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

[ynot]

In case you haven’t heard of the “Monty Hall” game it goes like this . . .

• You attempt to make the correct choice from three random options (only one is correct).
• After you make your choice one of the remaining choices is revealed as being incorrect.
• You are then asked if you want to retain your original choice or change to the other choice available.
• Your “final” choice is then revealed as being correct or incorrect.

The claim is that you are better of to change your choice because the original choice was made at 1 in 3 odds and the changed choice is at 1 in 2 odds. In other words your odds of winning improve from 1 in 3 to 1 in 2 by changing. If this is correct the improved odds should be evident in practical demonstrations of actually playing the game multiple times.

You can “play the game” very simply by using two black and one red playing cards. The red card is the correct choice. Get a friend to randomly place the cards face down so you don’t know where the red card is. Your friend knows where it is however. After you make your choice you friend turns over one of the other cards that they know to be black. Regardless of whether you then change your choice or not all testing I have done gives results that would be expected from 1 in 2 odds not 1 in 3 odds.

Is there something in my expanation of the game, claim or demonstration I’m getting wrong or is the claim proven wrong by practical demonstration?

 

[Baloney]

The trick is that it was 1 in 2 to begin with, because the test isn't double-blinded. Regardless of your first choice, it comes down to the card you initially chose and the remaining unknown (to the contestant) card.

 

[ddt]

Your description of the Monty Hall problem is right.

ETA: no, by switching your odds should improve from 1/3 to 2/3.

But you're basically saying that in your experiments, even not changing seems to give you 1 in 2 odds? Then you really should carefully review the setup of your experiment, or whether you inadvertently pick up some (non-verbal) cues from your friend.

Another point: how many turns did your experiment last? Was that amount statistically relevant?

 

[Jack by the hedge]

What were the actual results of your testing?

ETA: Either your testing sample was a fluke or there's a flaw in your method, because what you say you measured was that given a 33.3% chance, you won 50% of the time.

 

[ddt]

The trick is that it was 1 in 2 to begin with, because the test isn't double-blinded. Regardless of your first choice, it comes down to the card you initially chose and the remaining unknown (to the contestant) card.

No it's not. You might check out the other threads on the topic or the wiki page or any of the other bazillion write-ups of the problem.

 

[ynot]

Your description of the Monty Hall problem is right.

ETA: no, by switching your odds should improve from 1/3 to 2/3.

But you're basically saying that in your experiments, even not changing seems to give you 1 in 2 odds? Then you really should carefully review the setup of your experiment, or whether you inadvertently pick up some (non-verbal) cues from your friend.

Another point: how many turns did your experiment last? Was that amount statistically relevant?

I assure you I have gone to great pains to ensure there can be no cheating (even subliminally). When the friend places the cards I look away. The cards are in numbered positions and I just give the position number of the card I choose. I never touch the cards. When I make my choice the friend looks away so I can’t pick up any clues.

Have repeated the experiment over one hundred times. Please feel free to replicate it yourself and let me know your results.

ETA - My results so far are slightly in favour of not changing.

 

[sol invictus]

In case you haven’t heard of the “Monty Hall” game it goes like this . . .

• You attempt to make the correct choice from three random options (only one is correct).
• After you make your choice one of the remaining choices is revealed as being incorrect.
• You are then asked if you want to retain your original choice or change to the other choice available.
• Your “final” choice is then revealed as being correct or incorrect.

The claim is that you are better of to change your choice because the original choice was made at 1 in 3 odds and the changed choice is at 1 in 2 odds. In other words your odds of winning improve from 1 in 3 to 1 in 2 by changing. If this is correct the improved odds should be evident in practical demonstrations of actually playing the game multiple times.

You can “play the game” very simply by using two black and one red playing cards. The red card is the correct choice. Get a friend to randomly place the cards face down so you don’t know where the red card is. Your friend knows where it is however. After you make your choice you friend turns over one of the other cards that they know to be black. Regardless of whether you then change your choice or not all testing I have done gives results that would be expected from 1 in 2 odds not 1 in 3 odds.

Is there something in my expanation of the game, claim or demonstration I’m getting wrong or is the claim proven wrong by practical demonstration?

"After you make your choice you friend turns over one of the other cards that they know to be black."

That bolded "other" is critical. If that's what your friend is doing, switching will win with 2/3 probability.

Consider the possibilities:

2/3: You picked a black. You friend has no choice, and turns over the other black. If you switch, you win. If you stay, you lose.

1/3: You picked the red. Your friend turns over one of the black cards (makes no difference which). If you switch, you lose. If you stay, you win.

Therefore, with that ruleset switching has a 2/3 chance of winning, and staying has a 1/3. That can and has been verified both experimentally and numerically with computer codes, if you somehow don't believe the above argument.

If instead you're finding something else experimentally, there are several possibilities. One is you didn't do enough trials. Another is that your friend isn't actually following the protocol you described. Instead, he sometimes turns over the card you picked (when it's black), rather than one of the other two.

If he always turns over the card you picked when it's black, always switching is a losing strategy. If he randomly turns over a black card regardless of what you picked, always switching (and all other strategies) has a 1/2 chance of winning.

 

[ynot]

What were the actual results of your testing?

ETA: Either your testing sample was a fluke or there's a flaw in your method, because what you say you measured was that given a 33.3% chance, you won 50% of the time.

Do the experiment yourself and let us know your results.

 

[ravdin]

Actually, your odds change from 1 in 3 to 2 in 3 (not 1 in 2) by switching. In other words, switching doubles your odds of a successful result. Here's a thought experiment that might illustrate.

I spread a fair deck of cards on the table and tell you you'll win a prize for picking the seven of diamonds. You point to a card. I turn over fifty cards (the entire deck except for the card you chose and one other of my choosing), none of which is the seven of diamonds. Does it seem like it the odds will be 1 in 2 that your first choice was correct, or do you think you'd be better off switching?

The only way I could see the result ending up as 1 in 2 is if you never selected the red card as your first choice, which seems unlikely.

 

[AvalonXQ]

Do the experiment yourself and let us know your results.

There's no need to rely on offline testing; we can do the experiment right here -- you and me.

Tell me how many trials you would like to do.

 

[ynot]

"After you make your choice you friend turns over one of the other cards that they know to be black."

That bolded "other" is critical. If that's what your friend is doing, switching will win with 2/3 probability.

Consider the possibilities:

2/3: You picked a black. You friend has no choice, and turns over the other black. If you switch, you win. If you stay, you lose.

1/3: You picked the red. Your friend turns over one of the black cards (makes no difference which). If you switch, you lose. If you stay, you win.

Therefore, with that ruleset switching has a 2/3 chance of winning, and staying has a 1/3. That can and has been verified both experimentally and numerically with computer codes, if you somehow don't believe the above argument.

If instead you're finding something else experimentally, there are several possibilities. One is you didn't do enough trials. Another is that your friend isn't actually following the protocol you described. Instead, he sometimes turns over the card you picked (when it's black), rather than one of the other two.

If he always turns over the card you picked when it's black, always switching is a losing strategy. If he randomly picks a black card regardless of what you picked, always switching (and all other strategies) has a 1/2 chance of winning.

Yes one of the other cards is turned over, and no my results don't refelect 1 in 3 odds regardless of whether the choice is changed or not.

Please do the experiment using cards as I have and see if you get different results.

 

[sol invictus]

Yes one of the other cards is turned over, and no my results don't refelect 1 in 3 odds regardless of whether the choice is changed or not.

Please do the experiment using cards as I have and see if you get different results.

Can you tell us what the results of your trials were, exactly? Specifically, I'd like to know what happened when you followed the "always stay" strategy.

 

[ynot]

There's no need to rely on offline testing; we can do the experiment right here -- you and me.

Tell me how many trials you would like to do.

All testing relies on complete honesty. How can we rely on each other’s honesty online? Do the test yourself and rely on your own honesty. I’m thinking it might be fun to write a JavaScript version so it can be played online.

 

[Professor Yaffle]

So, you picked the red card 50% of the time even though there were 3 cards, only one of which was red? Quick, someone tell Randi - we have a contender!

 

[Professor Yaffle]

All testing relies on complete honesty. How can we rely on each other’s honesty online? Do the test yourself and rely on your own honesty. I’m thinking it might be fun to write a JavaScript version so it can be played online.

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html


ETA: just had a go myself:

Switching I won 19/25 times (76%)
Not switching I won 10/25 times (40%)

Not too far from predicted given the low number of trials.

 

[AvalonXQ]

All testing relies on complete honesty. How can we rely on each other’s honesty online?

I've already generated a list of 100 cases. Select your first door, 1 2 or 3, a hundred times.

I’m thinking it might be fun to write a JavaScript version so it can be played online.

That's been done.

 

[ynot]

Can you tell us what the results of your trials were, exactly? Specifically, I'd like to know what happened when you followed the "always stay" strategy.

In 115 “games” 111 “staying” would have been correct, and 104 “changing” would have been correct. Please do it yourself and let us know what results you get. Given it’s an experiment that can be simply replicated by anyone my results aren‘t as important as yours would be to you. I could be lying.

 

[Professor Yaffle]

In 115 “games” 111 “staying” would have been correct, and 104 “changing” would have been correct. Please do it yourself and let us know what results you get. Given it’s an experiment that can be simply replicated by anyone my results aren‘t as important as yours would be to you. I could be lying.

Up to 50 trials of each condition, using the above link, I'm getting

No switch: 16/50 = 32%

Switch: 40/50 = 80%

ETA, so you are saying that in 215 games (115 is a typo?), you initially chose the red card 111 times (regardless of the switching)? I'm suspecting your precautions to prevent leakage weren't as strong as you think. Or you just got really lucky.

 

[ynot]

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html


ETA: just had a go myself:

Switching I won 19/25 times (76%)
Not switching I won 10/25 times (40%)

Not too far from predicted given the low number of trials.

I have honestly related the results I got. Will try again using the applet. Perhaps there is a flaw in my human test that I'm missing.

 

[AvalonXQ]

In 115 “games” 111 “staying” would have been correct, and 104 “changing” would have been correct.

Wait, so your friend randomly dealt three cards 215 times, and you randomly picked the right card 111 of those times?
You don't see the statistical anomaly here? Forget turning over a black card and trying to stay/switch; just your selections themselves were anomalous!