"Monty Hall Problem" claim not verified by practical demonstration?

[Nova Land]

After my choice was made and the “other” incorrect card was revealed both remaining cards were turned over to see if “changing” or “sticking” would have won. Obviously only one needs to be turned over but there is no harm done in turning both over as long as the guessed card is correctly identified each time. A tally was kept of whether "sticking" or changing" was correct.

Then my guess was wrong. in the test you describe -- picking one of 3 cards, and seeing if it was red or black -- the card you initially selected should have been red about 1/3 of the time and black about 2/3 of the time.

Sorry to ask one more time, but I want to be absolutely clear. When you say A tally was kept of whether "sticking" or changing" was correct, the wording strikes me as less clear than A tally was kept of whether the card I had chosen was red or black.

That is what you mean, though? You checked to see if the card you had initially selected was red or black? And your choice was red half the time and black half the time.

 

[AvalonXQ]

Turning over one of the "other" cards happens after I make the first choice and before I make the second choice. As I don't know where the red card is I can't guarantee that the "other" card I'm turning over isn't the red card so when it happens I have to ignore that game. The revealled "other" card has to be a black card so one of the remaining cards is the red card.

Let me try to explain this again.

If you chose right the first time, then the "other" card will always be black. This means you will never throw out a game where you chose right the first time.

If you chose wrong the first time, then the other card will be red 50% of the time. This means that you will throw out half your games where you chose wrong the first time.

It's exactly the same as this:
1) I always choose A.
2) If the red card is A, I count it as a "stick".
3) If the red card is B, I don't count it.
4) If the red card is C, I count it as a "switch".
You should end up with 1/2 odds this way, but it's NOT the same as a Monty Hall game.

 

[Paul C. Anagnostopoulos]

Why don't you simply enumerate all possible games and count how many you would win by sticking and how many by switching?

~~ Paul

 

[Nova Land]

Turning over one of the "other" cards happens after I make the first choice and before I make the second choice.

Aha! There it is again. You're talking about a second choice.

In the test you described, there should be no second choice. You choose a card. Your friend turns over a card. You check the card you originally chose to see if it was red or black, and record that result.

Both in your OP and in this post, you indicate there is a second choice you have made. And if there is, that's your problem.

 

[Nova Land]

If you chose wrong the first time, then the other card will be red 50% of the time. This means that you will throw out half your games where you chose wrong the first time.

That's a good explanation of the design flaw in ynot's self-testing method.

But that doesn't account for the flaw in the actual results. Ynot claims that, out of 20 tests only once was the revealed card the red one. (Of the remaining 19, there were 10 times the card he chose was the red one and 10 times the card he didn't choose was the red one.)

I assume ynot actually meant there were 21 tests, with one discarded because the red card was turned over.

We'd expect the revealed card to be red about 7 times, the chosen card to be red about 7 times, and the unchosen card to be read about 7 times. Turning over the red card only once in 21 trials is even more unlikely than the original test results.

 

[ynot]

I read the rest of the thread now... this too will be a redundant comment, but the above is precisely as expected. The game you played works like this:

1/3: you picked red, switched, and lost

1/3: you picked black and turned over black, switched, and won

1/3: you picked black and turned over red, and ignored that game.

So as you can see, you should win half the time and lose half the time, which is precisely (by coincidence) what happened.

No - When playing by myself . . .

I picked one of the three cards as the contestant.

Then as Monty I revealed one of the other cards as being black. Trouble is being both Monty and the contestant I can’t know which of the other cards is black so I have to guess and some of the time I will guess wrongly and reveal the red card by mistake. This ruins the game and that game needs to be ignored. Only games where Monty reveals a black card are valid.

When I’m playing with another person that knows where the red card is no games need to be ignored because a black card will be revealed each time.

 

[ynot]

Let me try to explain this again.

If you chose right the first time, then the "other" card will always be black. This means you will never throw out a game where you chose right the first time.

If you chose wrong the first time, then the other card will be red 50% of the time. This means that you will throw out half your games where you chose wrong the first time.

It's exactly the same as this:
1) I always choose A.
2) If the red card is A, I count it as a "stick".
3) If the red card is B, I don't count it.
4) If the red card is C, I count it as a "switch".
You should end up with 1/2 odds this way, but it's NOT the same as a Monty Hall game.

So to make it the same as the Monty game only games where a black card is revealed (after the first choice is made but before the final choice and is made and result is revealed) can be used and games when the red card are wrongly revealed have to be ignored.

 

[AvalonXQ]

So to make it the same as the Monty game only games where a black card is revealed (after the first choice is made but before the final choice and is made and result is revealed) can be used and games when the red card are wrongly revealed have to be ignored.

No. If you do this, you will count all the games where sticking wins and half the games where switching wins, giving you the wrong answer.

This is like trying to count the number of pears and the number of apples in a basket, but every time the fruit is red you don't count it. You can't get an accurate fraction by selectively ignoring certain results.

 

[ynot]

I'd like to get this as clear as possible. The sentence I've colored blue is not entirely clear, plus it slightly contradicts the one in red.



Here's my attempt at writing what it sounds like you are trying to say.

• You selected three cards -- two black, one red.
  
• You dealt the cards out, face down.
  
• You chose one of the three cards.
  
• You turned over one of the other cards; if it turned out to be the red card, you discarded the test. That happened once in 21 tests.
  
• Of the remaining 20 tests, 10 times the card you chose (before turning over the other card) was red and 10 times it was black.

Is that correct? (If so, it strikes me as strange that only 1 time in 21 was the card you turned over the red card.)

It's that last bullet point I'd really like to be clear on. The results of the 20 games were exactly 10 in favour of changing and ten for not is much more open to misinterpretation than Of the remaining 20 tests, 10 times the card you chose (before turning over the other card) was red and 10 times it was black.

In testing to see if it’s better to change or not you don’t actually have to make a choice as you can clearly see the from the outcome of each game what the result of either choice would have been regardless. In some games choosing to “stick” would have won and in some choosing to “change” would have won.

Yes it is strange (against the odds) that only one game out of 21 games had to be ignored because the red card was wrongly revealed as being a loosing card (black card). It’s also strange (against the odds) that of the 20 valid games the results were exactly ten in favour of changing and ten in favour of not. It’s even more strange (to me at least) that I’m not able to get the same unusual results I get with actual cards when I use the online game.

 

[ynot]

No. If you do this, you will count all the games where sticking wins and half the games where switching wins, giving you the wrong answer.

This is like trying to count the number of pears and the number of apples in a basket, but every time the fruit is red you don't count it. You can't get an accurate fraction by selectively ignoring certain results.

Do you realise that the revealing of the “other” card (which has to be a black card) happens after the first guess is made and before the decision to change or not is made and the final result is revealed?

 

[Bob Klase]

No - When playing by myself . . .

I picked one of the three cards as the contestant.

Then as Monty I revealed one of the other cards as being black. Trouble is being both Monty and the contestant I can’t know which of the other cards is black so I have to guess and some of the time I will guess wrongly and reveal the red card by mistake. This ruins the game and that game needs to be ignored. Only games where Monty reveals a black card are valid.

When I’m playing with another person that knows where the red card is no games need to be ignored because a black card will be revealed each time.

Seems to me that a simple method would be to just pick your initial card. If you pick the winning card then switching would cause you to lose no matter what happens to the other 2 cards. If you pick one of the two losing cards then switching would cause you to win (because the other losing card would have been eliminated by Monty).

So you don't really have to go through the effort of picking a card, then turning over a card and then seeing if switching would cause you to win or lose. Just pick a card- if it's a loser then switching wins.

Odds of picking the winning card remain 1 out of 3, so odds of winning when you switch are the same odds as picking a loser in the first place.

 

[AvalonXQ]

Do you realise that the revealing of the “other” card (which has to be a black card) happens after the first guess is made and before the decision to change or not is made and the final result is revealed?

Yes. This doesn't change the fact that you're discarding a third of your trials, and every trial you're discarding is one where switching would win.

 

[ynot]

So you picked the right card 111/215=52% of the time, despite the fact that your true odds should be 1/3?

Obviously your game was rigged or in some other way invalid. In a fair game your results are essentially impossible - I can calculate the odds if you like, but even without doing so it's clear that they are basically zero.

Note that this has nothing to do with your switching strategy, it just means your game wasn't straight.

I don't need to do the experiment myself, ynot. It's obvious what the correct answer is.

It’s equally obvious to me what the results should be and I’m not surprised that these results are confirmed when I use the online game. What’s not obvious to me however is every time so far I have used actual cards (ether with some one else or by myself) I get results that aren’t obvious and are surprising.

 

[ynot]

Yes. This doesn't change the fact that you're discarding a third of your trials, and every trial you're discarding is one where switching would win.

Please explain how either switching or sticking could possibly win when the winning (red) card has been revealed before the choice to switch or not occurs? How do you pick a red card from two black cards?

In the 21 "self-games" I played I only had to ignore one game in practice. In the games with my grandson no games were ignored

 

[ynot]

Seems to me that a simple method would be to just pick your initial card. If you pick the winning card then switching would cause you to lose no matter what happens to the other 2 cards. If you pick one of the two losing cards then switching would cause you to win (because the other losing card would have been eliminated by Monty).

So you don't really have to go through the effort of picking a card, then turning over a card and then seeing if switching would cause you to win or lose. Just pick a card- if it's a loser then switching wins.

Odds of picking the winning card remain 1 out of 3, so odds of winning when you switch are the same odds as picking a loser in the first place.

As I’ve said, you don’t need to make a choice to find out which choice would have won or lost.

 

[ynot]

Aha! There it is again. You're talking about a second choice.

In the test you described, there should be no second choice. You choose a card. Your friend turns over a card. You check the card you originally chose to see if it was red or black, and record that result.

Both in your OP and in this post, you indicate there is a second choice you have made. And if there is, that's your problem.

There are definitely two choices in the game. First choice is one of three cards. Second choice is to retain that choice or change it. I have no problem in this regard.

 

[ynot]

Why don't you simply enumerate all possible games and count how many you would win by sticking and how many by switching?

~~ Paul

That's exactly what I did.

 

[ehcks]

There are definitely two choices in the game. First choice is one of three cards. Second choice is to retain that choice or change it. I have no problem in this regard.

But the only part that actually matters is whether switching would increase your chance of winning. You don't actually have to do that switch to test it.

An easy way to test it is to always assume you're switching. Pick one card, look at it, toss it out, then look at the other two cards. If you picked red, you're left with two blacks, and that's an automatic loss with a 33% chance to happen. If you picked black, the other black is discarded and you're only left with a red, so it's an automatic win with a 66% chance to happen.

EDIT: The easiest way is to assume you always stick. Pick a card. Is it red? You win 33% of the time. Is it black? You lose 66% of the time.

 

[AvalonXQ]

Please explain how either switching or sticking could possibly win when the winning (red) card has been revealed before the choice to switch or not occurs? How do you pick a red card from two black cards?

In the Monty Hall scenario, Monty would not have picked the red card -- he would have picked the other card instead. And then you would have won if you switched or lost if you stuck.

To make it easy, let's assume you start by choosing Card A and then you show Card C. To make the problem clear, let's go through the three situations and you can see how your solo game is different from Monty's game.

A) The red card is Card A.
Solo game: You choose Card A, you flip over Card C which is black. If you stick you win and if you switch you lose. (+1 sticking)
Monty: You choose Card A, Monty flips over either Card B or Card C at random, which is black. If you stick you win and if you switch you lose. (+1 sticking)
Result: the same, +1 for sticking

B) The red card is Card B.
Solo game: You choose Card A, you flip over Card C which is black. If you stick you lose and if you switch you win. (+1 switching)
Monty: You choose Card A, Monty flips over Card C which is black. If you stick you lose and if you switch you win. (+1 switching)
Result: the same, +1 for switching

C) The red card is Card C.
Solo game: You choose Card A, you flip over Card C which is red. You ignore the result (+0)
Monty: You choose Card A, Monty flips over Card B which is black. If you stick you lose and if you switch you win. (+1 switching)
Result: you discard a set of trials that would have been a win for "switching" if Monty were playing

Final result: Your solo game has a 50-50 expectation, while Monty's game wins by switching 2/3rd's of the time.

 

[Fredrik]

ynot, if you're getting the results you claim, then about 1/2 of your first guesses were correct, even though the probability to make the correct guess is 1/3. This would mean that you have a paranormal ability that could easily win you the Randi million (if that prize still exists...I didn't follow the developments after Randi started talking about discontinuing the challenge).

It's of course much more likely that either you or your friend is doing something very wrong that you haven't told us about.