Probability problems

[Subduction Zone]

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.

That "new information" does not change the original odds. It can't. What this information tells us is about the remaining two boxes. You are not making a new choice in that regards. Now you know that the one remaining one of Monty's is 2 out of 3 times the winner. The one box that you picked still has the same odds of being the winner. 1 out of 3. Switch.

 

[Subduction Zone]

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

It has been done. You are wrong. Do you need links? The math is clear and should not require "testing". The Mythbusters did an episode on this.


I would have thought that the example of 1,000 boxes would have made it clear.

 

[Subduction Zone]

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.

I disagree. When the original choice is made the odds are set. One simply does not know them One has to look at each population of choices as a whole. In the case of when you first chose a box your odds were 1 out of 1,000. No matter how many boxes Monty opens that contain duds do not change those odds. The idea of this puzzle is to mislead by implying that Monty showing a box changes the odds somehow. It can't. One is not starting all over again. Now it does not make nearly as much sense to switch if Monty opened up only one box out of his 999. If you switched your odds of winning would only marginally increase. But if by some miracle Monty did not know and managed to open up 998 duds in a row the odds that the last one would be a dud would be only 1 out of 999. The odds of the original choice are not affected by later actions.

 

[RestlessSoul]

It has been done. You are wrong. Do you need links? The math is clear and should not require "testing". The Mythbusters did an episode on this.


I would have thought that the example of 1,000 boxes would have made it clear.

Let’s put this another way. If I go into a casino, and place a bet on number 13 on the roulette table, my one off bet has a one in 36 chance of coming good. However, if I state my intention to place 360 consecutive bets on 13, my odds are much better than 35 to 1, right? Calculating them wouldn’t be that straightforward, but the chances move heavily in may favour, probably to something like 1/10. Maybe.

Anyway, I place my first 359 bets, and get nothing. Now what are the odds of the ball landing on 13 on my final spin? They’re 35/1, because the spin is completely random and unrelated to anything that’s gone before. What the odds were at the beginning of my session of at the table haven’t changed historically, but they are no longer relevant in this new scenario.

 

[Subduction Zone]

Let’s put this another way. If I go into a casino, and place a bet on number 13 on the roulette table, my one off bet has a one in 36 chance of coming good. However, if I state my intention to place 360 consecutive bets on 13, my odds are much better than 35 to 1, right? Calculating them wouldn’t be that straightforward, but the chances move heavily in may favour, probably to something like 1/10. Maybe.

Anyway, I place my first 359 bets, and get nothing. Now what are the odds of the ball landing on 13 on my final spin? They’re 35/1, because the spin is completely random and unrelated to anything that’s gone before. What the odds were at the beginning of my session of at the table haven’t changed historically, but they are no longer relevant in this new scenario.

No, that is a poor analogy. Each new spin resets the action. Your past wins or losses do not affect future wins or losses. With the boxes you make one choice. That action cannot be reset.


Once you make a choice with the boxes you have two populations. One population of two boxes. One population of one box. The odds are that the winner is still 2 out of 3 that it will be in the population of 2 boxes. Now one of the two boxes is revealed to you. That does not change the odds of the original choice. The odds are still the same that your original choice was 1 out of 3 a winner. The odds of the winner being in the other two boxes is still the same, but now you have the advantage of knowing that the winner is not in one of those two boxes, if it is there. The odds that it is there still remains the same.

 

[icehorse]

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.

Maybe this would help. Monte has more information than you do. He is NOT operating in the blind. For example, he does NOT have the option of revealing the 2nd door with the real prize.

 

[Subduction Zone]

Maybe this would help. Monte has more information than you do. He is NOT operating in the blind. For example, he does NOT have the option of revealing the 2nd door with the real prize.

Let's say that he did not know. That would only make a third of the cases trivial where switching would not make an difference. Monty's not knowing would only introduce a third possibility where it did not matter since we would know already that we lost. As set up we are only considering cases where Monty did not expose a goat so his personal knowledge does not even play a role.

 

[icehorse]

Let's say that he did not know. That would only make a third of the cases trivial where switching would not make an difference. Monty's not knowing would only introduce a third possibility where it did not matter since we would know already that we lost. As set up we are only considering cases where Monty did not expose a goat so his personal knowledge does not even play a role.

maybe I missed something - have you guys agreed to temporarily stray from the OP? if so, my bad.

 

[Subduction Zone]

maybe I missed something - have you guys agreed to temporarily stray from the OP? if so, my bad.

The Monty Hall problem was the second of two problems posted in the OP.

 

[Subduction Zone]

Stole this from elsewhere:

If you program a simulation of this in Python it becomes the Monty Python problem.

 

[It Aint Necessarily So]

I disagree. When the original choice is made the odds are set. One simply does not know them One has to look at each population of choices as a whole. In the case of when you first chose a box your odds were 1 out of 1,000. No matter how many boxes Monty opens that contain duds do not change those odds. The idea of this puzzle is to mislead by implying that Monty showing a box changes the odds somehow. It can't. One is not starting all over again. Now it does not make nearly as much sense to switch if Monty opened up only one box out of his 999. If you switched your odds of winning would only marginally increase. But if by some miracle Monty did not know and managed to open up 998 duds in a row the odds that the last one would be a dud would be only 1 out of 999. The odds of the original choice are not affected by later actions.

Hmmm.

Disagree.

There is a difference between Monty choosing at random and after looking. Only in the second case do the odds not change. Because if he can look, it's a sure thing that if he started with the big prize and showed you 998 losers, he still has it, whereas if he has been choosing at random, each reveal that is not the big prize changes the current odds.

What doesn't change is the original odds. Consider this: Monty has only one box left after revealing 998 empty ones, he shows it to you, and it's empty. Have your odds of winning changed then? Sure. Assuming that one of the boxes was the winner, you have it, sure thing, 100%, correct? When you started, your odds were 1/1000 = 0.1%, correct? Either the odds have been changing with each negative reveal, or the change only came with the last one, because somehow, we went from a 0.1% likelihood of holding the winner to 100%.

I say the odds have been changing with each reveal. If you disagree, but agree that they have gone from a 0.1% chance 999 reveals ago to 100% chance now, did the change occur incrementally throughout the process, or did the probability collapsed with just the last reveals? It happened somewhere.

What hasn't changed are the original odds. Yes, you won, and it was true all along that you began with 1 chance in 1000, even after you know you won. You can now say with 100% certainty after that 999th reveal that you had had only 1 chance in a 1000, and won. Only if Monty has been selecting empty boxes to show you that he knew were empty when he showed them does the odds remain 1 in 1000 for you when you are down to one box each.

It's my opinion that if you are down to one box each, and Monty has been looking, he still is 999 times more likely to be holding the the winner than you, but if his choices have been random like yours was when you chose it, you each have a box that started with 1 chance in 1000 of being the winner, but now, 998 reveals later, each of you have a 50/50 chance of winning, and neither of you know who chose the winner until after the 999th reveal.

Can you find a flaw in that argument? I can't.

 

[Subduction Zone]

Hmmm.

Disagree.

There is a difference between Monty choosing at random and after looking. Only in the second case do the odds not change. Because if he can look, it's a sure thing that if he started with the big prize and showed you 998 losers, he still has it, whereas if he has been choosing at random, each reveal that is not the big prize changes the current odds.

What doesn't change is the original odds. Consider this: Monty has only one box left after revealing 998 empty ones, he shows it to you, and it's empty. Have your odds of winning changed then? Sure. Assuming that one of the boxes was the winner, you have it, sure thing, 100%, correct? When you started, your odds were 1/1000 = 0.1%, correct? Either the odds have been changing with each negative reveal, or the change only came with the last one, because somehow, we went from a 0.1% likelihood of holding the winner to 100%.

I say the odds have been changing with each reveal. If you disagree, but agree that they have gone from a 0.1% chance 999 reveals ago to 100% chance now, did the change occur incrementally throughout the process, or did the probability collapsed with just the last reveals? It happened somewhere.

What hasn't changed are the original odds. Yes, you won, and it was true all along that you began with 1 chance in 1000, even after you know you won. You can now say with 100% certainty after that 999th reveal that you had had only 1 chance in a 1000, and won. Only if Monty has been selecting empty boxes to show you that he knew were empty when he showed them does the odds remain 1 in 1000 for you when you are down to one box each.

It's my opinion that if you are down to one box each, and Monty has been looking, he still is 999 times more likely to be holding the the winner than you, but if his choices have been random like yours was when you chose it, you each have a box that started with 1 chance in 1000 of being the winner, but now, 998 reveals later, each of you have a 50/50 chance of winning, and neither of you know who chose the winner until after the 999th reveal.

Can you find a flaw in that argument? I can't.

Yes, Monty's looking does not change the odds of the original choice. That is when the odds were set. If he has no clue too all that happens is that a third of the time that he will accidentally reveal a car. But we do not count those games due to the set up. As a result it will look as if Monty knew since we ignore the times that he made a switch pointless. Make a tree of all possibilities, but then ignore the tree where he reveals a car. The resulting trees will be identical to the ones where he knew and avoided the car on purpose.

Since the opening conditions state that Monty did not reveal a car that gives us a "do over" situation in the ones that he does.

Once again it is easier to see with larger numbers. With one thousand doors when one chooses one's odds of winning are One out of a Thousand. Monty opens up on the remaining 999 doors. How would that change the odds of your original choice? All that it means is that if Monty did not know his odds of getting through 998 doors, assuming it was in the group, is almost zero.

This puzzle can be confusing because it is so easy to forget that when one made one's original choice that it set the odds of having the right door, if one does not change, in stone. Revealing doors where the car does not exist does not change that choice.

 

[RestlessSoul]

Maybe this would help. Monte has more information than you do. He is NOT operating in the blind. For example, he does NOT have the option of revealing the 2nd door with the real prize.

And this makes all the difference. In the 999 box scenario, if Monte is making informed decisions then the process is deterministic - the past decisions have a bearing on the future. But if he’s playing blind, every decision is random, so the past has no bearing on the future. The last pick effectively happens in causal isolation from preceding events.

Either way, the jump from 3 boxes to 999 unbalances the calculation to such an extent that the illustration doesn’t work; there is no equivalence. In the three box scenario, the player is faced with a choice of one from two. One box contains a prize, the other doesn’t. That’s a 50/50 chance and remains so whatever Monte based his one previous decision on.

 

[Subduction Zone]

And this makes all the difference. In the 999 box scenario, if Monte is making informed decisions then the process is deterministic - the past decisions have a bearing on the future. But if he’s playing blind, every decision is random, so the past has no bearing on the future. The last pick effectively happens in causal isolation from preceding events.

Either way, the jump from 3 boxes to 999 unbalances the calculation to such an extent that the illustration doesn’t work; there is no equivalence. In the three box scenario, the player is faced with a choice of one from two. One box contains a prize, the other doesn’t. That’s a 50/50 chance and remains so whatever Monte based his one previous decision on.

That is incorrect. Think of them as populations. Monty revealing one of his doors does not affect the odds of your original choice. For now let's assume that Monty knows where the car is. He has two doors to choose from so he can always show you a goat. He still has a two out of three chance of having the car behind his two doors. You still have only a one in three chance of having a goat behind your one door.

 

[RestlessSoul]

That is incorrect. Think of them as populations. Monty revealing one of his doors does not affect the odds of your original choice. For now let's assume that Monty knows where the car is. He has two doors to choose from so he can always show you a goat. He still has a two out of three chance of having the car behind his two doors. You still have only a one in three chance of having a goat behind your one door.

Hm. The original odds remain unchanged, sure. But they no longer apply if we’re playing blind; he’s just thrown away his advantage, so new odds apply.

If, however, he made an informed decision, then yes - the odds remain in his favour, and I should swap.

Okay, got it now.

 

[ratiocinator]

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.

 

[Subduction Zone]

Hm. The original odds remain unchanged, sure. But they no longer apply if we’re playing blind; he’s just thrown away his advantage, so new odds apply.

If, however, he made an informed decision, then yes - the odds remain in his favour, and I should swap.

Okay, got it now.

One more note. The OP messed up. It is Monty, as in Python, Hall. Not Monte Hall. Didn't anyone watch the show?

 

[RestlessSoul]

One more note. The OP messed up. It is Monty, as in Python, Hall. Not Monte Hall. Didn't anyone watch the show?

No, I never watched the show.

I obsessively watched the first season of a U.K. show called Deal or No Deal, with Noel Edmunds, which may have been similar. Calculating or even estimating the shifting odds in that game was fiendishly difficult. Everyone was playing blind btw.

 

[Subduction Zone]

No, I never watched the show.

I obsessively watched the first season of a U.K. show called Deal or No Deal, with Noel Edmunds, which may have been similar. Calculating or even estimating the shifting odds in that game was fiendishly difficult. Everyone was playing blind btw.


View attachment 57634

We had the same show with Howie Mandel. I was wondering if we copied you or vice versa. It turns out to be neither. Though your version aired a couple of months before ours did:

Deal or No Deal - Wikipedia

 

[Mock Turtle]

In the Deal or No Deal scenario, it's surprising the numbers who are affected by what any previous holder had in their box - as if they didn't recognise it was a new game and so new odds apply, as for each new game, and being the same. Who holds what is irrelevant if the boxes are truly randomised. But then perhaps it was staged to some extent?