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

[jasonpatterson]

I had a very sharp student who wouldn't accept a written demonstration of this, so I wrote an excel spreadsheet and did this process something like thirty million times. It picked a random door for the prize, had a contestant pick a random door as a winner, etc. It then decided whether you'd win by staying or switching and totaled up the frequencies at the end. It worked out exactly as expected 1/3 vs 2/3 for stay vs switch. It was no great surprise, but for anyone who wanted more trials using an unbiased system, there ya go, I did 30 million of them. (As I recall I had to keep resetting the worksheet I did it in and adding up the results separately; would excel even let you make 30 million rows?)

I'm not sure why it didn't work for you and your grandson ynot, but I do know one way in which your solo method is flawed (though it wouldn't have changed your 21 trials by much.) If you are throwing out the results where you accidentally discard a red card in your Monty Hall role, then you're decreasing the frequency at which wins by switching will occur (since those can only occur when there are both a red and black card available, and half of the time you'd expect to ruin the test by discarding a red.) You can never eliminate a test in which winning by staying happens, as this only occurs when there are two blacks available for discard. I understand why you want to do it, but it invalidates your method.

You could do this much more quickly solo by pulling a card and checking it. Say that the first X attempts are going to be switches ahead of time. If you pick a card and it's red, then a switch will result in losing regardless of which you discard as Monty. If you pick a black, switching would have won, since you'd have had to discard the remaining black card in your role as Monty.

The next X attempts are going to be stays. If you pick a red card, you won, if you pick a black card, you lost.

Do this a bunch of times and calculate the win/loss frequency for each method and it ought to work out, assuming fair cards and randomness and the like.

 

[DavidByron]

While they are testing this piece of mathematics would anyone like to test that a flipped coin comes down heads about 50% of the time and a die roll comes out as a "6" about 17% ?

 

[phunk]

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?

Because monty would never have revealed that red card, he would have revealed the black one you didn't chose initially, leaving only the red one to switch to. Each time you 'accidentally' reveal the red one is a time where you absolutely would have won by switching.

 

[ynot]

Here’s the possible sequences of me playing both contestant and Monty . . .

(1) I as the contestant pick 1 card from 3

(2) I as Monty then reveal one of the other two cards (the ones I didn‘t pick). To replicate what the real Monty would have done I have to reveal a black card but there is no way I can always do this as I’m merely guessing. If I as Monty reveal a black card all is fine and the game continues and I then have the option as the contestant to choose to stick or change.

(3) If however I as Monty reveal the red card I have established that my contestant’s guess was wrong and the game ends before there’s stick or change choice.

Regardless of whether the stick or change choice is made in practice it has to be able to be made so it can be seen whether it would improve the odds or not with either choice.

 

[ynot]

Because monty would never have revealed that red card, he would have revealed the black one you didn't chose initially, leaving only the red one to switch to. Each time you 'accidentally' reveal the red one is a time where you absolutely would have won by switching.

When I’m playing both contestant and Monty I can’t always reveal a black card from the two unchosen cards. For this reason I have to ignore all games when I reveal a red card by mistake.

I ask you the same question - “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?”

 

[Dan O.]

ynot, you need to play this for yourself before you will believe it.

• Choose a number of games to play
• As Monty, Take three cards, one red and two black. Shuffle and lay them out so you don't know where the red card is.
• As yourself, Choose one of the cards and place a marker on top of it.
• As Monty, you need to remove one black card from the remaining two cards. But since you can't do this without revealing the cards to yourself we'll just delay this and pretend it hapened by sliding the two remaining cards together.
• As yourself, choose to change your pick by moving the marker to the other card or not.
• as the delayed action of Monty, remove one of the black cards from the stack with two cards.
• Turn all the cards face up leaving the marker on the last chosen card.
• Count your win or loss in separate tallies for changing your pick or not.
• repeat until you have played the number of games you chose to play.

 

[sol invictus]

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.

Yes, that's exactly what I said. The odds for that game are 50-50.

Perhaps try reading my post again, I don't know how to explain it any more clearly.

 

[ynot]

Yes, that's exactly what I said. The odds for that game are 50-50.

Perhaps try reading my post again, I don't know how to explain it any more clearly.

But the debate is about whether the final stick or change choice improves the odds, not an earlier prematurely revealed correct card.

All I’m saying is that a game that doesn’t allow for a final stick or change choice isn’t a Monty Hall game and should be ignored.

 

[sol invictus]

But the debate is about whether the final stick or change choice improves the odds, not an earlier prematurely revealed correct card.

And what I'm telling you is that in the version you played by yourself, it switching doesn't improve the odds. That strategy (and all others) has a 50% chance of winning.

All I’m saying is that a game that doesn’t allow for a final stick or change choice isn’t a Monty Hall game and should be ignored.

That's fine, but playing the game that way changes the odds (to 50%).

 

[ynot]

Double post

 

[ynot]

ynot, you need to play this for yourself before you will believe it.

• Choose a number of games to play
• As Monty, Take three cards, one red and two black. Shuffle and lay them out so you don't know where the red card is.
• As yourself, Choose one of the cards and place a marker on top of it.
• As Monty, you need to remove one black card from the remaining two cards. But since you can't do this without revealing the cards to yourself we'll just delay this and pretend it hapened by sliding the two remaining cards together.
• As yourself, choose to change your pick by moving the marker to the other card or not.
• as the delayed action of Monty, remove one of the black cards from the stack with two cards.
• Turn all the cards face up leaving the marker on the last chosen card.
• Count your win or loss in separate tallies for changing your pick or not.
• repeat until you have played the number of games you chose to play.

I can only assume you haven’t read my previous posts. I have played the game many times and I do “believe it” (have done so for many years).

I was doing the experiment as a practical demonstration of how it works to my grandson. Recent results I have achieved in playing the game with real cards however are either the result of some form of cheating without being aware of it, a flaw in the way I’m playing the game, lying to myself and/or forum members, or a spectacular run of luck that I would have as much difficulty believing as anyone. I don‘t think I have paranormal powers and won‘t be applying for the MDC anytime soon.

 

[AvalonXQ]

All I’m saying is that a game that doesn’t allow for a final stick or change choice isn’t a Monty Hall game and should be ignored.

But Monty wouldn't have ignored those games; he would have revealed the other card instead.

As my earlier list showed, and you appear to have ignored, every time you threw out a game is an example of a set up that you would have won by switching in Monty's game. By throwing out a specific subset of games, you're changing the rules and the odds.

 

[ynot]

And what I'm telling you is that in the version you played by yourself, it switching doesn't improve the odds. That strategy (and all others) has a 50% chance of winning.

That's fine, but playing the game that way changes the odds (to 50%).

Don’t know why you say the odds are 50%. The first choice is 1 in 3. If the red card is then revealed as being one of the unchosen cards then the game is over and there is no stick or change choice (the odds remain at 1 in 3). Surely the odds can only change if there is a second stick or change choice.

Please explain how the odds can be 50% if the red card is prematurely revealed or the first choice isn't changed.

 

[ynot]

But Monty wouldn't have ignored those games; he would have revealed the other card instead.

As my earlier list showed, and you appear to have ignored, every time you threw out a game is an example of a set up that you would have won by switching in Monty's game. By throwing out a specific subset of games, you're changing the rules and the odds.

I CAN’T do what the REAL MONTY can do as I CAN’T reveal a black card every time (I have no paranormal abilities). When playing with myself I HAVE to be both contestant and Monty. The ONLY way I can replicate what the REAL MONTY would do is by ignoring all games when I don’t do what the REAL MONTY would do (reveal a black card). If a red card is prematurely revealed before the switching option then the switching option effectively ceases to exist. Ignoring invalid games isn’t changing the rules and odds of the valid games. Not ignoring invalid games would be though.

I haven’t ignored what you have said, I don’t agree with what you have said.

 

[Pulvinar]

I CAN’T do what the REAL MONTY can do as I CAN’T reveal a black card every time (I have no paranormal abilities)...

You can by delaying turning over the card as Dan O said above-- just follow his instructions to the letter.

 

[GlennB]

I CAN’T do what the REAL MONTY can do as I CAN’T reveal a black card every time (I have no paranormal abilities). When playing with myself I HAVE to be both contestant and Monty. The ONLY way I can replicate what the REAL MONTY would do is by ignoring all games when I don’t do what the REAL MONTY would do (reveal a black card). If a red card is prematurely revealed before the switching option then the switching option effectively ceases to exist. Ignoring invalid games isn’t changing the rules and odds of the valid games. Not ignoring invalid games would be though.

I haven’t ignored what you have said, I don’t agree with what you have said.

It isn't an "invalid game", it's simply that your self-test has no way of dealing with it as would occur in reality (Monty would sort it out).

You may not throw out that trial. It's a real, live , 1/3 possible deal of the cards.

You've arrived at a point now where you're defending a flawed experimental design

 

[ynot]

You can by delaying turning over the card as Dan O said above-- just follow his instructions to the letter.

You're absoluetely right! I hadn't read and understood Dan O's method fully. It's a better and quicker solution than ignoring and one that can't be accussed of inadvertently cheating in any way. Thanks Dan O.

ETA - All I really need to do anyway is check if my first choice from 1 in 3 is correct or not. If it's correct I would have been better not changing. If it's wrong I would have been better changing. I should be incorrect twice as much as I'm correct. So far with cards I've been incorrect about as much as I've been correct. I don't expect anyone to accept I'm telling the truth (I wouldn't).

Hmmmm . . . Just realise that my last comment means that when I was playing Monty and revealed a red card it meant that my first choice was wrong and should have been counted as a loss.

Doesn't help to explain the results in the games with my grandson though.

Happy to report that when I check how well i can beat 1 in 3 odds using cards I can only do so in around 1 in 3 times.

Just checked the Lotto results and I'm equally useless at beating 1 in 3,83,8300 odds. Good thing that I only play for entertainment purposes

 

[Paul C. Anagnostopoulos]

That's exactly what I did.

Huh? Could you post your table of enumerated games?

~~ Paul

 

[Paul C. Anagnostopoulos]

I had a very sharp student who wouldn't accept a written demonstration of this, so I wrote an excel spreadsheet and did this process something like thirty million times. It picked a random door for the prize, had a contestant pick a random door as a winner, etc. It then decided whether you'd win by staying or switching and totaled up the frequencies at the end. It worked out exactly as expected 1/3 vs 2/3 for stay vs switch. It was no great surprise, but for anyone who wanted more trials using an unbiased system, there ya go, I did 30 million of them. (As I recall I had to keep resetting the worksheet I did it in and adding up the results separately; would excel even let you make 30 million rows?)

My favorite explanation for why it's better to switch:

When Monty opens a door with a goat and offers you the chance to switch, he has effectively offered you both of the other doors: one by revealing it and the other by letting you pick it. It is clearly better to take two doors than your original one door.

~~ Paul

 

[Galteeth]

While they are testing this piece of mathematics would anyone like to test that a flipped coin comes down heads about 50% of the time and a die roll comes out as a "6" about 17% ?

Not if it's "tails up" when you flip it