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

[phunk]

I just rolled a pair of dice 20 times. One die to represent the door with the prize behind it, the other to represent the contestant's guess. A roll of 1 or 2 means door 1, 3 or 4 means door 2, 5 or 6 means door 3. These are the results:

Prize  Guess   Stay  Switch     
1      3       lose  win
3      2       lose  win
3      2       lose  win
3      3       win   lose
3      3       win   lose
2      3       lose  win
1      3       lose  win
1      2       lose  win
2      1       lose  win
2      3       lose  win
3      2       lose  win
2      2       win   lose
1      1       win   lose
3      3       win   lose
3      1       lose  win
1      3       lose  win
2      3       lose  win
2      3       lose  win
2      2       win   lose
1      3       lose  win
Total:         6      14

6/20= 0.3
14/20 = 0.7

That's pretty close to the expected odds. With more trials it should only get closer.

 

[ynot]

Just did a quick 10 using the applet and got 8 in favour of changing. Wonder what I'm doing wrong with the humen test? Won't assume I have some paranormal ability.

 

[ynot]

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!

People do win Lotto. I don't expect I will be able to replicate my first results unless there is a flaw in what I'm doing.

ETA - It was 115 times.

 

[phunk]

Are you sure your friend is placing the cards randomly? Maybe you and him have similar biases.

 

[steve s]

The Mythbusters tested this and confirmed that it's best to switch.

Steve S

 

[wardenclyffe]

The Mythbusters tested this and confirmed that it's best to switch.

Steve S

Link: http://www.youtube.com/watch?v=Ytv-_RB4xWI

Ward

 

[ynot]

Are you sure your friend is placing the cards randomly? Maybe you and him have similar biases.

I guess it has to be some explanation like that.

I did this experiment to give my grandson a practicle demnstration how it's always better to switch and was gobsmacked by the results I got. Seems I had a freaky good-luck streak or was somehow "reading" my grandson (not paranormally). This is why I put a ? at the end of the title and asked others to replicate the experiment.

 

[tsig]

The Mythbusters tested this and confirmed that it's best to switch.

Steve S

They also confirmed that people wouldn't switch.

 

[wardenclyffe]

How old is your grandson? Perhaps you and he could watch the mythbusters episode together and then re-run your experiments to try and figure out where you went wrong (or if you just got lucky). Always better to figure out something together with your grandson rather than just showing him. The lesson will stick with him better if it's also a lesson for you.

Ward

 

[BillC]

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 got:

Always switching: win 68/100 (expectation: 67/100)
Never switching: win 27/100 (expectation: 33/100)

 

[Jack by the hedge]

Do the experiment yourself and let us know your results.

I don't have a friend handy to turn over a losing card so I wrote a program instead which chose the red card's position, made my first choice, discarded a black card I hadn't chosen and then decided whether I'd stick or switch, all at random, 1000 times over.

Results:

If I switch: Win = 330 Lose = 169

If I stick: Win = 160 Lose = 341

So I won about 66% of the times I switched and 32% of the times I stuck.

 

[Nova Land]

... 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?

Okay, I think I can explain this. I'm guessing -- but it's a reasonable guess.

Your explanation of what you're doing, while it may sound clear, actually isn't. I have bolded the key line. "Change your choice or not..."?

It isn't clear from your write-up, but it sounds to me as if

1. you initially chose a card
2. your friend turned over one of the other cards
3. you chose between the two remaining cards
4. you recorded whether your new guess was correct or not

Am I right that the test you ran consisted of making a choice whether to stay or switch after your friend turned a card over?

A good way to actually test the probability is to proceed as you began: you pick a card, and then your friend turns over one of the other cards. But at that point, there is no need for you to actually decide whether to switch or not. Simply turn over the card you chose, and record whether you would have been right or wrong. What you should find is that about 1/3 of the time the card you initially chose was red and about 2/3 of the time the card you initially chose was black.

But if your test was to make a new choice after your friend turned a card over -- which is what it sounds like it was -- then things get more complicated and it's easy to make a mistake without realizing it.

1/3 of the time you switched your guess should be wrong; 2/3 of the time you switched your guess should be right; 1/3 of the time you didn't switch your guess should be right; 2/3 of the time you didn't switch your guess should be wrong.

If you always chose to stick with your initial choice, then your results should have been that you were right 1/3 of the time and it would be easy to see.

If you always chose to switch, then your results should have been that you were right 2/3 of the time and it would be easy to see.

But if you chose by to stick sometimes and switch other times, you're going to get unnecessarily messy and confusing results which are easy to misinterpret. What you're likely to do in looking at the results is add apples to oranges and think you've discovered a new fruit.

If you kept careful records you should still be able to go through and detangle the results. But it would be simpler just to make a consistent choice -- stay or switch -- to start with.

 

[ynot]

This is getting a bit weird. I’m alone at present so can’t play more games with my grandson or anyone else so played 20 games by myself. I randomly shuffled and placed the cards and made a choice. If I then turned over the red card as the “other” card I ignored that game (only happened once out of twenty games). The results of the 20 games were exactly 10 in favour of changing and ten for not. I realise doing my own shuffling and placing isn’t exactly the best method but I did try to make it as genuinely random as possible. I know 20 isn’t a very large sample but I haven’t been able to get an equal result with 20 using the applet.

 

[ynot]

Okay, I think I can explain this. I'm guessing -- but it's a reasonable guess.

Your explanation of what you're doing, while it may sound clear, actually isn't. I have bolded the key line. "Change your choice or not..."?




It isn't clear from your write-up, but it sounds to me as if

1. you initially chose a card
2. your friend turned over one of the other cards
3. you chose between the two remaining cards
4. you recorded whether your new guess was correct or not

Am I right that the test you ran consisted of making a choice whether to stay or switch after your friend turned a card over?

A good way to actually test the probability is to proceed as you began: you pick a card, and then your friend turns over one of the other cards. But at that point, there is no need for you to actually decide whether to switch or not. Simply turn over the card you chose, and record whether you would have been right or wrong. What you should find is that about 1/3 of the time the card you initially chose was red and about 2/3 of the time the card you initially chose was black.

But if your test was to make a new choice after your friend turned a card over -- which is what it sounds like it was -- then things get more complicated and it's easy to make a mistake without realizing it.

1/3 of the time you switched your guess should be wrong; 2/3 of the time you switched your guess should be right; 1/3 of the time you didn't switch your guess should be right; 2/3 of the time you didn't switch your guess should be wrong.

If you always chose to stick with your initial choice, then your results should have been that you were right 1/3 of the time and it would be easy to see.

If you always chose to switch, then your results should have been that you were right 2/3 of the time and it would be easy to see.

But if you chose by to stick sometimes and switch other times, you're going to get unnecessarily messy and confusing results which are easy to misinterpret. What you're likely to do in looking at the results is add apples to oranges and think you've discovered a new fruit.

If you kept careful records you should still be able to go through and detangle the results. But it would be simpler just to make a consistent choice -- stay or switch -- to start with.

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.

 

[AvalonXQ]

This is getting a bit weird. I’m alone at present so can’t play more games with my grandson or anyone else so played 20 games by myself. I randomly shuffled and placed the cards and made a choice. If I then turned over the red card as the “other” card I ignored that game (only happened once out of twenty games).

That won't work. See, if you throw out every game where you turn over a red card at random, you'll end up throwing out half the games where you would have won by switching, but none of the games where you would have won by sticking.

In other words, using the protocol you just described above, you'll end up with a 50-50 record for switching/sticking.

 

[often mrunderstood]

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.

there's the flaw. You had three choices not two. You could have held, chose A, or chose B. Not just hold or don't hold. You left out the times you would have chose right.

 

[Nova Land]

I randomly shuffled and placed the cards and made a choice. If I then turned over the red card as the “other” card I ignored that game (only happened once out of twenty games). The results of the 20 games were exactly 10 in favour of changing and ten for not.

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.

 

[sol invictus]

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

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.

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.

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

 

[ynot]

That won't work. See, if you throw out every game where you turn over a red card at random, you'll end up throwing out half the games where you would have won by switching, but none of the games where you would have won by sticking.

In other words, using the protocol you just described above, you'll end up with a 50-50 record for switching/sticking.

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.

 

[sol invictus]

This is getting a bit weird. I’m alone at present so can’t play more games with my grandson or anyone else so played 20 games by myself. I randomly shuffled and placed the cards and made a choice. If I then turned over the red card as the “other” card I ignored that game (only happened once out of twenty games). The results of the 20 games were exactly 10 in favour of changing and ten for not. I realise doing my own shuffling and placing isn’t exactly the best method but I did try to make it as genuinely random as possible. I know 20 isn’t a very large sample but I haven’t been able to get an equal result with 20 using the applet.

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.