What may be tripping up people is that in A, sometimes Monty opens the door with the prize. You lose. But you you never get a chance to switch in those cases. In the cases that you do get the chance to switch, it really is a 50/50 choice. Those times that Monty opens the prize door should be exactly 1/3 [2/3 * 1/2].
The odds are that you still choose correctly 1/3 of the time. But only 2/3 of the time are you given a choice, and you’ve got a 50/50 chance at that point.
At least I think that’s all correct. I don’t swear by my reasoning.
Why not? Pigs are magical animals. (C.f. Lisa Simpson.)
Crap. After looking at the problem more carefully, but a little differently, I do see now that Monty’s policy would change your chances of winning. Depending on the probability with which he reveals the prize door (when he has the chance to), a player strategy of always switching will win anywhere from 0 up to 2/3 of the time.
In particular, if Monty opens the prize door with probability P on those trials when he has the choice to (which is 2/3 of the total number of trials), an always-switch strategy on the part of the player will win ( 2(1-P)/3 ) / ( 1/3 + 2(1-P)/3 ) = (2 - 2P) / (3 - 2P) of the time. In the “normal” case of P = 0, that comes to 2/3 of the time. In the case of P = 1/2 (the variant Xema describes), it wins half of the time. And for P = 1, it always loses.
If you adopt the switching strategy, the only way you can lose is by initially picking the prize door. Since the odds of initially picking the prize door are 1/3, the switching strategy will pay off 2/3 of the time.
Sorry, a computer simulation of a random-picking Monty would reveal a 50/50 chance for switching in this scenario. The thing is, in only one trial out of 51 would you ever get to the point where you have the option of switching, but in those that you do, it’s a 50/50 proposition.
I think of it this way: Monty is essentially asking you to bet on whether you picked right or wrong the first time. Since your chances of picking right were 1 in 3 the first time, you would want to bet that you picked wrong. The way you bet that you picked wrong is by switching.
While he’s doing that, could you tell us why you think that when 50 non-winning cards have been randomly selected and revealed, the remaining two should not have equal probability of being the winner? (If, as it seems, you do think that.)
Suppose instead of chosing a card from the (properly shuffled) deck, I say “I’ll just take the card on the bottom.” You leave it there and start dealing cards from the top, face-up. With 50 cards exposed (probably took many trials to reach this point), the Ace of Spades has not appeared. Which of the remaining two is more likely to be the winner? Why?
I apologize for the line by line quoting. I’m just stopping by quickly, and don’ty have time to form a coherent narrative…
Very good point; my apologies.
You make a convincing argument, but I still think the deck of cards simulation will show me to be correct. Using your logic:
Monty says he will only open a zonk, and he knows where the unpicked zonk is
Monty says he will open randomly, and happens to open an unpicked zonk
You are saying the probabilities are different. What if he was lying to you about his motivation? Does the probability then magically change?
1:52. Unless, of course, you are saying that your original pick had a 50% chance of being the ace of spades?
Correct. They all have a 1 in 52 chance.
This sounds very very similar to what I said above about the martingale. ie: 1/52% of the time we picked right, 1/52% of the time we picked wrong, and 50/5% of the time we never got here in the first place. I honestly don’t know who is right. Did anybody simulate this yet?
“1/52% of the time we picked right, 1/52% of the time we picked wrong, and 50/52% of the time we never got here in the first place. I honestly don’t know who is right. Did anybody simulate this yet?”
The odds are the same in both scenarios. (Though, to be pendantic, since you choose the first card, and I choose the remaining card, then I will always have the better odds).
Anyway, to illustrate, for clarity (though the exact nominal odds will be different), instead of 50 cards, use 100 (yes, I know this example has been posted already). The chances in each scenario of you picking the Ace of Spades is 1 in 100, or 1% (no repeating cards in this scenario).
Now, after you pick the card, and without dealing out any more cards and or anything else for that matter, on a scale of 1 to 10 (the scale isn’t important, it’s how you feel how likely is), how likely do you think you picked the A of spaces? Your answer should be very unlikely, or 1 (with 1 being the least likely that you picked it). Remember this last part, it is very important.
Now, let’s continue: you pick your card, I now look through the remaining deck and tell you that these 98 are not the A of spades, and I leave one down. More likely than not (in fact 98% likely, or a 9 out of 10, well 98 out of 100) the last remaining card I have down is the A of spades.
Why? Your initial pick has a 1% chance of being successful. I have just burned or turned over 98 other cards which are also not it. As a matter of logic, do you really think that you picked the A the first time around? 99 times out of 100 you did not. Mind you, we’re not saying that the last card down is the A, we’re just saying that it is 99% certain it is.
Someone with better math skills will come by and illustrate it for you. I just hoped that I was able to show the logic behind it.
I’m not buying the arguments here. When Monty opens one of the doors, the probability gets redistributed. The revealed door is no longer part of the problem. It’s 50-50.
I just wanted to thank PigBoy for this post. The scales have fallen from my eyes, and now I can see. A third of the time, you pick the right one initially, so if you switch you lose. 2/3 of the time, you pick the wrong one initially, so if you switch you win.
Simple!
If Monty was truly random of opening a door, then, as it has been posted before, he will pick the right door 1/3 of the time. Monty is a showman, so this is not going to happen. The audience wants to see the contestant sweat it out. If it did happen, and he did pick the right door, then the game would stop there and the contestant would know that he lost. His knowledge of where the prize is influences the odds.*
*- The odds become uncalculable if he is re-bluffing for any particular instance, I believe, but is still 2/3 over the course of the long haul (that is, to stretch the number of times this happens out to infinity).
program MontyCards;
var
firstpick,prizecard,trialpick,count :integer;
events, successes :integer;
deck :set of 1..52;
begin
Randomize;
events:=0;
successes:=0;
repeat
deck := [1..52];
prizecard := Trunc(Random * 52 + 1);
firstpick := Trunc(Random * 52 + 1);
deck := deck - [firstpick]; { remove my pick from avail choices }
{ now have Monty guess until only one card is left or he shows the prize }
count := 51;
repeat
trialpick := Trunc(Random * 52 + 1);
if (trialpick in deck) then
begin
count := count-1;
deck := deck - [trialpick];
end;
until (count=1) or (trialpick=prizecard);
{ did Monty not reveal the prize card? IOW, is the price card either
my card, or the one left? }
if (trialpick <> prizecard) or (firstpick = prizecard) then
begin
{ these are the only events that count }
events := events + 1;
if firstpick <> prizecard then successes := successes + 1;
writeln('events: ',events,' successes: ',successes);
end;
until events=100;
end.
It changes, but not through magic. If Monty says he’s picking a door at random but in fact is peeking and basing his selection on what he sees, then things change fundamentally.
I believe and hope I’ve said nothing to that effect.
If I understand correctly, you seem to be saying that the probability I’ve chosen the winner stays constant throughout this game. Whereas it seems to me that as cards are randomly chosen, exposed and shown to be losers, the chance that my card (and all other unexposed cards) may win is increasing. When 32 cards have thus been exposed, I’d say my odds are up to 1:20.
Can I note that this is quite a change (IMO, an improvement) from “patently false”. I’ll also note that this problem (like most others of its type) should be amenable to analysis; simulation need not be seen as the final arbiter of truth.