Explain to me this probability puzzle

[jerichothebard]

I think you are all making it too complicated.


If we are assuming that:
Monty will always remove one of the doors, and
that he will never eliminate the door with the car, and
that he will always give you a second guess,


Then there is actually only one round here. What happens with the first guess is entirely irrelevant. It's just showmanship.


Here's why:

It is inevitable [we assume] that Monty will eliminate a goat door. No matter whether you initially choose the door with the car, or one of the goat doors, Monty will ALWAYS eliminate a goat door. That's part of the game, and what makes it exciting.

But, since it's inevitable, that door doesn't matter.

No matter what happens, the first guess always has exactly the same result: Monty removes one (wrong) choice and gives you another chance.




The game doesn't start until that first door is out of the picture.

Which means that the game isn't about guessing one out of three doors, because you are always assured that one of those three possibilities will be removed and you will get to guess again.



Which means you are always guessing between 1 of two doors. A 50/50 proposition. Whether you "switch" or "stay", you really are just guessing between two doors.


so it doesn't matter. Stay, switch. It's irrelevant. You'll win half the time.





as for the one-in-a-million guesses, that's a totally different problem.

The difference is in the base assumptions - in the first case, 1/3 of the options are payoff. In the second, 1/1000000 options is a payoff. In this case, your switch analysis is correct - because he is narrowing the field by so many more than just 1/3 of the options.


In order for it to be the same problem, 333333 out of the original million would have to be payoff doors, and he would only get to eliminate 333333 of them - then you have a 333333/666667 chance - or just about 1/2 again.

jtb

 

[Pielorinho]

Jericho, you're wrong . The game begins before a door is eliminated, because your first choice is 2/3 likely to be wrong. Then MOnty essentially bundles the other two choices, and says, "If you chose wrong the first time and switch, you're guaranteed to be right: I'll eliminate all but one door from the remaining choices to guarantee that, but won't eliminate the right door."

If you don't believe me, have a friend test you on it. Here's the procedure:
1) Friend chooses a number at random, using a D6 and dividing in half, rounding up.
2) You guess a number between 1 and 3.
3) If you guessed right, friend chooses one of the numbers you didn't choose at random and eliminatees it.
4) If you guessed wrong, friend chooses the wrong number you DIDN'T choose and eliminates it.
5) You switch your choice to the remaining number your friend didn't eliminate.

Do this and keep track of your successes. Do it 100 times, and I betcha anything your successes are closer to 66% than to 50%.

Seriously, try it. This isn't just theory: this is falsifiable.
Daniel

 

[Gnimish88]

The problem with all this is...

I think someone already mentioned this, but I think it bears repeating. While it is true that your initial guess has a 1 in 3 chance of picking the car and that your second guess has a 1 in 2 chance of picking the car, you are forgetting a pretty important point:

The OTHER 1 in 2 chance is that the door you already choosen has car behind it.

If you like, by eliminating one of the choices, the probability that you made the right choice to start with went up from 1 in 3 to 1 in 2.

 

[jerichothebard]

Jericho, you're wrong . The game begins before a door is eliminated, because your first choice is 2/3 likely to be wrong. Then MOnty essentially bundles the other two choices, and says, "If you chose wrong the first time and switch, you're guaranteed to be right: I'll eliminate all but one door from the remaining choices to guarantee that, but won't eliminate the right door."

No, I'm not - what he says, in effect, is "Here's two doors. One of them has a goat. One of them has a car. Pick one."


No matter what happens, the outcome of the first round is identical. One wrong door is eliminated, and you are presented with two doors. One wins, one loses (unless you like goats as much as cars. In which case more power to you.) Therefore, the first trial is meaningless! It has no more bearing on the outcome of the game than the color of his tie. It's as much a premise of the game as the initial setup with a car and two goats.

Your first guess simply does not matter, because it always has the same result!





You have to consider the odds of the second round - the only round! - on their own merits, and the fact is that they are one-in-two. Two doors, one wins. The odds of the whole game can be boiled down to that, because the first round doesn't change anything.


Try it:
http://www.maestrakara.com/srjosh/chancegame.php

jtb

 

[Pielorinho]

I'm not sure what that game you linked to does; can you describe its coding in detail?
Let's try this a different way:

I've got three eggs in my hand. Two of them are hollowed out, but you can't tell the hollow ones from the real one, by the way I'm holding it. You're trying to choose the real egg.

You make your choice.
1) What are the chances that you chose correctly?
2) What are the chances that I'm holding the correct egg in my hand?

Now I hold two eggs in my hand. I choose one of the eggs in my hand that is hollow (I can tell them apart by their weight) and smash it against my forehead, demonstrating its lack of yolk to you.

3) What are the chances that I'm holding the correct egg in my hand?

If you gave a different answer to #2 and #3, please show your work .

The reason why your first choice matters is that it constrains me in which door/egg I remove.

Daniel

 

[Pielorinho]

Actually, I just played that game 30 times, and I was always within 1 game of winning 2/3 of the time (2 out of 3 games, 4 of 6, 11 of 18, 19 of 27, and 21 of 30). So it looks like it was programmed correctly and demonstrates that by switching, you'll win 2/3 of the time.

Daniel

 

[jerichothebard]

I'm not sure what that game you linked to does; can you describe its coding in detail?

$include=false;

if ($score=="") { $score=0; }
if ($game=="") { $game=0; }


if ($choice == ""){
	$winner=rand(1,3);
	//echo $winner;
?>
<form method='post' action='chancegame.php'>
<input type="submit" name="choice" value="1">Door #1<br />
<input type="submit" name="choice" value="2">Door #2<br />
<input type="submit" name="choice" value="3">Door #3<br />
<input type='hidden' name='winner' value='<?php echo $winner; ?>'>
<input type='hidden' name='games' value='<?php echo $games ?>'>
<input type='hidden' name='score' value='<?php echo $score ?>'>
</form>
<?php
} elseif ($over == false) {
	$count=0;
	echo "<form method='post' action='chancegame.php'>";
	for ($x = 1; $x < 4; $x++){
		if (($winner == $x)) { 
			echo ("<input type='submit' name='choice' value='$x'>Door #$x<br />\n");
			$count++;

		} elseif (($choice == $x) && ($winner != $x)) { 
			echo ("<input type='submit' name='choice' value='$x'>Door #$x<br />\n");
			$count++;
			
			
		} else {
			if (($include == true) && ($count < 2)) {
				echo ("<input type='submit' name='choice' value='$x'>Door #$x<br />\n");
				$include = false;
				$count++;
			}


			if ((rand(1,2)==1) && ($count < 2) && ($winner==$choice)) {
				echo ("<input type='submit' name='choice' value='$x'>Door #$x<br />\n");
				$count++;
			} else {
				$include = true;
			}
		}
		
	}
	echo ("<input type='hidden' name='winner' value='$winner'>");
	echo ("<input type='hidden' name='over' value='true'>");
	echo ("<input type='hidden' name='games' value='$games'>");
	echo ("<input type='hidden' name='score' value='$score'></form>");
	

} else {
	if ($choice == $winner){ 
		echo ("You Win!<br />");
		$score++;
		$games++;
	} else {
		echo "Sorry!<br />";
		$games++;
	}
	echo ("Your current score is " . $score . " out of " . $games . " games<br />");
	echo ("<a href='chancegame.php?score=$score&games=$games'>Play Again!</a>\n");
	
}
?>

In short, it picks a winning "door"
you pick a door
it takes away a door at random, making sure to leave the winning door and your chosen door (which may or may not be the same)
You pick another door
it scores you and you can start again.

Let's try this a different way:

I've got three eggs in my hand. Two of them are hollowed out, but you can't tell the hollow ones from the real one, by the way I'm holding it. You're trying to choose the real egg.

You make your choice.
1) What are the chances that you chose correctly?

1/3

2) What are the chances that I'm holding the correct egg in my hand?

1/1 - that's a given

Now I hold two eggs in my hand. I choose one of the eggs in my hand that is hollow (I can tell them apart by their weight) and smash it against my forehead, demonstrating its lack of yolk to you.

3) What are the chances that I'm holding the correct egg in my hand?
If you gave a different answer to #2 and #3, please show your work .

Well, your instructions are a little unclear, but assuming that the egg you smashed was one of the original three, and that it was empty, I have to say

3 eggs to start
-1 egg smashed
-------------------
2 eggs left


1 egg is full, 1 egg is empty (by definition)

1 full egg / 2 eggs total = 1/2, or 50:50.

See? First egg is irrelevant.





Another possible interpretation is that:

2 eggs to start
1 egg is empty and removed.

leftover egg must be full.

point?

 

[Pielorinho]

Jericho, did you play your game, using a consistent strategy? What results did you get?

1/1 - that's a given

D'oh! I forgot a line. After you made your choice, I put your chosen egg in front of you, so now I only hold the two eggs you DIDN'T choose in my hands.
Picking up from there, with one more change in italics:

2) What are the chances that I'm holding the correct egg in my hand?

[I see a 2/3 chance that I'm holding the correct egg]
Now I hold two eggs in my hand. I choose one of the eggs in my hand that is hollow (I can tell them apart by their weight) and smash it against my forehead, demonstrating its lack of yolk to you. I THEN PUT THE SMASHED EGG BACK IN MY HAND.

3) What are the chances that I'm holding the correct egg in my hand?
[given that I'm holding exactly the same eggs I was holding after #2, I see a 2/3 chance that I'm holding the correct egg]

Do you still think that the chances after #3 are 1/2? If so, how did my smashing an egg I knew to be false change the odds of my holding the correct egg? I'm still holding exactly the same eggs that I was holding after question #2.


Daniel

 

[jerichothebard]

Jericho, did you play your game, using a consistent strategy? What results did you get?

yes - within one win of 50%, using your switch strategy, then the same with a hold-em strategy. Although I only played about 20 tries on each, so I suspect that it wouldn't be exactly statistically valid.

D'oh! I forgot a line. After you made your choice, I put your chosen egg in front of you, so now I only hold the two eggs you DIDN'T choose in my hands.
Picking up from there, with one more change in italics:

2) What are the chances that I'm holding the correct egg in my hand?

[I see a 2/3 chance that I'm holding the correct egg]

right.

Now I hold two eggs in my hand. I choose one of the eggs in my hand that is hollow (I can tell them apart by their weight) and smash it against my forehead, demonstrating its lack of yolk to you. I THEN PUT THE SMASHED EGG BACK IN MY HAND.

3) What are the chances that I'm holding the correct egg in my hand?
[given that I'm holding exactly the same eggs I was holding after #2, I see a 2/3 chance that I'm holding the correct egg]

Do you still think that the chances after #3 are 1/2? If so, how did my smashing an egg I knew to be false change the odds of my holding the correct egg? I'm still holding exactly the same eggs that I was holding after question #2.


Daniel

Ahh... I see.


The odds change when you smash the empty egg, because you are changing the nature of the game.

Yes, you still have two eggs in your hand, but only one of them is a POSSIBLY FULL egg. One of them is provably not full. Even if you replace it in your hand, it is no longer part of the game - because the odds of my choosing the smashed, losing egg are ZERO, given the assumption that I am actually trying to win.

See, it isn't about whether you hold the winning egg in your hand - it's about whether or not I PICK the winning egg.


There are now two eggs which might be full. One of them is in front of you. One of them is in front of me.

So again, we're down to two eggs, one of which wins, one of which doesn't.

All your smashing and switching and putting down boil down to a simple coin toss.

So smash the other egg, collapse the waveform, and let out the cat.

jtb

 

[eris404]

I don't know if this is where your friend got it from, but this puzzle was featured in the book The Curious Incident of the Dog in the Night-Time, a rather excellent book that recently came out in paperback. I can't recommend the book enough. Maybe my favorite book of last year.

How odd, this the third time this week someone has brought up that book. I guess I should read it now.

I'm kinda surprised this thread is still going on. I've been enjoying the responses regardless.