.

 

[Subduction Zone]

Here is an extreme example There are 1,000 boxes. Your chance of winning is one out of one thousand. Monty has 999 boxes left. The odds are 999 out of 1,000 that the prize is in one of those boxes. Now Monty opens 998 of them. They are all duds. Monty's opening up of boxes does not change the original odds. The odds are still that the one box you have has only a chance of 1 out of 1,000 of being right. The one box that Monty has left has the odds of 999 out of 1,000 of being the prize.


This holds true even if Monty did not know where the prize was. Monty exposing a box does not change the original odds.

 

[icehorse]

Then you disagree with the analysis provided? If so, what part? I say that there is no way to randomly hit a number that is the chance of having hit that number.

Honestly, I found the problem so oddly worded that I wasn't sure the goal. I *guessed* that the goal had to do with the overall odds across all four squares. But I wasn't even sure of that guess

 

[It Aint Necessarily So]

Here is an extreme example There are 1,000 boxes. Your chance of winning is one out of one thousand. Monty has 999 boxes left. The odds are 999 out of 1,000 that the prize is in one of those boxes. Now Monty opens 998 of them. They are all duds. Monty's opening up of boxes does not change the original odds. The odds are still that the one box you have has only a chance of 1 out of 1,000 of being right. The one box that Monty has left has the odds of 999 out of 1,000 of being the prize. This holds true even if Monty did not know where the prize was. Monty exposing a box does not change the original odds.

Agree, provided Monty knows before revealing anything that it is not the big prize. I presume that you would agree that if Monty is revealing boxes without knowing if they are the big prize or not - if he is choosing them at random - then the odds for each player shift with each dud reveal, the chooser going up from 1 in 1000 to 1 in 999 with the first one, and Monty going down from 999/1000 to 998/999. 997 random dud reveals later, they each have a 50/50 chance. But if Monty always knows he will be revealing a dud, then 998 dud reveals later, it is still 999 times more likely that the holds the winner.

Honestly, I found the problem so oddly worded that I wasn't sure the goal. I *guessed* that the goal had to do with the overall odds across all four squares. But I wasn't even sure of that guess

The question is simply, what are the odds of your dart landing in a square that gives the odds of that numerical result? That cannot happen. You have a 25% chance of hitting 0%, a 25% chance of hitting 50%, and a 50% chance of hitting 25%. What's interesting is that if you answer that it is impossible by saying 0%, you might think, well, that could have come up, but it doesn't help. You have a 25% chance of that coming up, not a 0% chance.

 

[icehorse]

The question is simply, what are the odds of your dart landing in a square that gives the odds of that numerical result? That cannot happen. You have a 25% chance of hitting 0%, a 25% chance of hitting 50%, and a 50% chance of hitting 25%. What's interesting is that if you answer that it is impossible by saying 0%, you might think, well, that could have come up, but it doesn't help. You have a 25% chance of that coming up, not a 0% chance.

Got it! The first sentence in your last post was much clearer!

 

[RestlessSoul]

Here is an extreme example There are 1,000 boxes. Your chance of winning is one out of one thousand. Monty has 999 boxes left. The odds are 999 out of 1,000 that the prize is in one of those boxes. Now Monty opens 998 of them. They are all duds. Monty's opening up of boxes does not change the original odds. The odds are still that the one box you have has only a chance of 1 out of 1,000 of being right. The one box that Monty has left has the odds of 999 out of 1,000 of being the prize.


This holds true even if Monty did not know where the prize was. Monty exposing a box does not change the original odds.

It doesn’t change the original odds, but the original odds have no bearing on the outcome of the second decision, which the player takes independently of the first one. The original odds have no relevance to the new situation.

 

[RestlessSoul]

Once again, not so. When you make your initial choice the odds of that do not change. The odds were always one out of three that you would be right. The odds are that no matter what the odds were two out of three that the prize is in one of the remaining two boxes. Those odds do not change. Monty just exposes a box so that you have a clear choice.

Not sure about that. The ability to calculate probabilities accurately, depends in part on the information available when the calculation is made. Once Monte reveals the contents of one box, new information is available.

If one horse is scratched from a race before the off, the bookies adjust the market, at least in the U.K. they do. And if you are betting in-running on the Grand National, and the favourite falls at Becher’s Brook, the market adjusts then too, thanks to modern technology.

Look, in the final scenario, there are two boxes, and two possible outcomes. This is a stand alone contingency, therefore the odds are even, as to which box contains the prize. The original odds no longer pertain.

 

[icehorse]

Not sure about that. The ability to calculate probabilities accurately, depends in part on the information available when the calculation is made. Once Monte reveals the contents of one box, new information is available.

If one horse is scratched from a race before the off, the bookies adjust the market, at least in the U.K. they do. And if you are betting in-running on the Grand National, and the favourite falls at Becher’s Brook, the market adjusts then too, thanks to modern technology.

Look, in the final scenario, there are two boxes, and two possible outcomes. This is a stand alone contingency, therefore the odds are even, as to which box contains the prize. The original odds no longer pertain.

The good news is that the fact you're struggling with this puts you in good company. But the OP is correct. If you're a computer programmer, you can simulate this fairly easily.

 

[RestlessSoul]

On the other hand, in the 1000 box scenario, it does make a difference if Monte was playing blind. If he knew what was in the boxes, then yeah, he hung on to the winning box and the odds were stacked against me heavily all along, so there is a case for saying I ought to change. I do get that. He had a winning hand from the off, so now that he has played it and I have a chance to swap, I do see why I should probably do so.

However, if every decision Monte and I took up to that point was taken blind, ie random, then the final decision is random also, and independent of anything that occurred before.

 

[RestlessSoul]

The good news is that the fact you're struggling with this puts you in good company. But the OP is correct. If you're a computer programmer, you can simulate this fairly easily.

I was going to say that there’s only one way to find out for sure who is right here, and that’s to run the scenario 1000 times and see how it plays out. Probably that’s been done, in which case probably I’m wrong. I’d like to see the evidence though, I think I’m entitled to ask for that