The scams of Statistics...

[Scott Mayers]

I think stats are useful and most of it is not as such scams but have discovered a few famous ones that are and which I want to see if others can determine it for themselves here.

1) The Monty Hall Problem.

2) The Secretary (or Optimal Stopping)Problem.

Is anyone here familiar with these? I'm guessing this might be an interesting one for PhilX here? I have been challenged these before and find it odd that others can't get that the supposed claims on them as true. I was disturbed that these could and are used in practice. As such the second one seems to actually 'work' if and where people actually believe in them!! Can you figure out the fallacies in them?

[Scott Mayers]

The Monty Hall problem as from the Wikipedia link is summarized as follows:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Vos Savant's response was that the contestant should switch to the other door (vos Savant 1990a). Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their choice have only a 1/3 chance.

The page takes the opinions of Vos Savant here by asserting while they believe this is counter-intuitive, it is certain and has been proven. I beg to differ and is why I challenge others to try it to see first if they agree or not. I originally agreed until I re-examined it much later and figured out the problem. Can you figure what their error is?

[dionisos]

and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat.

I think it should be specified that the host opens another door, whatever the door you opened. Else he could by example open the door only when the door you choose is the door with the car, to give you a reason to change.
Apart from this, and other things a little off topic like "it depend if you want a car or a goat", i don’t see fallacy.

[Scott Mayers]

and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat.

I think it should be specified that the host opens another door, whatever the door you opened. Else he could by example open the door only when the door you choose is the door with the car, to give you a reason to change.
Apart from this, and other things a little off topic like "it depend if you want a car or a goat", i don’t see fallacy.

The fallacy regards their conclusion that you should switch. The host does know where the car is. But this is irrelevant.

If taken as a whole, you'd think that the person choosing has an option of 1/3 from start to finish. But this is not the actual case because the host already knows which door the car is in and will always be able to assure picking a door that has a goat. So the illusion is that the first choice actually counts when it doesn't. The host is basically restarting the game between two door options, not three, when he offers you to switch. So their error is to allow the first choice to be part of the statistics to determine the odds from start to finish. Thus, it does not actually matter whether you stay or switch as this will be fifty-fifty because the game was reset by the host by even asking whether you want to switch or not. Their conclusion is that you have a 2/3 improved chance by switching. But given their approach, you also have a 2/3 chance of losing if you take in the first decision. You just have to think of one purposely opting to have a goat from the start instead to see this as true.

[dionisos]

If taken as a whole, you'd think that the person choosing has an option of 1/3 from start to finish. But this is not the actual case because the host already knows which door the car is in and will always be able to assure picking a door that has a goat. So the illusion is that the first choice actually counts when it doesn't. The host is basically restarting the game between two door options, not three, when he offers you to switch. So their error is to allow the first choice to be part of the statistics to determine the odds from start to finish. Thus, it does not actually matter whether you stay or switch as this will be fifty-fifty because the game was reset by the host by even asking whether you want to switch or not. Their conclusion is that you have a 2/3 improved chance by switching. But given their approach, you also have a 2/3 chance of losing if you take in the first decision. You just have to think of one purposely opting to have a goat from the start instead to see this as true.

It is untrue, you dismiss information that you should not, and draw a bad conclusion. In fact, what you said is the mistake a lot of people do, and it is why this problem is know, if there was no trap it would be a uninteresting problem.
I will directly give you a example, i think it would be more useful than to explain it directly, and you could convince yourself after that, and search for your own mistakes.
X is the car, and O a goat, at first there are three possibilities:
XOO
OXO
OOX

Then imagine you take the first door (you get the car in the first possibility, and a goat in the other two).
After that, the host open a door with a goat, the three possibilities are transformed in:
XOO → XO
OXO → OX
OOX → OX

You could see that if you stay with the first door, you get 1/3 chance, but if you change, you get 2/3.

I could say more if you are still not convinced, but i hope it will do.

[Scott Mayers]

If taken as a whole, you'd think that the person choosing has an option of 1/3 from start to finish. But this is not the actual case because the host already knows which door the car is in and will always be able to assure picking a door that has a goat. So the illusion is that the first choice actually counts when it doesn't. The host is basically restarting the game between two door options, not three, when he offers you to switch. So their error is to allow the first choice to be part of the statistics to determine the odds from start to finish. Thus, it does not actually matter whether you stay or switch as this will be fifty-fifty because the game was reset by the host by even asking whether you want to switch or not. Their conclusion is that you have a 2/3 improved chance by switching. But given their approach, you also have a 2/3 chance of losing if you take in the first decision. You just have to think of one purposely opting to have a goat from the start instead to see this as true.

It is untrue, you dismiss information that you should not, and draw a bad conclusion. In fact, what you said is the mistake a lot of people do, and it is why this problem is know, if there was no trap it would be a uninteresting problem.
I will directly give you a example, i think it would be more useful than to explain it directly, and you could convince yourself after that, and search for your own mistakes.
X is the car, and O a goat, at first there are three possibilities:
XOO
OXO
OOX

Then imagine you take the first door (you get the car in the first possibility, and a goat in the other two).
After that, the host open a door with a goat, the three possibilities are transformed in:
XOO → XO
OXO → OX
OOX → OX

You could see that if you stay with the first door, you get 1/3 chance, but if you change, you get 2/3. I underlined the switch options and bolded the initial selections.

I could say more if you are still not convinced, but i hope it will do.

You make a mistake in the second part of the game. You have to illustrate it this way assuming the most left is the first choice:
XOO → XO(O) OR X(O)O
OXO → OX
OOX → OX

The error is to forget that the host in the first scenario can pick any one of the two goats to show if you select the car. They are distinct and real options that would have to be included in the statistic even if the value of both goats are the same. The above correction shows that you have 1/2 chance to win and 1/2 chance to lose considering ALL actual options.

[dionisos]

You make a mistake in the second part of the game. You have to illustrate it this way assuming the most left is the first choice:
XOO → XOO OR XOO
OXO → OX
OOX → OX

The error is to forget that the host in the first scenario can pick any one of the two goats to show if you select the car. They are distinct and real options that would have to be included in the statistic even if the value of both goats are the same. The above correction shows that you have 1/2 chance to win and 1/2 chance to lose considering ALL actual options.

Interesting, it seem you made a sort of false reasoning rule to "repair" the example, and move towards your conclusion. (conclusion you made for a completely different reason).
If i success to show you you are wrong (and i am relatively convinced i will), please think about it, the thing you just made here. (evidently if you success to convince me, it would be more interesting(for me at least)).

XOO → XOO OR XOO is true, but it is still only one case between three, it doesn’t change the probability to write it that way, if there is 1/3 to be in one situation, you could separate this situation in two sub-situation, it will not change the probability to be in it.
I am pretty sure you will not be convinced by what i just said, then let me make more clear explanation.
You make the experiment 12 time.
You choose door 1 each time
4 time it will be: XOO
4 time it will be: OXO
4 time it will be: OOX

Then, for the 4 time it will be XOO, 2 time the host will choose the door 2, and 2 time the host will choose the door 3 (2 time XOO and 2 time XOO)
for the 4 time it will be OXO, 4 time the host choose the door 3
for the 4 time it will be OOX, 4 time the host choose the door 2
Then we have:
2 time XOO
2 time XOO
4 time OXO
4 time OOX

If you choose to stay, you will have the car for the 2 time when we have XOO, and the 2 time when we have XOO, then you will have the car 4 time in the 12 time we do the experiment.

I hope i was convincing.

[Scott Mayers]

Your imagining something not there. The host in the second and third cases has only one goat to demonstrate because you already picked them as your first choice. Only where you select the car as a first option does the host require two possible goats to choose from. I haven't a clue of how you demonstrate your example above. It is nevertheless irrelevant.

See my edit above where I underlined the options available AND provided braces to show which ones were not selected of the two remaining options. The other cases only has one goat that the host is FORCED to select from uniquely.

[dionisos]

The host in the second and third cases has only one goat to demonstrate because you already picked them as your first choice. Only where you select the car as a first option does the host require two possible goats to choose from.

Yes, i agree, my explanation include it.

I haven't a clue of how you demonstrate your example above. It is nevertheless irrelevant.

I think my last explanation is pretty relevant, please show me in what step i am wrong, because if i don’t make one, you have 4/12 chance to get the car if you stay, and that is exactly what i want to show.
Yes when you choose a goat, the host have only one door to choose, but i don’t see how it change what i said.

See my edit above where I underlined the options available AND provided braces to show which ones were not selected of the two remaining options. The other cases only has one goat that the host is FORCED to select from uniquely.

Thx to improve the showing, i re-read it, and i still understand it the same way, in fact i think your explanation was already pretty clear.

If after reading what i said another time you still think i am wrong (or you don’t understand a step), i propose than you do the same thing that me, in your own way, you begin by saying we do the experiment 100 time, that when you choose the first door, 33 time you have XOO, 33 time you have OXO, and 33 time you have OOX, and you continues by saying what happen next.
Or whatever other way that permit to "simulate" the situation, and make it happen a lot of time.
The final step would be to write a program to simulate the situation, to execute this program with random choice from the player and from the host, and then see how many time the player get the car when he change, and how many time the player get the car when he stay.(this will not give you exact result, but it would show who is right).
But i don’t think we need to go so far, i think that if you try to do the same kind of things i did, you will see it.

[Scott Mayers]

The host in the second and third cases has only one goat to demonstrate because you already picked them as your first choice. Only where you select the car as a first option does the host require two possible goats to choose from.

Yes, i agree, my explanation include it.

I haven't a clue of how you demonstrate your example above. It is nevertheless irrelevant.

I think my last explanation is pretty relevant, please show me in what step i am wrong, because if i don’t make one, you have 4/12 chance to get the car if you stay, and that is exactly what i want to show.
Yes when you choose a goat, the host have only one door to choose, but i don’t see how it change what i said.

You appear to have a language barrier with English. You over-complicate it by adding multiples for no reason and it does NOT follow. If I were to delve into your further obfuscation, the discussion would not clarify my case but hide my argument among your posts as you digress unnecessarily. You require disproving my point first before I should take on your attempt to get me to interpret you.

[dionisos]

XOO
OXO
OOX
XOO
OXO
OOX
XOO
OXO
OOX
XOO
OXO
OOX
here are the 12 cases, the player choose the first door each time, and then stay, he will have the car 4/12 time. (that the host open the door will change nothing, because you stay, it is the same that if he has done nothing).

XOO or XOO
OXO
OOX
XOO or XOO
OXO
OOX
XOO or XOO
OXO
OOX
XOO or XOO
OXO
OOX

here are the 12 cases, the player choose the first door each time, then change, he will have the car 8/12 time.
Yes, 4 of this case have 2 sub-cases, but it change nothing to the cases themselves. You could separate XOO in (XOO and the host is nice, or XOO and the host is awful), or in whatever number of sub-cases, it is irrelevant.

[dionisos]

You appear to have a language barrier with English

Yes i have, it is inconvenient, sorry for that.

[dionisos]

If taken as a whole, you'd think that the person choosing has an option of 1/3 from start to finish. But this is not the actual case because the host already knows which door the car is in and will always be able to assure picking a door that has a goat. So the illusion is that the first choice actually counts when it doesn't.

If you choose one door, and you stay with this door whatever the host do, you will have 1/3 to get the car, the host could reveal one door, or reveal the two doors, or even add a new door with a car, or thousand doors with a car.
You had a door with 1/3 chance to have a car behind, and then if you choose 100 time a door and stay with it, you will have 33 car, what the host do with the other doors is irrelevant.

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

So their error is to allow the first choice to be part of the statistics to determine the odds from start to finish

.
No, it is not a error (or it depend what you mean exactly in fact).
Think not in probability, but in frequencies, probability change when you get new information, but frequencies don’t, and you deducts probability from frequencies.
If one in three time you get a car when you open the door A, then whatever the host will do with the door B and C, and D, and whatever, as long as he does not touch the door A, then one in three time you get a car when you open the door A.

Thus, it does not actually matter whether you stay or switch as this will be fifty-fifty because the game was reset by the host by even asking whether you want to switch or not.

No, the game have now only two possibilities, but these possibilities are not equivalent, it is here that you do your mistake.

But given their approach, you also have a 2/3 chance of losing if you take in the first decision.

You have 1/3 chance to win if you stay, and 2/3 chance to win if you change.

like you asked, i tried to show you where you was wrong, i hope it will do.

[Scott Mayers]

Slow down dionisos. I'm not going to bother arguing with you when you reintroduce data that doesn't exist.

You first need to recognize that no amount of repeating the whole game improves the odds. They are reset at each play from the start. If a coin toss is argued to be 1/2, you don't argue multiple cases as it doesn't improve the probabilities by repeating it. The odds are still 1/2 for each event. No need to introduce experiments that only act to confuse others.

Given the game on any try, it begins as 1/3 when you initially start. But when you have any other options as this game does, the most possible probabilities is 3! = 3 * 2 * 1 = 6. So you can't even go further than this to even make sense period! The first option allows you 3 possibilities and the second round only allows you 2. Thus talking of 12 or 30 tries appears as though you are purposely attempting to troll me here.

I showed that in given the conditions, after the first round, where the person chooses the car, the host KNOWS this and so opts to select one of two other available options where the goats reside. Thus, given

First Selection and Remaining Ones
1).....Car............Goat 1 AND Goat 2...
2).....Goat 1..........Car AND Goat 2...
3).....Goat 2..........Goat 1 AND Car...

You don't need to bother reversing the remaining ones above as the order is irrelevant. Then, you have the host eliminate a goat in only the remaining options:

..First Selection.....Host can show
1)....Car...................Goat 1
or....Car...................Goat 2
2)....Goat 1...............ONLY Goat 2
3)....Goat 2...............ONLY Goat 1

Now you have options to Stay, the above already shows you that of all first selections, there are two for Car and two for goats out of four. That is, you have 2/4 chance to win the car and 2/4 to win a goat. This is 1/2 for either overall possibilities.

If you Switch,

..First Selection = Stay.....Host can show.....Your options to Switch
1)....Car.......................Goat 1....................Goat 2 ONLY
or....Car.......................Goat 2....................Goat 1 ONLY
2)....Goat 1...................ONLY Goat 2.............Car
3)....Goat 2...................ONLY Goat 1.............Car

So this proves that even when you Switch, you only have 2/4 = 1/2 chance still.

[dionisos]

Slow down dionisos. I'm not going to bother arguing with you when you reintroduce data that doesn't exist.

I don’t introduce data that doesn’t exist, but ok i slow down.

You first need to recognize that no amount of repeating the whole game improves the odds. They are reset at each play from the start. If a coin toss is argued to be 1/2, you don't argue multiple cases as it doesn't improve the probabilities by repeating it. The odds are still 1/2 for each event. No need to introduce experiments that only act to confuse others.

Of course, i recognize that no amount of repeating the whole game improves the odds.
But you should recognize that if you don’t know the odds of a coin toss (and here we are in this kind of situation, we don’t agree with the odds), then a good way to find a approximation, is to repeat the coin toss, and to see what happen, to see how many time we have each events.
Then i hope you understand that my goal was not to confuse, but to make "the experiment", or a simulacrum of it, to avoid losing ourself in arguments.
(and yes, it was a failure)

So you can't even go further than this to even make sense period! The first option allows you 3 possibilities and the second round only allows you 2. Thus talking of 12 or 30 tries appears as though you are purposely attempting to troll me here.

I choose to make 12 time the experiment, because it was a multiple of 3, and a multiple of 2, i could have choose any other numbers, it was just convenient.
This 12 doesn’t mean anything.

I still don’t agree with what you said, you still make the same error, and then if i answer i would repeat myself. You also repeated yourself.
I think either you read too rapidly what i said, or my english is too bad.