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

[Jack by the hedge]

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.

(my bolding) That's the error in your method. The method selectively discards games which you would have won by switching if you had been playing with another person.

If you turn over red you must not discard that game. You must consider that if you had been playing against Monty he would inevitably have picked the other card and then you would always have won by switching or lost by sticking. So such games must count towards the total of those where switching wins.

 

[sol invictus]

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.

Here:

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.

In case you can't see, since you throw away games of the third type, of the games you keep, 50% are wins and 50% are losses.

Does that clear it up for you?

 

[malicus]

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

And extend this to a pack of cards. Ask someone to pick the Ace of Spades from a face down pack. Then take away 50 cards of which you know are not the ace. Happy to stick? 50/50 chance yeah? Feel lucky punk??

.....is how I saw the dawn of realisation.

 

[alfaniner]

The key is that Monty always reveals a loser -- he knows which is the winner. It is not a double-blind puzzle.

 

[GlennB]

The key is that Monty always reveals a loser -- he knows which is the winner. It is not a double-blind puzzle.

Yes. Some interpretations of the puzzle get mixed up on this, including (iirc) the original presentation to the unwary public

And the whole business of restricted choice has led me into many a 'vigorous discussion' with bridge-playing friends.

 

[Big Al]

I'll tell you what: let's play a little game. I'll put $100 in an blank envelope, and you put a $50 bill in another and a $10 bill in another. We'll leave the envelopes unstuck, and I'll walk into another room. I'll put the numbers 1, 2 and 3 on the envelopes and stick the flaps down. I will sign a legal document swearing that each envelope only holds one bill of either $10, $50 or $100 only in each of the three envelopes. Semantics are not involved. One envelope has $10, one envelope has $50 and one envelope has $100. I am allowed to know which bill is in which number envelope.

All you have to do is pick an envelope, and I'll open another one to reveal a $50 or $10 note. Then, you can switch or stick. Open the envelope you've chosen, and you win the $160 pot if the envelope holds the $100 bill. Good luck.

The point is, you have a 2/3 chance of being wrong when you first choose at random. My choice of an envelope (that I know is not the winner) is NOT at random, and so does NOT change your chance of having been wrong in the first place.

All I have done is to remove a wrong choice, reducing the chance that the envelope you didn't choose is wrong... but I know the chances are that you'll stick with your first (2:1 likely wrong) choice. People tend not to like changing their minds.

If this makes no sense to you, please sign an affidavit to play this game with me 250 times. If you agree, I'll happily pay first class travel for you and put you up in a five star hotel for the single day this will take.

Easy money.

 

[WhatRoughBeast]

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 - Actually, it's not all that strange. It's just that getting a random shuffle of 3 cards is really difficult. And yes, you really got hammered by your technique. The odds of revealing the red card once in 21 tries is about 498 million to one. I suggest you do this instead. Take a full deck, remove the aces, discard the ace of diamonds and shuffle the remaining 48 cards thoroughly. Make sure you cut after you shuffle. Have your friend/grandson go through it. Look at each card in turn until you come to one which is not a diamond. Put down the ace of that suit as Card 1. Look at successive cards until you find one of a different suit (but not a diamond). Put down the ace of that suit as Card 2. Put down your last ace as Card 3. You now have the aces of hearts, spades and clubs as your cards.

Now try your game. For each successive trial, reshuffle and cut your deck and select the aces as above.

And you can do this by yourself. Shuffle and select aces. Tally whether Card 1 is red or black. Now look at the other 2 cards. Is one of them black? Well, yes. Every time. This means that you can always show a black, so it tells you nothing about how the other two cards are distributed. The possible distributions look like this:

RBB - show either card 2 or card 3 - either way, holding wins
BRB - show card 3 - switching wins
BBR - show card 2 - switching wins

 

[often mrunderstood]

Try this. Since you will always switch. Make your first pick. Don't look.

Turn over the other 2 and if either one is red you win. Because Monty always throws out one loser of the 2.

Repeat. Record 20-30 results.

The game is the equivalent of asking if you want 2 cards or 1. If you ask for two and one is red, you win.

 

[Stomatopoda]

Play with 100 cards. 1 winner 99 losers. Pick a card. Dealer knows where the winner is and will toss out 98 losers, then ask you whether or not to switch.

Does that make the answer more obvious?