The scams of Statistics...

[Scott Mayers]

You don't need to experiment. I don't need to toss a coin a billion times to prove that on any one toss, it is always 1/2. This is the fallacy that gamblers think when they continue to purchase lotto tickets. The odds are reset each time one purchases a ticket for a each new draw. The only thing that applies is that by not playing, you don't have any chance at all to even play again. You can prove by practice that if you guessed heads should show up, that by continuing to play, you should be certain to eventually win. But if these wins require an investment for each draw, the average winnings would still reduce to 50% over the long run.

My argument above is perfectly logical and doesn't miss any interpretation upon the original puzzle as was given. It is presented in a way that only appears to make one think they have an advantage to switch. But if taken literally, it would imply being able to get something better for nothing in any game.

[dionisos]

You don't need to experiment. I don't need to toss a coin a billion times to prove that on any one toss, it is always 1/2. This is the fallacy that gamblers think when they continue to purchase lotto tickets. The odds are reset each time one purchases a ticket for a each new draw.

No, it is not the same thing at all, i know that the odds are reset each time we toss the coin.
Yes we don’t need experiment to know the odds, i should be able to convince you you are wrong, by logical arguments only.
The problem is that i don’t think you would try enough hardly, if you think you are right, you will not try hard enough to see your mistakes, then i thought that if we toss the coin many time, and you see it give head about 1995 time, and tail 1005, then you will reconsider your position, for real, and you will seriously search where you could have done a mistake.

My argument above is perfectly logical and doesn't miss any interpretation upon the original puzzle as was given. It is presented in a way that only appears to make one think they have an advantage to switch. But if taken literally, it would imply being able to get something better for nothing in any game.

But no, your argument is not perfectly logical, it is not foolish, but you make steps that you should not, and that break all the logical construction.
And these steps are pretty subtle, and i don’t think you will find and understand them, if you don’t look seriously where your argument is weak.
And the worst, is that you don’t seem to understand what i say (i think it is mostly my fault), and then you use you cognitive abilities to try to make me understand things i already understand, and prove false, things i never said.

I think, we could do it the hard way:
The first hard way, is to choose a protocol, make the experience many time, and see how many time each event happen, then we would empirically deduce a approximation of the odds, and it would give a good clue of who make a deductive mistake.
The second hard way, is to explicit the logical rules we use to make our conclusions, discuss this rules, and then make the conclusion using only this rules, in a true mathematical form.

I personally prefer the first way, because it is more simple, more fast.

[Scott Mayers]

Sorry, but if you follow my interpretation, it is clear-cut. What you'd have to do is to appeal to why you think I may have misunderstood the problem. And you are also making an error to assume that the statistics are sufficient to hint at any problem. You can actually coincidentally toss a coin and obtain a 'heads' one hundred percent of the time for just as many tosses. This, however, would only hint that there may be something wrong with the particular coin or to the one throwing it, not with the logic of a 'fair' coin under the given ideal assumptions.

I demonstrated that the original 'trick' of the puzzle was to dismiss the distinction between the two goats as distinct options. If you want, change them all to cars but have each one with a different color, like red, green, and blue, and then only aim to get, say, the red one. That is, assume that they are all different but only change one feature to which you find desirable. Then you'd see how the odds play out. The fact that the goats just happen to be representing the same kind of thing no one prefers to have, doesn't mean we can count all losses as equivalent.

[mickthinks]

Scott, I think dionisos gave a good indication of where he thinks you have misunderstood the problem, here:

The host is basically restarting the game between two door options, not three, when he offers you to switch.

Yes, but two doors with not equal chance to have a car behind

If you want, change them all to cars but have each one with a different color, like red, green, and blue, and then only aim to get, say, the red one. That is, assume that they are all different but only change one feature to which you find desirable.

I agree that this would be an equivalent game, and it has the same result: you would double your chance of winning the red car by "switching" after Monty Hall opens one of the doors to show a non-red car.

[Scott Mayers]

Scott, I think dionisos gave a good indication of where he thinks you have misunderstood the problem, here:

The host is basically restarting the game between two door options, not three, when he offers you to switch.

Yes, but two doors with not equal chance to have a car behind

If you want, change them all to cars but have each one with a different color, like red, green, and blue, and then only aim to get, say, the red one. That is, assume that they are all different but only change one feature to which you find desirable.

I agree that this would be an equivalent game, and it has the same result: you would double your chance of winning the red car by "switching" after Monty Hall opens one of the doors to show a non-red car.

No it wouldn't. Monty Hall is AWARE of which door contains the car. As such he also knows implicitly the doors containing the goats and always show ONLY the goats, never the car. However, the error by the authors of this missed the point that you have to count the options where both goats are available for him to reveal. But it gets counted illegitimately as one possibility and not two. As such there are two possible ways which one could switch and reveal a goat, not one!

If the first case is held, the left over possibilities are

1)Goat 1...Goat 2
2)Car.......Goat 1
3)Goat 2...Car

Then when the host, knowing which one has the car opts in case (1) to either pick Goat 1 to show to which leaves Goat 2 if he switches, or to pick Goat 2 to show but leaves Goat 1. These are two possibilities that have to go into the solution which is ignored by the solution.

In case (2), the host can ONLY pick Goat 1, not the car, to show and in case (3), the host must pick Goat 2, for the same reason. These are all four possibilities accounted for. The error was that the solution only picked case (1) as one possibility. I'd like to use a strike out to indicate this but this site disables the extra features of this software here to do it.

What is your misunderstanding?

[dionisos]

Car blue, car red, car green.(BRG)
There are six possibilities:
BRG
BGR
RBG
RGB
GBR
GRB

Now you choose the first door:
BRG
BGR
RBG
RGB
GBR
GRB
Here the probability to have the blue car is 1/3, because you don’t know in what possibility you are in.
The frequency in how many time you will get a blue car by doing it (choosing the first door), is also 1/3.

Now the host open randomly one green or red door:
BRG or BRG
BGR or BGR
RBG
RGB
GBR
GRB

Your mistake is to think that: P(BRG) = P(RBG)
This is false, the true relation is P(BRG or BRG) = P(RBG)
You do the mistake because you think about all cases, and believe all these case are equiprobable. (and then you think i miss some case, when i don’t)

And this is why i propose to you to stop thinking about probability, but to think directly about frequencies:
1) BRG 2 time in 12
2) BGR 2 time in 12
3) RBG 2 time in 12
4) RGB 2 time in 12
5) GBR 2 time in 12
6) GRB 2 time in 12

Then when you divide the case 1) in two sub-case, the case 1) still happen 2 time in 12, and if you divide this case in 2 equiprobable sub-cases, each of these sub-case happen 1 time in 12. (here you could see why i choose 12, because it is convenient, i could have chosen 100, or even 1, but it would end up with decimal numbers and confuse the argument)
Then P(BRG) = P(RBG) / 2

Now i propose to you to try to demonstrate that "P(BRG) = P(RBG)" if you are still not convinced. You will see that you fall at it, or that you will need to take it like granted that when you divide the problem in N independent cases, all these cases will be equiprobable, what is false.

Imagine i roll a dice, i have these equiprobable cases to start with:
d1 (i get a 1 on the dice)
d2
d3
d4
d5
d6
Now if get 1, i will toss a coin:
(d1 and head) or (d1 and tail)
d2
d3
d4
d5
d6
You could see i have now seven case, but P((d1 and head) or (d1 and tail)) = P(d2), and P(d1 and head) = P(d2) / 2

Now i hope, i really hope you see your mistake.
Forget the things i said to you about making a experiment, it was to show you the empirical data seem to contradict you, and then you would rethink seriously your reasoning, but it only seem to confuse you, and to make you believe i think empirical data constitute proof, or other crazy things, like thinking that i believe the different experiment aren't independent.

[HexHammer]

Monty Hall

Never believed in switching would do any difference. I can't make a 100% increased chance to win just by switching.

Just because the host allows me to switch, what difference woudl that do if I allowed myself to switch, before I made my final desition? ..NOTHING!!!

[dionisos]

Monty Hall

Never believed in switching would do any difference. I can't make a 100% increased chance to win just by switching.

Just because the host allows me to switch, what difference woudl that do if I allowed myself to switch, before I made my final desition? ..NOTHING!!!

Imagine there are 99 goats and 1 car.

You choose one door, then i reveal to you 98 goats, do you think that by doing it, i increase your probability to get a car, from 1% to 50% ?

Then when you are in a game like it, just ask the host to first open the doors with a goat when he will show you if you win or lose, he will have to open at least 98 doors with a goat, and by doing it, your probability to get the car will increase to 50%. (i hope you see the nonsense)

[HexHammer]

Monty Hall

Never believed in switching would do any difference. I can't make a 100% increased chance to win just by switching.

Just because the host allows me to switch, what difference woudl that do if I allowed myself to switch, before I made my final desition? ..NOTHING!!!

Imagine there are 99 goats and 1 car.

You choose one door, then i reveal to you 98 goats, do you think that by doing it, i increase your probability to get a car, from 1% to 50% ?

Then when you are in a game like it, just ask the host to first open the doors with a goat when he will show you if you win or lose, he will have to open at least 98 doors with a goat, and by doing it, your probability to get the car will increase to 50%. (i hope you see the nonsense)

You are speaking nonsense. You don't understand how the mechanics of the problem works.

[dionisos]

On the contrary i understand it pretty well.

[Scott Mayers]

Now the host open randomly one green or red door:
BRG or BRG
BGR or BGR
RBG
RGB
GBR
GRB

Your mistake is to think that: P(BRG) = P(RBG)
This is false, the true relation is P(BRG or BRG) = P(RBG)
You do the mistake because you think about all cases, and believe all these case are equiprobable. (and then you think i miss some case, when i don’t)

No, what you did above only adds force to what I said by adding the probabilities for the variation of what's behind the doors. This is alright. It just doubles the possibilities including which doors they are behind. Can you NOT count? In the above, if you count what the odds are for any color, they are all distributed evenly and equal 1/2. Just count the underlined term for instances of a B, for instance. you will count four out of a total of 8 which is 4/8 = 1/2! Do this for any color and either for a Switch or a Stay. They always equal 1/2.

The reason I asked you to do this is to show you how by interpreting each thing as distinct, you can see that any selection is arbitrary and fairly distributed. You cannot think you can get something 'free' (some presumed benefit of Switching as 2/3) for nothing. And the equivocation error I mentioned that YOU made was NOT to include at least the 'or'ed options.

I can't follow your own later argument further than this as you introduce the whole set of odds as 12. Where do you get this number? If you count the above, there are 8 maximum possibilities (double the 4 I used for simplicity without concerning probabilities of doors but only results).

[vegetariantaxidermy]

The thing about the Monty Hall paradox is that pretty well no one ever changes doors, so there is no way of ever testing it accurately. You can do all the mathematical equations and computer simulations you want, but it's something we can never know for certain.

[dionisos]

Do you think P(BRG) = P(RBG) ?
Or do you agree P(BRG or BRG) = P(RBG)

You could replace "2 time in 12" by "1 time in 6" or by "1.333 time in 8", or by "16.666 time in 100", this have no importance.

[vegetariantaxidermy]

Clearly this is a nerd-only thread.

[dionisos]

Ok, i propose anybody that think changing the door don’t improve probability, to play a similar game with me:
You take a card game, the ace of hearts is the car, all the other 31 cards the goats.

take one card, then ask somebody to remove 30 no-ace of hearts cards (goats).
You stay with the same card, i change it for the remaining card.

if i get the ace of hearts, i earn 100€, if you get it, you earn 200€.

We play this game at least 10 time, if you think staying or changing don’t change the probability to have the ace of hearts, you could see that it is lot of free money for you