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View Full Version : Let's Make a Deal (Paradox Followup)

Moozh
10-12-2004, 02:05 PM
This is inspired by RocketManJames' Paradox Post. To be honest, I tried going through that thread, but the language in there got too messy. I think this is a similar problem, but I find it easier to visualize.

You're on the show Let's Make a Deal and you've been given the chance to select a prize from one of three doors. Only one door holds a prize, the other two are empty.

You select a door (at random). After you choose, the host goes to one of the two remaining doors and opens it, revealing it to be empty (regardless of what you choose, the revealed door will always be empty).

You are then offered a choice to select again. Do you

B) Pick the opposite door
C) Choose again randomly

Does it matter, and why? I'm guessing that many of the people who frequent this forum have heard of this problem and know the answer. I've done simulations for answers A and B, but not C.

EnderW27
10-12-2004, 02:29 PM
Simulation or no, I can't imagine C would have a greater EV than B.

JFB37
10-12-2004, 03:12 PM
It's an oldie but a goodie. The answer is that you always switch. Here is why. . .

You have a 1/3 chance to have selected the prize on your first pick. If you switch, you get nothing. Of course, if you picked a blank and switch you get the prize. Because one door gets taken out after you pick, it changes your odds.

Here is what happens if you don't switch:
1/3 chance you get the prize. End of story.

Here is what happens if you do switch:
1/3 chance you picked the prize and switch to nothing.
2/3 chance you did not pick the prize and switched to it. There is nothing else to switch to, because he showed you the open door.

In short, switching gives you a 2/3 chance of winning. Standing pat gives you 1/3.

TomCollins
10-12-2004, 03:57 PM
If there were 50 doors, and Monte opened every door but 1, and the one that you chose, it is incredibly obvious that you switch. Can you see why? I'll let others elaborate.

Moozh
10-12-2004, 06:19 PM
Yes, that's a nice extreme example that illustrates the point. Now, obviously choice A gives you a 1/3 chance of winning. Not so obviously, choice B gives you a 1/2 chance of winning (now I'm not so sure...). I'm not sure about C, but I would hazard a guess that it would be halfway between the other two options, or about 5/12ths.

The part I've always struggled with is the proof for B being 1/2.

Assuming you pick one at random, 1/3 of the time you will guess correctly. Thus, the two remaining doors will be wrong, so when you switch, you will guess wrong. Thus, of the 1/3 you guess right, 100% of the time you will lose.

2/3 of the time, you will initially guess incorrectly. Once you do, the other wrong one will be revealed, and when you switch, you will guess correctly 100% of the time.

Thus, it seems like you should win 2 out of 3 times when you automatically switch. But, if I remember correctly (an now I'm not so sure), the value for B is 1/2.

The more I think about it, the answers should be 1/3 for A, 2/3 for B, and 1/2 for C. So is the 'proof' correct?

34TheTruth34
10-12-2004, 06:39 PM
[ QUOTE ]
Here is what happens if you don't switch:
1/3 chance you get the prize. End of story.

[/ QUOTE ]

maybe you should reread the story and take a closer look at the ending???

ALav10
10-12-2004, 06:46 PM
Moozh, I'm not sure where you got B = 1/2, but your last line is correct. There are many different "proofs", but the easiest is the simple logic you worked through. (I tought a course on basic game theory this summer and we looked at this problem extensively) Heres another way to look at the problem:

Essentially, by the host opening doors, the probability of those doors being correct transfers to the remaining closed door. This is just another way to think about it, so in the original problem (3 doors; called the Monty Hall Problem) the probability of being correct when you switch is 2/3, since the door that the host opens transfers its 1/3 probability to the other. Similarly, in the 50 door problem, the probability of being correct if you switch is 49/50, since all 48 opened doors transfer their 1/50 to the last.

I hope that this alternate way of looking at this problem helps some and confuses none. Like I said earlier, there are many different ways to solve the problem.

ALav10
10-12-2004, 07:08 PM
The 50 door probelm makes the situation simpler. To test your understanding, heres one that makes the situation harder:

You have seven doors and behind one door is a prize. You get to pick three doors. Of the four remaining, the host opens three to reveal nothing. This leaves four doors closed, the three you initially chose and the one remaining that the the host didn't open. So your decision is to:

a) stay with all three of the doors you initially chose, or

b) switch to the one remaining door.

Which do you choose?

pudley4
10-12-2004, 07:24 PM
Why does this question show up here every month?

And why do people still get it wrong?

And the answer to the follow-up (w/7 doors) is:
<font color="white">you thought I'd give you the answer to such an easy question? yeah, right... </font>

Moozh
10-12-2004, 07:34 PM
I think I was confused because when I ran my simulations, I actually did it for A and C, not B. C is 1/2, which makes perfect sense. Ah well, case closed in my book.

10-12-2004, 08:31 PM
[ QUOTE ]
The 50 door probelm makes the situation simpler. To test your understanding, heres one that makes the situation harder:

You have seven doors and behind one door is a prize. You get to pick three doors. Of the four remaining, the host opens three to reveal nothing. This leaves four doors closed, the three you initially chose and the one remaining that the the host didn't open. So your decision is to:

a) stay with all three of the doors you initially chose, or

b) switch to the one remaining door.

Which do you choose?

[/ QUOTE ]

You switch, because 3/7 is less than 4/7.

mike1212
10-20-2004, 11:55 AM
[ QUOTE ]
[ QUOTE ]
The 50 door probelm makes the situation simpler. To test your understanding, heres one that makes the situation harder:

You have seven doors and behind one door is a prize. You get to pick three doors. Of the four remaining, the host opens three to reveal nothing. This leaves four doors closed, the three you initially chose and the one remaining that the the host didn't open. So your decision is to:

a) stay with all three of the doors you initially chose, or

b) switch to the one remaining door.

Which do you choose?

[/ QUOTE ]

You switch, because 3/7 is less than 4/7.

[/ QUOTE ]

That sounds right, any confirmation from anyone?

TomCollins
10-20-2004, 12:38 PM
He is correct.

SossMan
10-20-2004, 01:00 PM
Jesus, I thought we were going to set a record and go a whole month without seeing the goddamn Let's make a deal riddle

Now to go search for the "Should I fold AA preflop" thread and the "is party poker rigged" thread, and the "look at this bad beat" thread...

AceKQJT
10-20-2004, 01:20 PM