The Monty Hall Problem
The Monty Hall problem goes like this: You are presented with three identical doors. Behind one of them is a car and behind the other two are goats. You want the car. Monty Hall tells you to choose one of the doors. Regardless of which door you choose, at least one of the two remaining doors will have a goat behind it. Monty Hall, who knows where the car is, then opens one of the doors that has a goat behind it. He then gives you the option of either sticking with the first door you chose, or switching your choice to the other unopened door.
Question: What should you do? Should you stay where you are? Swtich? Does it make a difference?
This problem is a staple of courses in elementary probability theory. Virtually everyone, upon hearing this problem for the first time, reasons as follows: After Monty Hall opens one of the doors, there are only two doors remaining. Therefore, regardless of which door you choose you have a 50-50 chance of being right. So it doesn't matter whether you stay where you are or switch to the other door.
This argument is clear, convincing and wrong.
I suspect that every mathematician who has ever presented this problem to a lay person has had the following experience: Mathematician explains problem. Lay person reasons as above. Mathematician explains that, actually, you double your chances of winning by switching doors (more on this in a moment). Lay person gets annoyed, agitated and belligerent. Lectures mathematician on the subtleties of the problem. Repeats, ad nauseum, that after Monty Hall opens one door there are only two, equally likely, doors left!.
I was reminded of this when I came across this blog entry, from The Daily Howler, posted on March 31. The blogger, Bob Somerby, was commenting on this review of a new book about probability theory intended for nonmathematicians. The reviewer wrote:
Before scoffing, chew on the now famous Monty Hall problem, named after the host of “Let's Make a Deal.” A contestant knows that concealed behind three doors there are two goats and one new car. The contestant chooses Door No. 1. The beaming host opens Door No. 3 to reveal a goat, and then asks the contestant if he would like to change his choice to Door No. 2. Two doors add up to a 50-50 proposition, obviously. So why bother? Because the odds have actually shifted. The chances are now two out of three that changing to Door No. 2 will obtain the car. (Emphasis Added by Somerby)
Somerby replies:
Say what? We don’t know what the Kaplans wrote to provoke that highlighted sentence. But for the record: If the contestant changes to Door No. 2, he’ll obtain the car half the time—and “half the time” is not “two out of three.”
Ugh. That's totally wrong, I'm afraid.
The following day Somerby revisited the topic. Apparently some e-mailers informed Somerby that he was mistaken. Sadly, Somerby decided to dig in deeper:
Many e-mailers wrote to insist that there is a counterintuitive “Monty Hall problem” of the type we discussed yesterday (see THE DAILY HOWLER, 3/31/06). We haven’t had time to review this in detail, but:
We didn’t dispute that there’s some such effect—an effect which the Kaplans describe in their book. What we said is this: Whatever that counterintuitive “Hall effect” might be, the Times review doesn’t seem to describe it. We’ll persist in our statement about the situation as described in the Times review: In that situation, it just isn’t true that the contestant would gain an advantage from switching his guess. We’ll grudgingly try to sort through the matter. But what a bad time for this storm to reach land—on a weekend when tyrannical guvmint allows us just 47 hours.
Oh my. The description of the problem given in the Times may not be a model of clarity, but it is clear enough to make the point. Somerby is wrong and the reviewer is right.
I don't know about any counterintuitive “Hall effect”, but the mathematics of the situation are not especially complicated. When the contestant makes his initial choice he has a probability of 1/3 of being right. In other words, he will get it right 1/3 of the time and get it wrong 2/3 of the time. Nothing Monty Hall did in opening one of the doors changes that simple fact. By sticking with your initial choice you will be wrong 2/3 of the time. By switching, you will win 2/3 of the time.
Or think of it this way: Your initial choice is correct 1/3 of the time. Does it really make sense to say that when Monty Hall opens one of the remaining two doors your chances of having made the correct choice magically jump to 1/2?
Or try it this way: Suppose you choose door number one, and Monty Hall opens door number two. The choice Monty Hall then gives you is not really Door Number One vs. Door Number Three. Really the choice is Door Number One vs. Not Door Number One. Since door number one is only correct 1/3 of the time, surely it makes sense to swtich.
Still not convinced? Okay. Suppose you had 100 doors. The doors conceal one car and 99 goats. You choose door number one. Monty Hall then opens 98 goat-bearing doors. Are you seriously claiming that in this situation it makes no difference whether you switch or not? If you are not seriously claiming that, then explain to me how this situation differs from the original version. (In my experience, this way of putting it usually gets people to realize that things are not as simple as they originally thought. One time, though, I had a student in my office who was absolutely convinced that it made no difference whether or not you switched. No matter how I tried to explain it he woudn't give in. So I whipped out this example and smiled. Without missing a step he informed that even in this case it would make no difference whether you switched. I stopped smiling.)
Want more? Try this computer simulation. Do it many times and keep track of your statistics.
Think the computer simulation is rigged? Fine. Get out a pad and a pencil. Make a list of every possible scenario. (For example: Car behind door number one, you choose door number one, Monty Hall opens door number two.) It's a little tedious, but there aren't that many possibilities. Then put a little mark next to all the scenarios in which you will win by switching. I think you will find that by switching you will win 2/3 of the time.
Somerby's a smart guy, and I suspect he will eventually come to realize that he has made a mistake here. I've been a big fan of his blog, which is mostly devoted to exposing the blatant, jaw-dropping insanity that spews forth from our nation's press corps and political pundits, for a while now. But lately he's been annoying me a bit by hammering various liberal pundits for what strike me as pretty minor sins. So consider this brief essay a small measure of payback.
Let me close with two amusing variations on the Monty Hall problem.
Variation One: Monty Hall does not know which of the three doors contains the car. You choose one of the doors. Then Monty Hall chooses one of the remaining doors and opens it. It contains a goat. He then gives you the option of sticking with your original door or switching. What should you do?
Variation Two: This time there are five doors, concealing one car and four goats. You choose one of the doors. Monty Hall, who knows where the car is, opens one of the remaining goat-bearing doors. He then gives you the option of switching. You make your choice, after which Monty Hall again opens a goat-bearing door. Again you have the option of switching. This process continues until there are only two doors remaining. What is your best strategy?
Answers in a subsequent blog entry.
44 Comments:
I dunno much about big city lawyerin', but it seems to me that actual numbers are gonna be better to clarify this.
We got us two different events that we need to take into consideration. First pick and second pick.
Odds of my first pick being right are 1/3. No disputin' that.
For my second pick, I gots two doors to choose from, so I reckon the odds of that pick, taken as an independent event, are 1/2.
To win if I stick, both my first pick and my second pick must be right. .3 * .5 = .15
To win if I switch, I'm saying my first pick was wrong, and my second pick is right. .7 * .5 = .35.
.35 > .15, so switch.
Interestin' observation is that my total chance of winnin', considering both possible victory conditions, is .15 + .35 = .5. Huh.
Math is hard.
Jason, I was wondering if you could answer a few questions for me?
From purely mathematical prospective is the 3 to 4 million years of man existence enough time for man to evolve from ape like to present genetic makeup?
Also I found this on sciam.com http://www.sciamdigital.com/index.cfm
The Recent African Genesis of Humans; Special Editions; New Look at Human Evolution; by Rebecca L. Cann and Allan C. Wilson; 8 Page(s)
In the quest for the facts about human evolution, we molecular geneticists have engaged in two major debates with the paleontologists. Arguing from their fossils, most paleontologists had claimed the evolutionary split between humans and the great apes occurred as long as 25 million years ago. We maintained human and ape genes were too similar for the schism to be more than a few million years old. After 15 years of disagreement, we won that argument when the paleontologists admitted we had been right and they had been wrong.
Once again we are engaged in a debate, this time over the latest phase of human evolution. The paleontologists say modern humans evolved from their archaic forebears around the world over the past million years. Conversely, our genetic comparisons convince us that all humans today can be traced along maternal lines of descent to a woman who lived about 200,000 years ago, probably in Africa. Modern humans arose in one place and spread elsewhere.
Is 1 million years enough time?
How about their claim of 200,000 years?
Thanks, Mark
I love the Monty Hall problem because it makes me feel so smug to know the right answer. Nevertheless, every time I run across it I have to re-analyze it to convince myself of the correct answer. I'm beginning to suspect that there is a flaw in the wiring. Maybe the situation of facing an individual who is obviously trying to trick us changes the way we calculate the odds. The deceit analysis circuits kick in.
I had a hard time swallowing the Monty Hall enigma, so I wrote a computer program to test it, with a slight twist. Monty goes through the door-picking/goat showing routine any number of times--thousands, if you like. The twist is that there are two contestants: Switcher and Sticker. Before Monty opens a door and shows the goat, Switcher and Sticker agree on a door, which they pick at random (but they both pick the same door). After Monty shows the goat, Sticker always sticks with the door that was originally chosen, whereas Switcher always picks the other door. After about 10,000 iterations, Switcher has way more cars that Sticker has. And there's no way to rig the simulation, because Switcher and Sticker are both subject to the exact, same conditions except for the fact that one switches and the other doesn't.
It's still counterintuitive, but it does indeed work that way.
Okay, I'll attempt to spoil tomorrow's blog entry by giving the correct answers today...
In the first problem, stay or switch has no impact: Monty's pick gives you no info. You know there's a 1/3 chance you picked right, and, if so, there's a 100% chance Monty's subsequent pick would be a goat. If you picked wrong, there's only a 50/50 chance that Monty's subsequent pick would be a goat, but it was. So there's two possible, equally likely situations: I picked right, which happens 1/3rd of the time, or I picked wrong, and so did Monty, which happens 1/3rd of the time (the remaining 1/3rd of the time I pick wrong, and Monty picks right, and the game ends prematurely). So after my pick and Monty's, the chances are 50/50. Staying or switching don't affect the odds.
For the second problem, I would stay until we're down to two doors, then switch. This ensures that when there's 2 doors remaining, one (the one I picked initially) has a 1/5 chance of being right, and the other one has a 4/5 chance. So 4/5 of the time I'll win. Any other strategy reduces the certainty that the set of unopened, not currently chosen doors contains the prize, and raises the probability that my currently chosen door contains the prize. I.e. if after 1 round I switch, I'm picking from a set of 3 doors that have a 4/5 chance of containing the prize. Thus, my new choice has a 4/15 chance (greater than my earlier 1/5 chance) of hiding the prize, and the unopened doors (now including my original pick) have an 11/15 (less that 4/5) chance of hiding the prize. Continuing the process I think, (if I'm doing the math in my head right) that I end up with something like an 11/30 chance that if I stay with my last choice I'll win, and a 19/30 chance that if I switch I'll win. I'm better off with the 4/5 chance of winning-if-I-switch I get by following the stay-until-theres-only -two-doors-left strategy.
now, the problem in the first "user exercise" is that extrapolated to a statistical exercise, it creates the unlikely scenerio that Monty will actually get the Goat every time. The reality is that Monty should get the car some of the time if he doesn't know.
His odds of getting a goat are 1/3 * 1.0 (i have the car already, so he's guarenteed a goat) + 1/3 * .50 * 2 (2 50-50 shots of getting the goat), meaning 1/3 + 1/3 = 2/3rds of the time he'll get the goat. It makes sense, he doesn't know and to him there are objectively 2 goats.
So IF he gets a goat, you have the 50-50 independent second guess that makes more sense. Realistically, your odds across the whole are back to 1/3, not 1/2, because 1/3rd of the time you wouldn't get that second guess.
Interestin' observation is that my total chance of winnin', considering both possible victory conditions, is .15 + .35 = .5. Huh.
which is the answer to user exercise 1 - if monty doesn't know, the first guess doesn't matter.
I remember this one from a while ago, when Marilyn vos Savant (could that really be her real name?) ran it in her column, explaining it in a not-entirely-helpful way (yours is/are much better!). The next week she ran a whole bunch of letters from professional mathematicians berating her for getting the "wrong" answer and/or making entirely irrelevant or bizarre quibbles about how the problem was stated. Naturally, at first I too thought she had made a mistake (thinking, as you say, that the odds were 50/50); but I thought about it some more and figured it out. I tried it on my housemates, one of whom held out until we ran the simulation. It's not even the result (that switching means that you win 2/3 of the time) that does it, it's the mechanism (that first you're right and switch to the goat OR first you're wrong and switch to the car). After this happens to you a few times you see how it works. Interestingly, I had a co-worker who didn't even get the right wrong answer. Oh well.
Anonymous is confused, but I'm not sure how to get him from where he is to where he needs to be (better start over). Same with Joe.
JY is right about the second variation, in which his strategy (staying until we're down to two doors - that is, my original choice and the one remaining closed door) indeed gets you a 4/5 chance of winning. At first I thought that it didn't matter what you did until you get down to two doors, while you should then switch; but if you switch earlier on that means your odds decrease to 2/3 instead of 4/5. That is, the problem resets, as if the earlier picks had not occurred. Nice catch JY!
However, he is wrong about the first variation. Whether Monty knows where the goat is makes no difference. It just means that in the 2 out of 3 times that you do win (by switching, as before), half of the time you switch to a closed door behind which there is a car, and half of the time you switch to an open door, through which the car is plainly visible. As told, the story concerns the former situation (it stipulates that he shows you a goat); but that doesn't change anything.
In thinking about this, it's very tricky to keep track of when exactly Monty is giving you information and how to take advantage of it. That was my mistake in thinking about the second variation. In order for him to have given you information in the earlier rounds, you must stay; while in order for him to give you information in the last round, you must switch.
Thanks Jason, this is fun!
I spent a month trying to understand why I was wrong after some poor explanation I saw failed to tell me that Hall knew where the goat was. Without that info, I was looking at the cases where Hall had opened doors at random, and hadn't found the goat in the process. In which scenario the odds of the two remaining doors are indeed equal.
-steve s
Someone should explain to Mark the difference between "most recent common ancestor" and "point of divergence".
The "only" changes from the most recent common ancestor are going to be those relating to the differences seen between humans NOW.
200,000 years is probably sufficient to evolve different pigmentation levels and suchlike, wouldn't you think? They're not very large changes.
Also, Mark, it helps to realize that evolution proceeds at different rates at different times, depending on the selective pressures at work.
If Monty knows which one has the car, then the way it works is simple:
2/3 you pick a goat door, 1/1 he reveals a goat.
2/3, then, the car is behind the unpicked door.
The remaining 1/3, it's behind your door.
If Monty DOESN'T know, it's a bit more complicated.
2/3 you pick a goat door. if you pick a goat door, 1/2 that he reveals a goat, and 1/2 he reveals the car. If you pick the car, of course, 1/1 he reveals a goat.
therefore, it's 2/3*1/2=1/3 that he reveals the car, 2/3*1/2=1/3 that he reveals a goat and you should switch to the unopened door, and 1/3 that he reveals a goat and you shouldn't switch.
So given that the first third didn't take place... it's an even chance as to which it is.
If he revealed the car, well, if you can switch to it, you should. If you can't, then it doesn't matter what you do.
If he didn't reveal the car ... it doesn't matter what you do.
Michael is correct.
The original description of the problem (in the book review) failed to mention the vital fact that Monty Hall knows which doors have goats behind them and always picks one of those. Without that information (or an assumption that this is the case) it's impossible to say what the probabilities are.
The computer simulation link has an excellent explanation that uses a graph to enumerate all the possibilities. That's what really brought it home for me. Plus, it shows how the chances become 1/2 if Monty opens a door randomly.
Good stuff, Jason.
-JC
It took me a while to figure out the Monty Hall problem, but once I did I was able to come up with a nice generalized solution that works for any number of doors. The key to the 'classic' puzzle is that it's an iterative puzzle -- that is, the analysis of each round depends on what happened in the previous round(s). Each new door Monty opens gives you a new bit of information. Therefore, no matter how many doors there are you should always stay til the last round (two doors left), then switch.
As for variant 1, there's a way you can make it even more convoluted. Break it down to two subvariants: in 1a, Monty opens a door at random and if he gets the car, you lose on the spot. In 1b, he opens a door at random, and if he gets the car you can still switch. Now what's your best strategy, and how often will you win?
Duck, a flaw in your analysis: you are given the opportunity to switch to the unopened door.
1/3 of the time -> you picked correctly, switching bad.
1/3 of the time -> you picked incorrectly, Monty shows goat, switching good.
1/3 of the time -> you picked incorrectly, Monty shows car, stay or switch doesn't matter (neither unopened door has the car).
So half the time Monty reveals a goat switching will win, the other half, sticking will win. You have to look at the frequencies!
Another way to look at the original version and the first variant is this: What does Monty's 'pick' tell you about your original pick? In the original version, it tells you nothing: Monty will always reveal a goat, regardless of what you picked. In the first variant, Monty has a lower chance of picking a goat if you picked correctly than if you picked incorrectly. So if Monty does reveal a goat, that raises the odds that you picked correctly: he has provided you a little bit of information about your pick.
My responses...
Variation 1: Switch doors. It doesn't really matter whether Monty knew; the choice is still between "door 1" and "not door 1", and there's still a 2/3 chance that "not door 1" has the car.
Variation 2: Switch when there are only two doors left. That would make the choice between "door 1" (with a 1/5 chance of a car) and "not door 1" (with a 4/5 chance of a car).
Michael "Sotek" Ralston said...
Someone should explain to Mark the difference between "most recent common ancestor" and "point of divergence".
Please do. That is why I asked. I am trying to understand.
I relies this is off the current topic, but point me in the right direction and I will go read.
Mark
JY, thanks for your response. This is giving the brain cells a workout.
I was indeed playing what [commenter] wolfwalker calls subvariant 1b, where you are playing subvariant 1a (where if Monty reveals a car you lose). (And I agree with wolfwalker's first paragraph.) And you had me going for a bit.
But I still think, even in 1a, that Monty's ignorance doesn't matter, and that you will win 2/3 of the time by switching and 1/3 of the time by staying as the scenario is described by Jason, i.e. in which Monty reveals a goat. Lord runolfr's explanation is correct but a bit brief. Let me try (again, this is for variation 1a).
2/3 of the time your first pick is a goat. Half of those times Monty will then reveal a car (you lose). But ex hypothesi that didn't happen. So the described scenario concerns what happens the other half of the time: he reveals a goat, you switch to the car. So when you first pick a goat, either (something that didn't happen) or (something good). And that happens 2/3 of the time.
Concerning the information conveyed by Monty's pick: the point is not what I learn about my original pick (which is indeed nothing, as you say), but instead about which door to pick should I be given the opportunity to switch.
Consider yet another variant: as before, three doors, you pick one. Now Monty offers you the chance to switch to another door, only without showing you one. Now you really are just flipping a coin. The odds of winning are 1/3 (staying with a car) vs. half of 2/3 (switching to a car) = 1/3. What Monty's pick does is eliminate one of the choices of where to switch if you switch, doubling your odds to 2/3. But you have to switch to take advantage of it (just as you had to stay in the opening rounds of variant 2).
Doesn't your reasoning (that knowing picks of goats tell you nothing) apply to variant 2 as well? But your answer to that one is correct.
I hope Jason will explain whichever fallacies of probability are at work here, lurking behind the doors....
Duck, there's a really simple way to
prove to yourself that my interpretation (or yours) is the valid one -- try it! Take, say, an Ace and two Kings, shuffle 'em, you pick 1, then have 'Monty' randomly turn up 1. If 'Monty' turns up the Ace, start the round over, because that's not the scenario we're interested in.
You'll see that, if you do this say 20 or 30 times, ignoring the rounds where 'Monty' turns up the Ace, the card you picked is the Ace half the time.
Let's see again why this is:
You deal:
Card 1 = A, 2 = K, 3 = K
You pick a number from 1 to 3, and 'Monty' flips a coin to decide his pick: Heads (H) -> choose lower numbered card, Tails (T) -> choose higher numbered card. So the pick/flip scenarios are:
1. 1/H -> Sticking is best
2. 1/T -> Sticking is best
3. 2/H -> Invalid Scenario! (here, the Heads flip forces Monty to turn over card 1, the ace).
4. 2/T -> Switching is best
5. 3/H -> Invalid Scenario!
6. 6/T -> Switching is best
As you can see, there are six possible sequences for this particular deal, and all are EQUALLY likely. Two of them are invalid. By the time Monty shows you his pick, you know you aren't playing scenario 5 or scenario 3. But that's all you know. Scenario's 1, 2, 4, and 6 are all equally likely. In half the scenarios, sticking wins, in half, switching wins.
Contrast this with the original Monty Hall problem. Monty still flips the coin, but most of the time his pick is forced upon him:
1. 1/H -> Sticking is best
2. 1/T -> Sticking is best
3. 2/H -> Switching is best (Monty has to show you card 3)
4. 2/T -> Switching is best (Monty still has to show you card 3)
5. 3/H -> Switching is best
6. 3/T -> Switching is best
Again, six equally likely scenarios. In 2/3, switching is best.
Six equally likely scenarios, and in 4 of them, switching is best.
Duck, on the other hand, I can now make an argument from authority. Jason's posted his followup.
Thanks again, JY! I think I get your answer (and Sotek's) - now all I need to do is figure out my answer. (Some of it is right, after all...). For example I don't get the first sentence of your 8:57. Let me go read the wiki thing.
You know, I just thought of something. If you were trying to explain this problem to Kenneth, you might end up saying: "Kenneth, what's the frequency??" =8-0
I would like to offer a more transparent explanatin for the original Monte Hall (MH) problem in which you pick one of three doors, one of which hides the car and two of which hide goats, followed by MH opening a goat door (he knows where the car is) followed by you sticking with or switching your coice.
Consider a slightly different game in which you are given two options: you can either pick any one door and win the car if its behind that door; or you can pick any two doors and win the car if it is behind either of them. It is easy to see that with the first option your chance of winning the car is 1/3 and with the second option (which you would certainly pick unless you enjoy a somewhat unusual life style involving goats) your win-car chances are 2/3.
But this is really exactly the same as the MH problem (isomorphic to it). The strategy of sticking to your original choice is the same as selecting option one. It makes no difference what MH does. The strategy of switching is the same as option two. Let's say you want to pick doors 1 and 3 in option two. Then make your first choice door 2. Now doors 1 and 3 will be opened; one by MH and the other by you and since MH gets the goat, you will get the car if its behind one of these two doors.
This works as well for more than three doors. First consider a reordering of events in the original MH problem: MH opens a goat door before you do anything leaving you with two doors to choose from. You then choose one door and then can stick with that door or make a change. Either way, you have a 50/50 chance of winning the car. While this game seems rather silly, its what essentially takes place with the five door problem if you change your choice after your first door selection. MH has simply opened some of the doors and reset the game to fewer doors. Your best bet is to stick with your first choice until the last round and then switch. In the five door game, this in effect allows you to choose four doors and win the car if its behind any one of them. I. e., it changes the game to one in which your options are to choose any one door, any two doors, any three doors, or any four doors, depending on how long you stick with your first selection. Lord runolfr gave a similar explanation in an earlier comment, but my explanation is the sole of clarity (hence the fishy aroma).
Well, I guess you learn something new every day.
Just to note, the pencil-and-paper strategy can be deceptive to those who aren't mathematically minded.
Also, for the variant in which MH doesn't know where the car is, *and* you get to switch if he reveals the car...
There's two primary cases.
1) 1/3 the time, he reveals the car. Switch to this door, you moron.
2) 2/3 the time, he reveals a goat. 1/2 of these times, you want to switch, 1/2 you don't.
All in all, 2/3 of the time you win, as long as you don't do the moron strategies (ie, switching to a known goat door, or not switching to the known car door).
If you always stay if he reveals a goat... then you get it the 1/3 of the time you were right the first time, and you still get it the 1/3 of the time he reveals the car, for a total of 2/3rds.
If you always switch, then the 2/3rds of the time you were wrong at first (per the original problem), you win, as long as you're picky when you switch.
In the "instalose" or "reset" variant, where if he reveals the car, they rewind the cameras and try again, then it doesn't matter.
and yes, at this point I'm being redundant.
PS, Mark: Given that you're actually asking, as opposed to the way you originally presented it which sounded like a Creationist "gotcha" question along the lines of Hovind, I'll go into a bit more detail.
200,000 years ago, give or take a bit, was when the most recent individual who was in the (direct maternal - they use mitochondrial DNA for this) line of descent for all modern humanity was alive.
About 4 million years ago was when the individual who was in the line of descent for both humanity and the most closely-related apes was alive.
All those who weren't on the line of descent from the second to the first, and aren't in the line of descent from apes, died off without leaving offspring who made it.
For instance, the ancestor of the Neanderthals - that individual was more recent than 4 million, but maybe not as recent as 200 thousand.
That make it a bit clearer?
Michael "Sotek" Ralston
Yes it make since.
Just to clarify. You believe that all the people on earth are decended from one women. And that all the veriations of skin color and such are from 200,000 years of evolutonary change.
Also that all people diverged from the ape about 4 million year past.
To go back to the original question. Again just a question.
Is this enough time?
Thanks Mark
Mark,
Keep in mind that the way that these estimates are made is by working backwards based upon rates of genetic divergence derived from other measures. Those rates are not unreasonably high. So in short, yes, it is enough time.
Mark: Right. And yes, given the way these dates are calculated, they're pretty much by definition enough time.
(And I wouldn't use the word "believe". It's what the evidence suggests. If you accept evolution, it's obvious that all members of any given species have a common ancestor far enough back. If you don't, well, most religions also claim that, just at a bit more recent date.)
Heh. This problem has been discussed for a long time. I had a Turbo Pascal program to illustrate it and some text. This was posted in the Fidonet Evolution Echo back in 1995, and that was a repeat of an earlier post to the Science Echo:
Post
Here's the text part. About the only thing here that may go any beyond the explanation of the opening post is my discussion of the idea that once one is left with only two doors to choose between, that there has to be a probability of 1/2 in there somewhere...
===
The analysis is pretty elementary for probability theory, although extremely non-intuitive.
Problem statement:
Monty Hall, MC for "Let's Make a Deal", would present contestants on the program with three doors, behind one of which would be a prize worth having, and behind the other two would be negligible consolation prizes or humorous booby prizes (perhaps a Thunderbird for the win, a toaster oven for the consolation, and a goat-drawn cart for the booby prize). The contestant would be asked to pick one door. Monty would then have one of the two remaining doors opened to reveal either the consolation or the booby prize. Monty never revealed the real prize at this step. The contestant was then asked whether he or she wished to stick with their original choice of door, or switch to the sole remaining unselected door.
The question, then, is whether the best strategy is to always stick with the original choice, to always switch to the remaining choice, or whether it makes no difference if one chooses randomly between the two remaining choices.
Making this general is very easy. We will consider probabilities for the class of cases of n doors, where n is greater than or equal to 2.
Let us call the user's original choice A, and the remaining choice
at the end B.
When the winning door is determined randomly, the user's original choice can be seen to have P(win(A)) of 1/n. The probability that the user chose wrong is then 1-P(win(A)), or (n-1)/n.
Monty reveals n-2 non-winning doors. This action has no effect upon the probabilities given previously.
The user can select between their original probability of winning, or the complement of their original probability of winning. In all cases where n > 2, it is to the user's benefit to take the complement, where P(win(B)) = 1-P(win(A)). In other words,
switching makes the best strategy for n > 2. At n = 2, P(win(A)) = P(win(B)).
But what about selecting randomly between the two choices? Half of the time the user chooses A, and half the time the user chooses B, so the probability of payoff is
1/2 * P(win(A)) + 1/2 * P(win(B))
Since there is a winner behind either A or B,
1 = (P(win(A)) + P(win(B)))
and the probability of a payoff reduces to 1/2, regardless of the
number of doors originally in the problem. However, for n > 2, P(win(B)) > 1/2. Switching is still the preferred option.
In summary, there is no case in which switching is not at least as good a strategy as sticking, and for all cases where n > 2 there is a probable benefit to switching, switching is identified as the best strategy for maximizing payoff.
Hopefully, this thread will now die its well-deserved death.
===
Wesley R. Elsberry
tgibbs, thank you. I will study on the topic more.
Michael "Sotek" Ralston, thank you also. It is interesting to me that both side of the debate beleive some of the same things.
Mark
"Behind one of them is a car and behind the other two are goats. You want the car"
no no no my friend I would much prefer the goat......
first I always picked the center and stayed with it 9 of 20
then I always picked the center and always switched....
13 out of 20.....
again 9 out of 20
when I swtiched it was 14 out of 20...very odd
Here's an easy way to understand the problem assuming Monty knows where the car is and never opens that door:
Suppose the first choice Monty offers you is an opportunity to choose any two doors, promising that if the car is behind either one he'll eliminate the other one for you and you'll win. It should be clear that you'll win 2 out of 3 times this way.
Choosing one door and then switching is the same as choosing the other two in the first place.
As steve s points out, the problem with this problem is that the questioner sometimes forgets to point out that Monty KNOWS where the car is, and Monty deliberately opens his door to reveal where he knows a goat will be. If Monty doesn't know, then the problem can be understood as Monty just happening at random to open a goat door, and it's a different problem.
Dave S.
The Monty Hall problem solution depends on what Monty knows before the contestant even selects a door. In the three door problem as stated, Monty knows what is behind each door and then selects a goat door. Monty's selection is not a random event and he, in effect, changes it to a two door problem where the chances are 1/2 for each of the remaining doors. Consequently, the contestant always has a 50 percent change of choosing the car door and no basis for switching doors.
If Monty does not know what is behind the doors, his choice of a door to open is a random event and if the car is behind it the contestant wins. If there is a goat behind the door that Monty randomly selects, then the contestant still has a choice of the two remaining doors. While there is no basis for the contestant to switch doors in this instance either, he has another chance of winning the car if it is behind the door he selects. Consequently, the contestant now has 2/3 chances of winning the car.
The probability of winning is not determined by whether the contestant switches doors or not. But it is determined by Monty's foreknowledge, or lack of it, of what is behind the door that he opens.
If Monty knows what is behind the door he selects to open, the contestant has 1/2 chances of winning and 2/3 chances if Monty doesn't know.
This typifies why educating people can be so hard. Lots of seemingly great explanations can be wrong. And lots of right ones can be good but still not effective. The original post IS correct. And it explained it well. And even gave people the means and method to show themselves.
Whether or not Monty knows clouds the issue. For the sake of the game he HAS to know. Because he ALWAYS reveals a non-winning door.
For the masses who say the chances are now 1/2 that would only apply IF the contents of the doors were shuffled after a door was opened.
Unfortunately, throwing numbers and probability theory at people who don't understand, isn't going to help. People are notoriously bad at probilities. You're already talking above them. That's just talking even higher.
Nice to see that the people who are trying it are seeing the results. Your chances of winning are twice as good if you switch.
The door you picked initially has a 1/3 chance of being correct. The only way that changes is if things are moved. They aren't.
How about you don't open a door. in which sitruation do you have a better chance of winning... picking one door or two? Sounds stupidly simple, but if you pick one and keep it every time. And someone else picks both other ones every time...
It was both refreshing and annoying to see this pop up. Good luck for those of you that want to know how it works. TRY IT. Forget the math you don't know how to use.
The Monty Hall Problem
----------------------------------
Hello,
Let's put it this way:
The number of possible cases is 12, out of which 6 are winning cases.
By symmetry we can reduce the number of cases to 4 by assigning the car to door No. 1.
If the car is behind door No.1 the 4 possible cases are:
1: Player picks door 1 - Monty opens door 2 - door 3 is empty
2: Player picks door 1 - Monty opens door 3 - door 2 is empty
3: Player picks door 2 - Monty opens door 3 - door 1 has the car
4: Player picks door 3 - Monty opens door 2 - door 1 has the car
If the player decides to switch doors after Monty opens the door, he/she wins in cases 3 and 4 and looses in cases 1 and 2.
If player decides to stick with his first choice, he/she will win in cases 1 and 2 and loose in cases 3 and 4.
Therefore, the odds of wining are the same whether the player decides to switch doors or not.
GS
Well... ooops...
All that is true except the last sentence.
Cases 1 and 2 have a probability of 1/3 x 1/2 to happen, the pick of the player (1/3) followed by the pick of Monty (1/2).
Cases 3 and 4 have a probability of 1/3 to happen because Monty has only one choice if the player picks a door with a goat.
Therefore the best strategy is to switch doors.
GS
I see it like this. Initially you pick one door, so your chance to win is 1/3. Monty gets the other two doors so his chance is 2/3. What he then does is essentially invite you over to his side, to the side with 2/3 chance to win. He opens one duff door on this side, eliminating the risk that you choose incorrectly once you made the switch.
I don't think it matters if he knows what is behind the doors. He basically offers you the contents of his two doors instead of the content of your one door, so by going over to his side your chance of winning has doubled. I would switch even if he did not open a door first.
I have been working on a variation of the problem. This concerns what happens when the host randomly eliminates a door.
On average we know that if the host’s eliminations are random, the game will be voided 1 out of every 3 games. By voided I simply mean that the host will reveal the car and nullify the game, leaving you with no decision to make. And, on average, the two other games will see a win by swapping and a win by sticking. For this reason, the 50/50 probability holds up.
Let's imagine it with three playing cards, two Jacks and one Queen. Below are the six possible outcomes, where A is the card you first chose, B is the card that is randomly eliminated and C is the remaining card:
---A-B-C---
1. J1 J2 Q
2. J2 J1 Q
3. J1 Q J2 VOIDED
4. J2 Q J1 VOIDED
5. Q J1 J2
6. Q J2 J1
This is mathematically the most likely outcome if you were to play the game six times. In that scenario, two games are disqualified. Out of the other four games, two would be won by swapping, two by sticking.
Now consider: the game is played as above six times. To make it simpler, there are six rows with three cards. And one of the cards in each row is a Queen, the other two are Jacks. So there are six Queens to be found.
The player chooses a card from each row. Question: at this point, how many Queens is the player most likely to have found out the six cards chosen? The answer here is not controversial - it is of course 2. That is the most likely outcome.
Ok, so to complete the game the host turns over a remaining card in each of the three rows. If the host discovers a Queen in any row, then that particular row is disqualified and you hand over the card you chose from that row.
Assume that the host DOES NOT know where the six Queens are. Nevertheless he successfully avoids eliminating any of them.
So you are now left with a choice of keeping your originally chosen six cards or swapping to the remaining six cards.
What is your best option to increase the number of Queens you end up with? Does it make no difference?
One thing we can agree on: if, as expected, two of the six games had been voided by the host eliminating a Queen - then for the remaining four it would have been 50/50 odds of getting a Queen. In that case, the most likely number of Queens you end up with is still two, whether you swap or stick.
However, even though the host’s eliminations were random, we know that none of the six games were voided. Are the odds stilling 50/50 for all six games, whether you swap or stick? If so, then the expected number of Queens among your first selections is probably 3, though you originally estimated 2.
It has been claimed that so long as the all host’s eliminations were random, the probability is 50/50 for the non-voided games. Is this true, regardless of how many games were voided?
In the case of no games being disqualified, has the probability moved away from 50/50, and if so in which direction - swap or stick?
Bear in mind that when you first selected six cards, the odds were 1/3 in each row, and you estimated you would probably have two of the six Queens. After the host has randomly eliminated a Jack from each rows, does that give you grounds to change your first estimate? If your first estimate still holds, then the odds - as with standard Monty Hall - are 2/3 in favour of swapping. If by sticking you ended up with the 2 Queens, then by swapping you would end up with 4. The interpretation here is: no games being voided is exactly the same as the host knowing where the Queens are and avoiding them.
On the other hand, if the eliminations were random but no games were voided, this might be a clue that you originally chose the Queen more often than you first thuoght. After all, a game can only be voided if you originally chose a Jack. Normally you would expect two out of six games to have been voided. Instead none were. Are you now likely to have selected more than your previous estimate of 2 Queens. If your first estimate of 2 was correct before, based on the information you had then, is there good reason to change it now. If so, is the best estimate now 3 Queens? That would bring us back to 50/50. Is it higher? Might you now be more likely to have 4 Queens? If so then the odds are 2/3 in favour of sticking.
So there we have it - conflicting answers. Which is correct?
Can't wait to hear what others up with!
Enjoy!
This is A LOT of fun. I am tutoring a class where this problem is used to arouse student's suspicion of their intuitive "logical" reasoning. For it, I wrote a simulation of both the knowledgeable (regular) problem and the amnesiac (where Monty forgets which door hides the money) variations. Get it here: http://discursively.org/index.php?entry=entry060809-182859
The real difficulty is finding an argument which correctly predicts the knowledgeable case, but does yield the same answer to the the amnesiac variation.
Simon
sorry: that last line should read:
does NOT predict the same answer to the amnesiac variation.
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