dvk@mi-cec.UUCP (Dan Klein) (10/03/83)
Scott- Sorry. While your math is correct, your assumptions are wrong. Indeed, the probability of any door being the winner is 1/3. Having ruled out the door Monty shows you gives a 2/3 probability of *either* of the two remaining doors as being the good one (WRT the set of three). Given that the probability of *any* door being good as 1/3, you seeing a bad door does not change the probability of either remaining door being good. Once you know that door #1 is bad, the probability of doors #2 & #3 being good (or bad) is now 1/2. It don't make no nevermind what you do! Stay or change, your chances are the same, and you'll look just as foolish when he shows you the cow and milking stool whether you changed or not. -Dan Klein, Mellon Institute, Pittsburgh
asente@decwrl.UUCP (Paul Asente) (10/04/83)
This argument seems pretty watertight: If you were correct in your original choice of doors, then you win by not switching & lose by switching. If you were wrong in your original choice, then you lose by not switching and win by switching (since Monte has thoughtfully eliminated the other booby prize). You are correct in your original choice 1/3 of the time and incorrect 2/3 of the time. Therefore, you lose by switching 1/3 of the time and win by switching 2/3 of the time. So you should always switch, and you'll win 2/3 of the time! (Amazing) -paul asente (decvax, allegra, ucbvax, ihnp4)!decwrl!asente
mcewan@uiucdcs.UUCP (mcewan ) (10/05/83)
#R:mi-cec:-16500:uiucdcs:28200022:000:1133
uiucdcs!mcewan Oct 4 15:45:00 1983
*sigh* I'll try again.
Wolog, assume that you chose door #1. Let W(i) be the event that door i
is the winner, and let S(i) be the event that Monty shows you door i.
Then
P(S(1)) = 0
P(S(2) | W(1)) = 1/2 P(S(3) | W(1)) = 1/2
P(S(2) | W(2)) = 0 P(S(3) | W(2)) = 1
P(S(2) | W(3)) = 1 P(S(3) | W(3)) = 0
a priori P(W(i)) = 1/3 for i=1,2,3
Applying Bayes formula
P(S(2)|W(3))P(W(3))
P(W(3) | S(2)) = -------------------------------------------------------------
P(S(2)|W(1))P(W(1))+P(S(2)|W(2))P(W(2))+P(S(2)|W(3))P(W(3))
1*1/3
= ---------------------
1/2*1/3+0*1/3+1*1/3
1/3
= ---------
1/2
= 2/3
Similarly, P(W(3) | S(2)) = 2/3.
The trick is that no matter what door you chose, Monty is going to show you
a losing door. Since this is independent of the door you chose, being shown
a losing door gives you NO information about your door, and cannot affect
the probability of that door being the winner.
Scott McEwan
uiucdcs!mcewanlaura@utcsstat.UUCP (Laura Creighton) (10/05/83)
Sorry folks, but I have a real problem with the question. What is a "Monty Hall"? (i assume from most of the discussion that he is a fictitious person, but I can't be sure) and for what reason is he making deals? Is this some underground classic that I have missed? laura creighton utzoo!utcsstat!laura
dmmartindale@watcgl.UUCP (Dave Martindale) (10/06/83)
I disagree with Dan Klein. His argument is essentially that knowing which door is bad leaves the remaining doors with equal probability of being good. Another way of thinking of this is that the door he chooses gives you no new information, it just reduces the number of possible choices and the remaining ones are equally likely. But this isn't true. You will pick the good door 1/3 of the time; in that case Monty is showing you one of the two bad doors but since both are bad his choice contains no useful information. In this case, switching doors is guaranteed to give you a bad door. But with probability 2/3, you pick a bad door, and Monty shows you the other bad door. Here, he is TELLING YOU WHICH DOOR IS GOOD, which is useful information, and switching doors is guaranteed to give you a good door. Thus the strategy of switching doors gives you the good door 2/3 of the time, while the strategy of keeping your original choice only gives you the good door 1/3 of the time. Another way of looking at it is that Monty is NOT picking one of the bad doors at random, which is an assumption built into your argument. He is picking at random 1/3 of the time, but 2/3 of the time he is picking the ONLY remaining bad door. Dave Martindale decvax!watmath!dmmartindale allegra!watcgl!dmmartindale
dvk@mi-cec.UUCP (Dan Klein) (10/07/83)
1) Laura - Monty Hall was the moderator/huckster of a popular U.S. TV show called (you guessed it) "Let's Make a Deal", wherein housewives and sometimes their hubbies dressed in riduculous outfits (a giant tomato, for example), and got to be ridiculed on TV while they jumped up and down in an apoplectic moronic ecstasy for the chance to "win" fabulous prizes. (Just what I *need*! A Sheep!) The "big event" was the "Door #1, door#2, or door #3 (or the box that Carol Merrill is standing in front of...). 2) At the instigation of another netter, I ran a simulation of the problem. Believe it or not, Ripley, it pays to always change. Follow whos ever argument you please. The probability of winning if you *always* change is 2/3. If you never change it is 1/3. Try it. I was amazed, too. -Dan Klein, Mellon Institute, Pittsburgh
tjt@kobold.UUCP (T.J.Teixeira) (10/08/83)
Having read Dave Martindale and Dan Klein's responses (always switching
doors gives you a 2/3 chance of winning), I don't believe it.
Firstly, if you hypothesize a second contestant who will win what is
behind the other door, both contestants can increase their chance of
winning to 2/3 by swapping doors!
Secondly, I did my own simulation:
================================================================================
int wins; /* number of times I win by switching doors */
int losses; /* number of times I lose after switching */
int monty; /* number of times Monty shows me the good door */
int doors[3]; /* count how often each doors is a winner */
/* to verify the random number generator */
main()
{
register int i, door, mydoor, montysdoor;
for (i = 0; i < 10000; i++) {
/* pick a winning door at random */
door = rand() % 3;
doors[door]++;
/* I pick a door at random */
mydoor = rand() % 3;
/* Monty picks a door, but not mine! */
while ((montysdoor = rand() % 3) == mydoor);
if (montysdoor == door)
monty++;/* Monty shows me the car */
else if (mydoor != door)
wins++; /* I win by switching */
else
losses++;/* I feel like a sap! */
}
printf("monty %d wins %d losses %d\n", monty, wins, losses);
printf("door #1 %d #2 %d #3 %d\n", doors[0], doors[1], doors[2]);
}
================================================================================
monty 3302 wins 3258 losses 3440
door #1 3330 #2 3419 #3 3251
================================================================================
If Dave and Dan are right, why are the wins and losses roughly equal,
and what's wrong with my simulation (no flames please on coding style
-- it's just a 60-second hack)? I showed you mine, now show me yours!halle1@houxz.UUCP (J.HALLE) (10/12/83)
This is very similar to the coin&chest problem of earlier in the year. It also is a Restricted Choice problem. Assume you pick door one. Now, if you are wrong, Monty's choice is restricted; he must show you the one he does. If you are right, his choice is random. Thus the fact that he showed you a particular door implies that he had to, making the switch proper.
thh@kcwin.UUCP (10/13/83)
I think I've come up with a simple explanation of why it pays to switch: First let's assume that I always switch. If I choose the correct door (probability 1/3) and then switch I will always end up with the wrong door (no matter which door Monty shows me.) However, if I choose the wrong door (probability 2/3) and then switch I will always end up with the door with the grand prize (because Monty will have exposed the door without it). So I'll win 2/3 of the time if I always switch. If I never switch then my chances of winning will never change from the original 1/3. Tom Hadley, Western Electric, Kansas City we53!kcwin!thh
mcewan@uiucdcs.UUCP (10/14/83)
#R:decwrl:-343000:uiucdcs:28200025:000:284 uiucdcs!mcewan Oct 13 12:54:00 1983 <FLAME ON> Will everyone please learn how to read. Monty does NOT chose a door a random! He ALWAYS choses a losing door. That fact is what gives you the information that the third door has a 2/3 probability of being the winner. <FLAME OFF> Scott McEwan pur-ee!uiucdcs!mcewan