sleestak, that logic reflects a scenario where a contestant walks into the studio after a zonk is revealed. In that case, conditional probablity would increase the winning odds of both remaining doors to 50%, as your father has calcualted.
However, by being in the studio and selecting a door before a zonk is revealed, the contestant is revealed additional information about which door likely hides the prize. At the outset, the odds of choosing the correct door are 1:3, while the odds that the prize is behind one of the other two is 2:3. This does not change even after a zonk is revealed in the set of remaining doors; this prevents the problem from collapsing to 1:2 odds, and is the logic behind switching by RJKUgly and others.
It might be easier to think about this in terms of actual trials rather than probabilities. First, assume that Monty decides to tape 300 shows. The door assignment of the winning prize is random for each show, with the caveat that it appears behind each door 100 times. Furthermore, assume that the contestant always picks Door #1, and Monty always reveals a zonk.
In 100 cases out of 300, the winning prize will be behind Door #1. Monty will reveal Door #2 in 50 of these cases, and Door #3 in the remaning 50. The contestant will win only if he doesn’t switch.
In 100 cases out of 300, the winning prize will be behind Door #2. Monty will reveal Door #3 in all 100 cases. Likewise, in 100 cases out of 300, the winning prize will be behind Door #3. Monty will reveal Door #2 in all 100 cases. The contestant will win only if he switches.
That leaves 200 cases out of 300 in which the contestant will win only if he switches, and 100 cases out of 300 in which he’ll win only if he does not switch.
He didn’t explain it in the email. He just did the math and expects us mere math mortals* to figure it out the rest. The part I quoted was his whole reply.
This is what I think is going on and causing the confusion. When Monty opens the door w/o the prize, the probability is changed. What people are arguing is that the 1/3 probability from the door Monty checks automagically goes to the door the contestant did not pick. So if the contestant picked door 1, Monty opens door 3, then the odds behind door 2 having the prize gets the additional 1/3 from door 3.
The problem with this is that probability from door 3 will not go to just one unopened door. It will spread between the two un-opened doors evenly. Thus it becomes 50/50.
At least that is my guess to the reasoning.
I’ll send him an email with dwalins list and see what he says. It might take him a bit to respond. It might be Tuesday or Wednesday before I hear from him.
Slee
*My Dad has been known to say things (without trying to be snarky) like “Calculus<or linear algerbra, bascially insert any higher math here>, that’s kindergarden work”)
I recall reading an interesting variation of this puzzle:
Suppose there were 5 doors, and you get to pick two. There are zonks behind three of the doors, and great prizes behind two. So, you have a 70% chance of getting at least one door right on your first guess.
After you select your two doors, Monty (as always) shows you one of the doors you did not pick, containing a zonk.
Now you have the option of changing one…or both of your doors for one or both of the others. Now how many doors, if any, should you switch?
The problem with your dad’s formulation of the problem is that the conditional probability was not in effect when you chose your original door. The odds of your initial choice being right do not change as more information is revealed, because the agent revealing the information is not honest. In no cases will Monty reveal the actual prize behind a door not chosen. In other words, since you will never gain information that the chosen door was incorrect, you in fact gain no additional information at all concerning your initial choice. The odds you were correct remain 1/n no matter how many doors (up to n-2) get opened to reveal nothing. It was always gauanteed that (n-1) doors were wrong. Seeing specific doors opened grants no additional insight because the opening of doors is not a “fair” activity.
Biotop – I switch both doors. The odds are only 1/10 that I chose both prizes initially. They are 3/10 that I missed both initially. So, in 6 cases I would not change my outcome (assuming that both prozes are equivalently desirable). In 3 cases I improve. In only 1 case would I lose.
(By the way, I love the Monty Hall puzzle. I think Cecil taking an interest in explaining it was one the main reasons I first came to the SDMB!)
+++++
But Spiritus Mundi, you have a 70% chance of having at least one door right and you are still switching both doors?! Seems counterlogical. And suppose you had to have both doors right in order to get any prizes? Would you still switch both??
In case you’re wondering, switching 1 door only in the “both doors to win” improves my chances to 15% (I think – didn’t bother to actually do the math, but it seems to be a 50-50 shot at a 50% win that holds in 6 out of 10 cases.)
If he wants to see a more rigorous analysis based on conditional probabilites, you might send him to this page, or this one. There is also this analysis from the American Mathematical Monthly.
The key issue in this (endless) debate is what constraint, if any, there is on Monty. The normal assumption in the Monty Hall Problem — not always stated, unfortunately — is that Monty will reveal a door, and it will be a blank door, after the player has made his first choice. Given this constraint, the player has a 2/3 probablility of winning if he switches doors.
And I’ve just come across some more discussion, and more academic journal citations, here. This page also addresses Monty’s choice of policies and how it affects the player’s strategy.
Marilyn vos Savant in “The Power of Logical Thinking” devotes a chapter to the Monty Hall paradox explaining why you should switch, and includes some of the letters she received claiming she was wrong (including from a number of statisticians). Makes for amusing reading, and demonstrates that not even “experts” (i.e. in this case, statisticians) are always right! Some of her respondents did the obvious thing and simulated the problem on a computer, and, as expected, demonstrated that you win 1/3 of the time if you don’t switch, and 2/3 of the time if you do switch.
I just look at it as - before things start you can choose to pick 1 door (by planning to point at one and saying “I want that one”), or 2 doors (by planning to point at one and then saying “I want the other two doors”). If the latter, clearly at most one of the two doors has the prize, so the fact that Monty points at a door that HE KNOWS does not have the prize and you know that he knows that, does not convey any additional information. Hence the probabilities don’t change from the initial situation. No conditional probabilities apply.
I thnik it is clear that **sleestack’s[b/] dad understands conditional probability; he just mistakenly applied it in this case. If the additional door had been revealed through a “fair” mechanism, then his analysis would be correct. This is (I feel) the point that confuses most folks who fai to intuit this problem.
Sorry to “bump” this when the discussion seems to be pretty much complete, but folks that are interested in similar problems might like to take a look at Puzzling Adventures in Scientific American
Fun stuff, but I thought I should point out that the page has a typo in their answer for problem #3. When the opponent reveals 1 box your odds after a switch increase from 2/5 to 8/15 (6/15 => 8/15). This is a 33% better proposition, but it does not quadruple your odds."
The correct analysis of this problem throws up a lovely nuance which properly explains the furore between rival advocates of one third and one half.
Those dissenters who find the solution counter-intuitive are correct when they say the contestant has a 50% chance of winning.
Assuming the choice not to switch or switch is made equally:
(1/2 x 1/3) + (1/2 x 2/3) = 1/2
This isn’t the question being asked though.
The question is when the 50% of instances the contestant won are studied retroactively, what proportion of that 50% occured when the contestant switched her original choice.
The answer to this question is 2/3.
That the contestant has a 50% chance of winning if she chooses one of the two remaining closed doors at random should never have been in doubt, yet none of the “correct” answers above point this fact out.
The issue is whether adopting a policy of ALWAYS changing or ALWAYS sticking changes your odds, which of course it does, because the additional information available renders random choosing an act of folly.
One more time, it must be stressed:
100% switch = 2/3 chance of winning
50% switch = 1/2 chance of winning
0% switch = 1/3 chance of winning
The middle strategy is nearly always missing from explanations I have seen of the problem, probably because the puzzle’s clever construction excludes it from linguisticaly making any sense as a condition in a one shot scenario. Skipping over this factor is not just the crux of why some people don’t intuitively understand the solution even after it is explained, it is also the crux of why a lot of people who do “get” the answer don’t know how to explain it properly to those who don’t. This hidden switch variable is what is responsible for creating the illusion for many that the monty hall problem is in someway paradoxical. It’s also the blind spot which causes people who should know better to hesitate initially in trusting their own correct answer when it’s assailed by others.
Well, yes, if you decide randomly whether to switch or not, then you have a 50% chance of winning. I don’t think anyone disputes that, which makes it rather boring to discuss. What does it add to the discussion to note that?
Nope. The question posed is “should you switch”? Not, “should you flip a coin and switch if it comes up heads?” (unless it is a two-headed coin, of course, in which case the answer is yes.)
The reason I find your “third option” to be unhelpful in illustrating the answer is because you can find the odds of winning with a “random switch” only after you have already found the odds of winning with a switch and with standing pat. If someone can get that far, then they don’t really need the illustration, do they?