The "Let's Make a Deal" Paradox
 

OK - This is driving me bananas so I'm challenging you guys to explain this to me so I can convince myself that it's really true.

I was listening to NPR yesterday (weekend edition saturday) and they interviewed some math guy who explained the probability of Let's Make a Deal. The contestant chooses one of three doors, then Monty reveals an empty door and gives the contestant a chance to change their choice. It seems obvious to me that the chances at this point are 50/50. Even chances for switching or not switching. This guy says that's not the case. The chance of getting the right door if you switch is double that if you don't switch (2/3 to 1/3). But he didn't go on to explain any further. ARRRGH

So this morning I woke up Griff cause it was driving me crazy. Then I went online to look it up and found this link. I think I understand the way thry're explaining it but my intuition is still fighting me.

Basically, your chances of guessing the wrong one are 2/3. If you guess the wrong one, Monty must pick the other wrong one so by switching, you get the right one. I guess it all boils down to the fact that you decrease Monty's choices by picking the wrong one. How could you explain this to someone without p-ing them o?

Now let me answer my own question. I think I like this one. if we just increase the scale, it seems much more obvious. We have 100 lottery tickets (and one is the winner). I choose one. The probability that one of the remaining 99 is the winner is 99/100. Then someone removes all but one of those 99 saying that they are all losers. Then it seems obvious that you'd want to switch. I don't have the exact math at my fingertips for this one but my intuition feels much better about it. sigh.

BTW here's where I found this.

We did this one on the Cellar mk 2!

How do you know the applet doesn't cheat?

Well, strangely enough in 5 runs of the applet I stayed with my original guess all five times and won the $ 4 of those 5...

5 runs is not statistically valid and yet the initial intuition answer is reinforced not disproved.

Update:
I ran the simulation 12 times; 6 times without switching and 6 times with switching. Result: I won 4 times out of 6 in either strategy.

Applet schmapplet. I'm willing to make you a deal right now.

You put down $100, and I'll put down $500.

I'll lay out 52 cards, face-down, and I'll look at them while I lay them down. You have to pick the 3 of clubs. You pick your first card, but don't turn it over. I'll take away 50 other cards and show them to prove that they aren't the 3 of clubs. Now if you have the 3 of clubs, you win the pot, while if I have the 3 of clubs, I win the pot.

Agreed?

Methinks this be the fly in the ointment. The odds are 51 to 1 against me. No thanks.
After you remove the 50 cards, let me choose which of the 2 I want. And one of them is the 3 of clubs, your on.

Point: Missed
 

The point of my previous comment is that the demonstration applet doesn't help reinforce the argument on the website since most people won't run it a statistically valid number of times to confirm the non-intuitive strategy.

Were I that guy I'd remove it since it's not helping.

Re: Point: Missed
 

I, for one (not a statistically valid sample) felt compelled to run the applet a statistically valid number of times. Maybe you should try Vegas cause I think you just got lucky.

Hey Toad - Thanks for the card example - that's a good one. I think my intuition is catching on. Griff's Dairy farmer uncle will like this one at the next family gathering. He asked me one Thanksgiving what was the maximim amount of money you could make playing Jeopardy. Cool guy. Can't remember the answer tho.

Ummm... let's see...
You'd need to get all the questions in Jeopardy except a $100 one... that would be $8900. The last $100 question would be the Daily Double... you'd bet the whole shebang and enter Double Jeopardy with $17800.

You'd then proceed to get all the Double Jeopardy questions except for two $200 ones. That would be $17600, added to your take from the first round would give you $35400.

You'd then hit the penultimate question, a Daily Double, bet it all, and have $70800. And then the last question, the final Daily Double, another bet of everything, and you enter Final Jeopardy with $141600.

In Final Jeopardy, you bet it all once again and leave with a cool $283200.


But that's the best possible placement of the Daily Doubles. (I'm not even sure if they ever put them in the top row.) If you had the worst case scenario, you'd hit it on a $500 question and it would be your first question... so you'd end up with $500 after answering it correctly, then run the table, and enter Double Jeopardy with $9000. You'd then hit the two daily doubles under $1000 spaces as your first two Double Jeopardy questions, and have $36000. All the rest of the squares are worth $16000 for a total of $52000. Then you have $104000 after winning your Final Jeopardy bet.

What did I miss?? :3eye:

what you missed...
 

TAXES!!!!!! :D

Brian

For me, the "Let's Make a Deal" question took a few minutes of thinking the first time I heard it before I grasped the way the odds work. None of the explanations above would have helped me much though, so I thought I'd offer an explanation from a different angle for those whose brains work like mine.

It helps to turn the question around in your head a little bit... let's assume there are two players, one who plans to stick with their original door, and one who plans to switch.

The one who sticks with their original door is effectively trying to pick the right door from the getgo, and has a one-in-three chance of being right. But the player who intends to switch doors is basically trying to pick one of the wrong doors from the getgo (since they don't intend to stick with that door), and this player has better odds... a two-out-of-three chance of successfully identifying a wrong door.

Even if the players don't decide whether they're switching until the question is posed, the odds don't change.... if you switch, then your original selection was effectively trying to identify one of the wrong doors, and your chances of successfully selecting a wrong door is about 33.3% better than a player who sticks with their one-in-three chance of guessing the right door originally.

Make sense? It can be a hard one to wrap the brain around.

And it has never made sense to me, for a simple reason:

I pick Door #1. Monty opens Door #2, and Rosie O'Donnell is standing behind it, so obviously I didn't want that one. He offers me the choice of keeping Door #1, or switching to Door #3.

What I do not, and probably will never understand is how the second choice is not _independent_ from the first one. Yes, Monty has information that I do not as to which door has the prize -- but the mere fact that I do not have that information makes it irrelevant to my choosing process. I don't know whether Monty is trying to help me or screw me over by making his offer, and I have no way of finding out short of making a blind choice; all I _do_ know is that between my first choice and my second, the situation has changed and my options are different.

When I make my second choice, I am making an independent choice between two doors. One has a prize. One has a goat. A third door has a goat, but that door is _no longer an option_, so I don't grasp why it is figured into the odds when I choose again. (Really, if Monty always offers you the choice to switch, the first choice is irrelevant; he will always open a goat door, so you'll always have a final choice that narrows it down to two.)

I've taken math courses through Calculus 3. I've taken Probability and Statistics courses. I've read lots of explanations of the "real" odds for each door... and I _just don't get it._

I don't get it either, but such a thing does not prey upon my mind. I, personally, might be excited to win the family of goats.

My enjoyment of the show was to see what goofy costumes the contestants were wearing, and I also loved the segment where Monty would look around the audience for an assortment of bizarre objects that women might have in their purses ... "I'll give you $50 for a hot dog ..."

In the two player version they ran sometimes, it was best to _NOT_ switch. The way that one worked, two contestants would each pick a door. Monty would then open a losing door which belonged to one of the contestants, and offer the other contestant the chance to switch to the remaining door.

Er, ok... let me try another way of explaining. This is fun, because it's challenging to explain it, and bends my brain like a pretzel.

I think we can agree that before the player selects a door, each one has a 1/3 chance of being the winning door. When a player selects a door, they basically divide the doors into two groups... the door they have chosen, and doors they have not chosen. The door they have chosen has a 1/3 chance of being a winner, and the doors they have not chosen, as a group, has a 2/3 chance of containing a winner.

So Monty opens one of the doors in the "doors the player did not choose" group. Now, even though you know what is behind one of the doors in that group, the "doors the player did not choose" group still has a 2/3 probability of containing the winning door, you just know not to choose the one that had it's contents revealed. So you go with the remaining door in that group, and use that 2/3 odds to your advantage.

Intuition wants us to break down the problem and say that after a door is opened, and we have one winning door and one losing door before us, we have a 50/50 chance. But it's better to think of the do-I-switch choice as "Was I right on my initial guess?" Chances are, you weren't. It's a 2/3 chance you were wrong, as a matter of fact.

Any better?

OK, I follow your logic that the door I chose is 1/3 and the other is 2/3 *IF* you consider the *WHOLE* game. But I still don't understand why the odds are figured on the whole game and not on the final choice.
Can you explain to me why if I win the "Daily Number" and I bet that same number tomorrow, the odds are still 1/1000. Why is the previous day disregarded in this case?:confused:

Wait..I see..If you take the day before into account it is 2/2000 which is 1/1000. Nevermind.:rolleyes:

Well, the problem with the 50/50 intuition is that it assumes that the first choice has no bearing on the second choice. But it does. Consider the two possible scenarios for the initial door selection:

1) You select the door that does contain the prize, which will occur one in three times. Monty can open either of the other two doors at random. His choice is unaffected.

2) You select a door which does not contain the prize, which will occur two in three times. Monty is forced to open the other non-winning door.

Scenario #1 will happen about 33% of the time, and scenario #2 will happen about 66% of the time, right? So, after you make your initial choice, it is safer to bet that #2 happened. So accordingly, it is safe to bet that he was forced to open the other empty door, and therefore it follows that the third door, the one you did not pick, contains the prize. I hope I explained that well, my brain is getting pretty tired.

The reason the Daily Number's odds don't change is that one day's selection has no effect on the other days' numbers, it's always a random draw. Statistics don't always follow intuition... for instance, if you flip a normal coin 100 times and it surprisingly comes up heads every time, what are the odds that it will come up heads again on the next flip? It seems outlandish, like it would become less and less likely for it to turn up heads. But against (my) intuition, the chances are still exactly 50%. That's because each flip of the coin is a random result independent of all the other flips. Wierd.

Anyway, have a good weekend everyone.

Good lord I leave for a month and the cellar is suddenly discussing things so far over my head I didn't feel it go by. Well maybe it's not that bad but still way to deep for me.

by the way Hi everyone nice to be around again.

Welcome back, Cam.:D

Don't let the Monty-knows-something-I-don't issue trick you. The fact that he knows where the prize is only means that he will never accidentally open the door with the real prize when he's showing you one of the "other" doors.

I've seen several explanations that say something like "but Monty knows where the prize is, so he picks the other one." This is just worded poorly. He isn't trying to help you or screw you, he's just playing the game too, and it has nothing to do with the mathematical odds.

Monty's knowledge is critical. If he did not have it, he would sometimes open the door with the prize. In those cases he did not, you would have a 50/50 chance whether you switched or did not.

But... as always, I don't grasp the causal relationship between the two choices.

Let me phrase it this way:

Game #1: Monty shows Joe Contestant three doors. Joe picks Door #1. Monty says "Here's what was behind Door #2," and opens it to reveal a goat. Monty says "Would you like to switch to Door #3?"

Game #2: Monty shows Joe Contestant two closed doors and one open one. The open one contains a goat. (Alternatively, he shows three closed ones, but opens one goat door before Joe makes a choice.) Monty says, "Which one will you choose?"

In my mind, these two games are identical in terms of probability; Game #1 merely has additional non-binding gamesmanship before it's time to choose "for real."

In Game #1, you make an initial choice, but it is in no way binding; the only effect it has (other than suspense for the audience) is to help Monty select which goat door he's going to open. (If you guess the prize door initially, he can open either; if you guess a goat door initially, he will open the other goat door.) Either way, the ambiguity has been removed, and you are now faced with a situation equivalent to Game #2 (two doors that you know nothing substantial about, one prize, one final choice to make).

This doesn't change the fact that there's a 2/3 chance that the prize will be behind one of the two doors you didn't initially choose; that much is simple mathematics. But the situation _after_ Monty opens the door is not the same situation as _before_ Monty opens the door; to claim that the second choice (viewed in a vacuum) isn't a fifty-fifty chance in and of itself is to play clever semantic games with the problem as a whole. That 2/3 chance is tempered by the 100% certainty that one of its two options will be eliminated before the actual choice is made.

Dammit, that's exactly what I said. Quit trying to confuse the issue.

Vsp thought Monty's knowledge was coming into play in that he is trying to influence the outcome of the game. That is not the case. He simply follows the rules for his role. A trained monkey could do his job and it wouldn't affect the coutcome.

Part of the problem is that the scale is so small that intuition misleads us. Here's what basically happens in the three-doors scenario.... the contentant picks one door, then the host opens all but one of the remaining doors. That remaining closed door was unopened because:

A) it contains the prize (likelihood: x-1/x), or
B) you selected the door that contained the prize, so an empty door was randomly opened from those that were left (likelihood: 1/x).

...where x=3 in the Let's Make a Deal context.

So imagine that there are not three doors, but a thousand (x=1000). The contestant picks a door, and like in the three-door game, Monty opens all of the other doors except one... 998 doors spring open to reveal goats. Now, like in the three door game, there are only two closed doors to choose from. But are the chances 50/50? No, because that fact that there are only two doors now does not change the fact that your chances of guessing right were one in a thousand before those 998 doors opened. You can still bet that one of the 999 doors you didn't pick is the winner (likelihood 999/1000, or 99.9%), you just happen to know what is behind 998 of them, so you know not to pick those.

Any better? I know my explanations aren't the greatest.

EDIT: I had the little formulas reversed in my A/B scenarios... oops

The causal relationship you are looking for is this: the door Monty opens is affected by the door that you choose.

You can't view the second half in a vacuum, because your initial choice determined which door he would open (or, 1/3 of the time, it at least narrowed down the choices).

In Game 2, the odds are indeed 50/50, because the door that Monty opens was chosen between the two losers at random. But in Game 1, 67% of the time Monty <B>is forced by the rules of the game</B> to open the only remaining losing door.

It still doesn't entirely make logical sense to me. x can be 3, 1000 or 3.4985767E+12, but if (x-2) are eliminated before you make a choice, you're still choosing between two options in the end, one of which is an unknown and the other one of which is (998 losers + one unknown). You make a choice earlier in the game, of course, but that choice doesn't determine whether you win or lose.

If I was applying for a job with 999 others, and only two of us were brought back for a second interview (myself being one of them), I'd be a hell of a lot more confident after the reduction than before...

Well, that's not an effective parallel, since it's a decision based on information about all of the choices, and there is no random element.

Say you're outside with some friends, staring at a beautiful starry sky with a new moon. Your friend tells you that she's chosen one star from the sky, and wants you to guess which one she is thinking of. You contemplate the question for a moment, then, from the thousands of visible stars, you indicate a star in the western sky. She responds with, "Ok, I was thinking of the one you pointed out, or..." she points to another, near the south horizon... "that one." Which is more likely? There are only two stars now to choose from, so you might think the chances are 50/50, but in order for it to be the star you pointed out, you would had to have correctly guessed from all the stars in the sky.

What are the chances that you guessed right, and she just picked some random star as an alternative? And what are the chances you guessed wrong, and she's pointing out her real star as an alternative?

Another example... say you entered a contest for a $1 million prize. The company conducting the contest is bringing every entrant into their office, so you go wait your turn in line. Once you get in, they tell you that the winner of the $1 million was either you, or some lady named Blarda in Columbus, OH. You have to pay them $1000 to find out, or keep your money and risk losing the $1 million. Do you think you ave a 50/50 chance of being the winner, or do you just keep your $1000?

I don't know about that -- I've been involved in some seriously random-seeming job interview processes...

Chick's head games have no place in a discussion mathmatical odds.:p