# The Monte (Door # X) Dilema

We all know the drill. You pick door 1. Monte opens door 3, which is a loser. Are the odds in your favor if you switch picks to door 2, or do you stick with door 1?

The tons of rhetoric behind this issue is ludicrous. Everyone seems to be missing the simple answer.

First, we must know UNDER WHAT CIRCUMSTANCES Monty shows the losing third door. There are three possibilities:

1. He only opens the odd (extra losing) door when the contestant is correct.

This info would soon be detected and exploited by the public, so it can be dismissed as a viable option.

1. He only opens the odd door when the contestant is wrong.

Ditto

1. He opens the odd door as dictated by the hair that happens to be up his ass at the moment (randomly), OR he ALWAYS opens the odd door (since, after all, there is ALWAYS a loser door to open at this point).

Question: Do I change or stick with my original pick?

Answer: It doesn’t matter. Hell, if for no other reason, think about this…if a certain “pattern” of revealing doors in certain circumstances existed, then the “Let’s Make a Deal” staff would’ve figured it out long ago…

Reasoning:

Once Monte opens one of the loser doors, your odds of winning BECOME 50/50. Your OVERALL odds of 33.3% haven’t changed, only NOW you know that AT THIS POINT you have a 50/50 shot. One door wins, one door loses. Period. Initially, your odds are 1/3. One door wins, two doors lose. HOW THE WINNER IS REVEALED has NO EFFECT on your initial odds of winning. Monte can always open one loser door and make it APPEAR that you now have a 50/50 shot, but in reality, your initial 1/3 odds haven’t changed. This is so painfully obvious that I can’t believe the amount of discussion that this topic has generated!

Case in point:

A coin flipped 100 times will, on average, land on heads/tails with a 50% rate. Say I flip a coin 10 times, and every time in comes up heads. Would you put your money on tails for the next flip? You might, but you’d better understand that there is still a 50/50 chance that it will come up heads again. The coin DOESN’T KNOW that it just came up heads for the 10th time, now does it?

Monte has 11 doors. You pick one. Your odds are 1 in 11. You’ve got one door, he’s got 10. He opens 9 of his 10, all losers. Now, he’s got 1, and you’ve got 1. One wins, one loses. Your odds AT THIS POINT are 50/50. Do you switch doors with him? Well, you can if you want, but it still doesn’t increase your OVERALL odds of winning. Your OVERALL odds of winning are still 1 in 11! Your OVERALL odds of winning do not change simply because you know how many losers there are AFTER THE FACT.

You buy a raffle ticket at a school. 1000 tickets were sold. Your odds of winning are 1 in 1000. You go to the raffle event. They call out 500 losers. You’re not one of them. Your odds are NOW 1 in 500 that your name will be called as a winner. This does NOT change the fact that your OVERALL odds are STILL 1 in 1000. Even if they call out all but you and someone else remaining to win. Your odds of winning NOW are 50/50. But, your OVERALL odds of winning in the first place are still 1 in 1000.

It doesn’t matter whether or not Monte reveals a loser door, as long as he does it without bias or circumstance. Your OVERALL odds of winning are 1 out of 3, no matter which way
you slice it. One door wins, two doors lose AT THE TIME YOU PICKED. NOTHING CHANGES THIS.

Those other arguments about “A boy shows up one day, and with a birth rate of 50/50, what are the odds that a girl will show up the next day” are ALL BS!! The coin didn’t remember what it “flipped as” last, why should a previous gender determine which gender should show up next?

50/50 is 50/50 is 50/50 no matter what happened in the past. The past doesn’t mean shit. The same can be said for any other win/lose ratio.

How do you think casinos stay in business? They stack the odds slightly in their favor, and live fat. If past occurrences had any actual effect on present odds, then the casinos would change their odds daily.

Coins don’t remember. Roullette wheels don’t remember. Past occurrences mean squat.

“A girl showed up yesterday, so odds are a boy will show up today”. Bullshit. If the birthrate is 50/50, then the odds of a boy/girl showing up today is 50/50, no matter who showed up before.

This seems so obvious…how can intelligent people get confused here???

JJ Richard

a) You talk a lot without saying much.
b) Your terminology is such that I have no idea which side you stand on. (What is “Your odds” vs. “Overall odds”?
c) I’m fairly sure you’re on the 50/50 side, which is the wrong side.
d) Everybody with half a brain has discussed this to death.
e) Our Host has covered this as well.

JJ, I’ll confess I sort of lost the thread of your argument about halfway through. So I’m just going to work off the original problem.

Now assume, Monty always picks a door, regardless of whether your original pick is right or wrong (this is the traditional method and you made this assumption as well).

Now if your original pick was right, you should stick with it regardless of what Monty does. That’s pretty obvious.

But if your original pick was wrong, then Monty opens the other wrong door and offers you the last closed door. If your first pick was wrong, this last remaining door will always be right. You follow me?

Now go back to the moment of your original pick. You have no way of knowing what door has the prize. Therefore you have a two out of three chance that your first pick will be wrong. And by the logic I demonstrated above, you should always change choices if your first pick was wrong. So two out of three times, changing picks is the correct choice.

Little Nemo: You are correct, of course, assuming you know a priori that Monty always opens an incorrect door after you guess. This is the premise of the original logic puzzle, not the strategy for optimimun performance actually playing
Let’s Make A Deal.

Of course, the optimum strategy for LMAD has much more to do with the quality of your costume and the contents of your purse than your door-picking techniques.

If Cecil Adams did not exist, we would be obliged to create Him.

[Moderatot Hat ON]

Sounds like something that can have a factual answer, so I think I’ll let the GQer’s have a hack at it.

[Moderatot Hat OFF]

How about this? We’ll call the correct door G door and the two incorrect doors X1 door and X2 door.

After you have selected (a 1 in 3 chance of being correct), Monty opens either X1 or X2.
Now there are two doors closed, one of which you selected, one G and one X (50/50). Your odds of being right have improved without you raising a finger.

Now you are given the option to change your selection. If you do not change your selection you have a 50/50 chance of winning. If you decide to change your selection and it was was X1, you win. If your selection was X2 and you switch, you win. If your selection was G door and you switch you will lose. It seems that, if you switch, the odds of winning become 2/1 in your favour.

Actually, Cecil has covered this whole schtick. Here’s the definitive answer:

http://www.straightdope.com/classics/a3_189.html

This question is a dead horse we’ll not be beating in GQ. If you’d like to argue your take on the issue, please continue your flogging in Great Debates.

``````Nickrz
GQ Mod``````

Woohoo! The Battle of the Moderators! Who will win, GQ or GD? Or will this end up in MPSIMS, under the benevolent tutelage of Eutychus?