Game Show Probability: Newsletter Challenge (05/01/07)

I'd pick the same door again. My reasoning is this: If I had have picked a door with the goat behind it the first time, then the host would have just opened that door. The host's action suggests that he wants me to change my mind.

Or has the host already thought that I would think this? Or did he think that I would think that he....

Is there a twist? Should I pick the "third door" since that is where the car is?

Its all just to build the tension before the advertisement break.

I agree with Davo on the premise that if the wrong door was opened intentionally.

Therefore, the door requested to be opened has the car behind it.

Ok,

There are 2 sets at the start of the puzzle:

Set one: The door you pick

Set two: The other two doors.

Each door has a 33% chance of having the car so

Set one: has a 33% chance of containing the car and

Set two: has a 66% chance of containing the car

The host opens one of the two doors in set two, showing it doesnt have a car, and removing it from the set.

The probabilities of the sets containing the car have not changed, however there is now only one door in set two.

So you should switch to set two as it improves your odds of getting the car from 33% to 66%.

Questions?

Hi Guest,

The probabilities of the sets containing the car have not changed, however there is now only one door in set two.

Yes, the probabilities of the sets containing the car have changed.

The second "set" had a probability of 2/3 because it contained 2 doors out of the total number of doors (3 doors). After eliminating one of those doors the set has one door out of the total number of doors remaining (2 doors), so it has a probability of 1/2 of having the car.

See also my post #52

So my chance of winning the lottery tomorrow change if I win it today. Future probability changes ? C'mon !

Hi Guest,

I don't know if you're the same guest I replied to before or not. Subscription is free .

I'm sorry I don't see the connection between our arguments. I am saying that the probability of the car being in a particular 'set' = (number of doors in that set)÷(total current possibilities).

It is not about the future, and equally it is not about the past. It is about the present.

Yes. What is 1/3 and 2/3

1/3 ≈ 0.333333

2/3 ≈ 0.6666667

So 1/3 and 2/3 'is' 1.

(It would have been better grammar to write "What are 1/3 and 2/3?")

(It would have been better grammar to write "What are 1/3 and 2/3?")

Well, not exactly. If the answer you are seeking is a definition or description of those entities, then you are correct. Example, "What are 1/3 and 2/3?" They ARE fractions.

However, if you are talking about the SUM of those two numbers, and I believe that is the intent here, the word SUM or some other singular word (such as "total", etc.) is implied, so "What is (the sum of) 1/3 and 2/3" IS correct grammar.

"What is (the sum of) 1/3 and 2/3" IS correct grammar.

I assume you meant to say: "What is (the sum of) 1/3 and 2/3" ARE correct grammars.

Most English speakers around the world would say "Do the bloody maths, darn it!" In the US, we'd say "Do the blasted math @)(*&head!" So I propose that we write "grammars", instead of "grammar", to suggest that we make not just one decision in writing such a sentence, but many.

No, I stand by what I said. Notice I used quote marks. It was a SINGLE quotation, therefore IS is correct grammar.

Regarding the use of the term "math", it is short for the word "mathematics" is it not? Is "mathematics" a singular word, or a plural word? Singular I believe, as in a (singular) branch of academic studies. Is there any such thing as one "mathematic"? Perhaps "Do the bloody maths" is more of a colloquialism, and as such, does not need to be grammatically correct.

I love this admonition, " 'Ain't' ain't a word, and you ain't gonna use it!"

So is this ' 2 out of 3 buyers agree ' ? Maybe that's ' 666 out of 1000 '

The devil you say! 666 is "the number of the beast"!

Actually 2 out of 3, being .6666666666666... should be rounded to 667 out of 100, since the exact number would be a compound fraction, 666 and 2/3.

Hi All, Grammar is the study of the rules governing the use of a language. Grammars are the studies of rules governing the use of multiple languages. Davo

I was just being pedantic about 33% etc Davo . Bit hypocritical really as I do it as well to save typing time.

Hi Kris, This thread started off a bit slowly, but we seem to be forming two sides with irreconcilable differences quite nicely now. I defected from the 50:50 army, but really I think it depends on the host's intentions, Davo

Agreed Davo. I am looking carefully at the wording .Kris

Here's a quote that may clear things up.

• "But," you might say, "none of this shakes my belief that 2 and 2 are 4." You are quite right, except in marginal cases -- and it is only in marginal cases that you are doubtful whether a certain animal is a dog or a certain length is less than a meter. Two must be two of something, and the proposition "2 and 2 are 4" is useless unless it can be applied. Two dogs and two dogs are certainly four dogs, but cases arise in which you are doubtful whether two of them are dogs. "Well, at any rate there are four animals," you may say. But there are microorganisms concerning which it is doubtful whether they are animals or plants. "Well, then living organisms," you say. But there are things of which it is doubtful whether they are living organisms or not. You will be driven into saying: "Two entities and two entities are four entities." When you have told me what you mean by "entity," we will resume the argument.
  Quoted in N Rose Mathematical Maxims and Minims (Raleigh N C 1988).

That should help. Or not.

Hi Ken, You have covered all bases by saying it may help or not.

Hi Davo,

Actually, I was being facetious re "2 and 2 is 4" vs "2 and 2 are 4". In the US, "is" is more often used. I suppose the reasoning is that it is a shorthand for "The answer to the question (what is 2 and 2) is 4."

"Maths" sounds funny to us, but makes sense: we would not say "do the mathematic" (unless "mathematic" was the name of a dance step). "Grammars" sounds bizarre in this context to everyone, I suspect.

I'll admit my intention here and perhaps elsewhere is as much to obfuscate, as clarify.

Hi Guest,

By your reasoning, if the host opened the second door in your "set 2" and revealed a second goat, then 'set 2' would still have a 2/3 chance of having the car even though there were no doors left in 'set 2'!

"I'm making assumptions here about the "format" of the end game, and, the host's knowledge of the car's location."

Try it with a pack of cards; select nine red and one black; get your friend to shuffle them (the ten cards) thoroughly; choose one at random and place it face down without looking at it; tell your friend to look at the remaining cards and remove eight red ones placing them face up on the table. Now, what are the chances that the card you first removed is the black one, and, what are the chances that the card your friend still has is the black one?

Reply to post #52.

Different scenario: the host has no knowledge of the viewers (original) choices.

Reply #46 covers the options in simple detail.

Hi Davo,

I'm having second thoughts about your reasoning. If I select one door and stick to it every time then I have a 1/3 chance every time, regardless of which order the host opens the doors.

But if the host deliberately selects a door with a goat behind it every time, and if he never selects the door I choose, then he is acting as an intelligent 'filter' for the other two doors and stacking the odds in favour of the door he leaves closed.

Disregard prior comments. I am converted. That's IF the host is doing this.

Regards, Davo

Suppose you have seen (by watching prior contests) that the host acts randomly. For example, once you saw him revealing which door hides the car, even though the contestant's guess was wrong. When he asked you "Do you want to stick with your original answer?" the contestant got this odd look on her face -- like teenagers do when they say "Well, DUH."

But at other times, you've seen him reveal only a goat. Also, you've noticed that he flips a coin frequently. Every time you've seen him do this, it comes up "heads."

Now you're on stage. Will what you have observed re the host's behaviour affect your decisions?

I hope this serves to clarify or obfuscate.

Hi Ken,

Suppose you have seen (by watching prior contests) that the host acts randomly. I don't think I could observe randomness. I would have to interpret his actions as random or guided.

"Once you saw him revealing which door hides the car." Once is not very many times. How many times have I watched the show to see the host reveal the car only once?

"Every time you've seen him do this, it comes up "heads."" Now I really think your trying to obfuscate.

If the host's choice is random (perhaps 'uninformed' is a better word) then he has 1/3 chance of finding the car and you have 1/2 x 2/3 =1/3 of finding it.

BTW, I enjoyed your post to Davo from Davo.

re Random: good point. I should have written "that the host appears to act randomly".

re "Once you saw him revealing..." Let's say you've watched the show ten times, and he revealed the car once.

re "Every time you've seen him ..." There is not much obfuscation going on here. I think it is at about this (referenced) post where I began to obfuscate less, and became vaguely helpful. Maybe. Here, the point is that what the host does, aside from having shown you a goat's location, makes no difference in the probabilities as presented to a given contestant.

re "If the host's choice is random..." I think this gets to the crux of the issue. When you get up on stage as the contestant, it really doesn't matter what the host has done in past contests. From what you can observe on stage, you can only say that you picked a door, and that he opened a different door to show you the location of a goat. Did he do this intentionally? Who knows. Does he always do this? Who knows. You also know he is saying that you can change your original guess. Did you learn anything that would make you want to change your guess or hold with the first guess? I think so.

Put another way: If this is the only time the show has ever been run, and you are the only contestant, and the show is going to close (from bad ratings) immediately after your time on stage, you can still, with confidence, say what your probabilities* are of winning the car if you 1) stay with your original guess or 2) switch. (This is making the reasonable assumption that the host is not moving objects around, etc. It is also making the reasonable assumption that, given that he has opened, for example, door 2 to reveal a goat, that you do not then switch to door 2.)

As this probability question is usually presented, you are told that the host always knows the location of the car, and always opens a different door than the contestant chooses, and always reveals a goat. (That would mean that, sometimes, he has to choose between two goats, and other times he chooses between the car and a goat.) But none of that really makes a difference for the guy on stage.

So, your final statement is not precisely true as regards the current contestant's decision. Whether or not the host acted randomly, he has already acted: whether by random act, guided act, or divine intervention, he has shown you the location of a goat. How or why he did that does not matter to the current contestant. It might matter to the show's accountants. If he makes a habit of (whatever), then future contestants might have an easier or harder time of it.

* You, of course, cannot say if you will win or lose. You can only assess the probabilities.

BTW, in light of your conversion, I think you probably are already aware of much of what I've said above. I've included the (perhaps excessive) detail for the benefit (or am I evil incarnate?) of others.

Hi Ken

I think you are being mischievous.

If there is a consistent history of a particular behaviour, the probability of that continuing must be high, and should be accounted in the decision.

Fyz

I'd agree that I have been mischievous.

However, re history influencing your decision to switch or not, I am not being mischievous in the post referenced. Unless the question spelled out in some detail what the history actually has been, then you have to make the assumption that you do not know that history, and are faced only with the instant.

Maybe, the host will, for several weeks, reveal a goat (the other goat) only when the contestant picks a goat. If contestants recognized that as a pattern, then they could wind lots of cars. Likewise, if the host reveals a goat only when the contestant picks the car, then again, the contestants win lots of cars. So you can reasonably assume that he is not doing something that would obviously erode the game. The only behaviour like the one described in the question (that he shows a goat) that can be extended reasonably to a consistent behaviour (without giving the game away) is the expected one -- that he always shows a goat, and that goat is behind a door other than the contestant's first pick.

But maybe he throws an odd behaviour in every now and then. You can't know that from the question, so you have to resort to simple probability. So, regardless of reasonable history, the probabilities work out just as you have worked them out. The assumptions you may want to annunciate are pretty straightforward, and no one would crucify you for failing to make them explicit.

In that sense, the history doesn't matter -- you can infer it from the data, and if you feel the need, you can make it explicit in your answer assumptions.

If there is a consistent history of a particular behaviour, the probability of that continuing must be high, and should be accounted in the decision.

Actually, in this case, consistent behaviours that give away the game would lead to just the opposite. If the host continues too long with such behaviours, then everyone learns how to win a car every time, so there will be a strong incentive to change behaviours.

But it wouldn't be every time - and some game shows are designed to reward particular characteristics - knowledge being the most common, but intelligent behaviour could be another.

See another of my replies to you to see that, if the selection is random there is no benefit (or downside) in swapping - because the cases that create the benefit in swapping have been excluded - host would have displayed the car in these cases, rather than transferring to the door with the goat. If in doubt, try it out (doggerel accidental, but left anyway).

N.B. If you or I were to go on a game show (please note subjunctive), I'm pretty certain we'd assess it first to assess optimum strategy. Most contestants watch them anyway - so something would be known about history - which I also imagine becomes part of the format.

Fyz

Hi Gang:

The car AND the goat are behind BOTH doors! Once a door is picked and opened, we collapse the wave function and get only one choice!! Hee-Hee.

Hank

Wave your prize goodbye?

When given a choice of 3 and one is eliminated, and you have an option to "reselect" a door you now have a 50/50 chance of selecting the right door any way you look at it! Those odds are undeniable. This is 5th grade math. Are you smarter than a 5th grader?

Depends on the fifth grader. In this case you are talking about RANDOM reselection - whereas you would have additional information to improve your chances if the game-show host is known to behave consistently. (To extend your argument to the absurd limit - would you still say you had 50:50 chance if the doors were transparent)

The host and the guest are in adverse positions. At first, the guest has 1 in 3 chances to win. After the host reveals a goat, the guest stands a better chance, 1 in 2 to win, so he should simply flip the coin, i.e. pick again on a random basis. But the guest is smart enough to know that the host wouldn't put him in a better position, hence revealing the goat it is a last chance diversion. But the host knows that the guest is smart, hence him revealing or not revealing the goat is a random act. Again, the guest is smart enough to know that revealing the goat is a random act. Hence, his best chance is to pick again between the remaining doors, i.e. whether he keeps his first choice or not should be decided by the flip of the coin. For the host, in the long term, the pseudo-random revealing of the goat pays because on ocasions he skews the guests choice towards missing the car (the IQ of the guests follows the Gauss "bell") .

What nonsense. A flip of the coin can never the best strategy. If the odds are 50:50, it won't change them. If they are not 50:50, there is a better strategy.

Being the brilliant actor I am I would ponder a bit, look confused, scratch my head, look over to my family in the crowd (who will cheer... naturally) and then hit the gameshow host on the chin a couple of times until he tells me which door to open.

Now, to maximise the propability all you have to do is choose a door and open it yourself (after already having incapacitated the host) because he was clearly rigging the show. There is no reason to believe that he will stay true to form and not open the door you chose, nor is there any reason to believe that he will try the double-bluff and actually open the door you chose. With all that said propability is useless and I return to my original concept of brute force solving all problems. (Especially problems involving pansy game show hosts bah...)

The better question would rather be... Howd they get a car behind a door... A curtain I could understand, but a door? I think the show is pulling a fast one... There are goats behind all three doors.

BWAHAHAHAHAHA!!!! That's gotta be the best answer I've ever seen to one of these challenges! Good Show, Good Show!

Switch. In the first "round", there was a 1 in 3 (33%) chance of finding the car behind the door you picked. After the host has eliminated one of the doors, the unopened, unpicked door has a 1 in 2 (50%) chance of finding the car.

I'm not sure that I fully believe this explanation, since in my mind, the chances of the car being behind the door you originally picked has gone upto 50% also. But this was the answer given last time this question appeared as a Challenge , so it must be right.

hi ER....just a feeling so do not switch...game host is baiting you.....but no matter what it is still 50/50 at this point.

This is a poorly (presumably intentionally) worded question regarding conditional probability, and a key consideration--the specific conditions the host uses to select a door to open--is key to the decision process & outcome. Depending on the host's method for choosing a door to open, a player's odds of winning may increase OR be unaffected by switching to another door.

At the outset there's a 2/3 probability that a player selected a goat (either "goat A" or "goat B")--which means a select & hold strategy wins a goat 2 of 3 times. Thus, the odds of having selected the car after a door is opened to reveal a goat remains 1/3 (even though the odds of the car being behind one of the two doors is 50% the odds a player missed picking it are still 2/3).

A) IF the host will ALWAYS and KNOWINGLY open a door to reveal a goat (and never a car) then switching increases one's chance of winning to 2/3! There are three possible outcomes at this point in this scenario:

-- If you've picked goat A you'll switch to a car (when goat B is revealed)

-- If you've picked goat B you'll switch to a car (when goat A is revealed)

-- If you've picked the car you'll switch to & win the other goat

B) IF the host opens an unselected door at random--pure guessing--then switching will make no difference and the odds of winning are 50% if a goat is revealed. In this scenario there are four possible outcomes (given the problem boundary conditions that the host didn't accidentally open a door to reveal an unselected car after the player picked one or the other of the goats):

-- Player picked goat A, goat B is revealed, switching wins car

-- Player picked goat B, goat A is revealed, switching wins car

-- Player picked car, goat A is revealed, switching wins goat B

-- Player picked car, goat B is revealed, switching wins goat A

C) If the game is structured such that the host has the option to offer to open a door presumably the game is "rigged" to entice the player to change their selection from the car to a goat (by conveying the illusion of the first scenario described above). In this scenario switching is not a built-in random act but is rather a manipulated event designed to entice the most profitable outcome for the host and the player's odds of winning the car after switching are likely 0.00%.

Surely in situation A, there are FOUR possible outcomes, too. Namely:

-- If you've picked goat A you'll switch to a car (when goat B is revealed)

-- If you've picked goat B you'll switch to a car (when goat A is revealed)

-- If you've picked the car you'll switch to & win goat A (when goat B is revealed)

-- If you've picked the car you'll switch to & win goat B (when goat A is revealed)

i.e it is not just picking a door which is an event here, but also the revelation of one or other goat. Thus chance of winning the car are 1 in 2. Chances of winning the remaining goat are also 1 in 2.

What leads to confuses the situation here is that both "booby" prizes are goats. But they are still two separate, individual (but indistinguishable) goats. If one was a goat and the other a sheep, then the analysis becomes clearer.

Also, the motives of the host are unknown. The only "known" element regarding the host is that he reveals one of the booby prizes.

This changes the probability of either of the remaining doors revealing the car to 1 in 2. Probability of an event only depends on the current situation - not on past events (or their related probabilities).

Most of the answers on this blog reveal a poor understanding generally of probability. I suspect the answer will do likewise.

If you've picked the car initially, when a goat is revealed (A or B) you can only switch to the other goat (B or A). So the last two options are really the same relative to getting/not getting the car.

I agree with you 100%!

By 100%, do you mean "all the time" i.e., given a sample of 100 responses by this guest, you would agree 100 times?

Or do you mean that you agree with everything the guest says in this post? What if something written in the post is incorrect? Then you would have to accept his mistake as your own. Not that there necessarily is anything grossly wrong... or maybe there is.

Is this guest really "aiming" for the correct answer, or merely waving the gun in the general direction of a correct answer? Put another way, if his 30,000 cfm blower, (propelling him along on this line of reasoning) fails, do you want to be stuck in the middle of the lake with him? Are there Gremlins in his logic?

Oh, Ken! You crack me up 100%!

Me too. 1/3 of the time and 2/3 of the time in any order consecutively or at the same time concurrently.

You are wrong yet again. Post 46 is correct, as was 41 (approximately) and some others.

Fully listed as suggested in post 74, there would be six options, a pair for each of the three in section A of post 46. BTW, I'm still awaiting your bb apology - I only indicated that I intended to accept it...

Re-expressing the classical answer without redividing the cases:

The door you select initially will prove car-full on one occasion in three. Similarly, the car will be behind one of the other two doors on two occasions in three. If the host always opens a door that does not conceal the car, that does not change the chances that the car was behind one of those two other doors. Nor does it change the chances that the car was behind the door you selected originally. So the chances are still one in three that is behind the original door, and two in three that is is behind the third door.

Got it now? If not, get an acquaintance to help you with an experimental check (two reds and a white?). Then argue with mother nature as to whether she understands statistics.

Fyz

How about this: The contest involves looking at 1 double-door garage and 1 single door garage.

Yes, we could think of the game-show host always opening a door as being equivalent to him always giving you the option of opening two doors instead of the original one, and keeping the car if it's behind either. Obviously, the order of opening the doors makes no difference, and it doesn't matter who opens them.

BTW, I have a large lawn, so perhaps more use just now for a goat than for another car.

Fyz

Have a BB gun ready for the Goat. Only joking - some of my best friends are Goats.

As long as they're not goat-boaters...they're just a menace. They are dying out though, so it's less of a problem these days.

Fyz - LOG IN!! (and yes, I am shouting )

Sorry, I usually try to remember.

Fyz

We like to know you're there! <hug>

Is there an emoticon for increased embarrassment?

↑

Fyz,

Well, I have read all the postings so far as well as the web-pages referenced for "Monty's Dilemma" but I remain unconvinced. This enigma is still a conundrum for me, or is it the other way around? I used to read Vos Savant's column all the time and found her to be only about 99% correct. She was not infallible.

Also, I do not accept computer simulations as proof, since they may be based on the same faulty "logic" which compels people to arrive at the some conclusion that you and Vos Savant (and many others) have.

Let's examine your "re-expression": Yes, the door you select will prove "car-ful" 1 out of 3 times. Yes, if the host always opens a door that does not conceal the car, that does not change the chances that the car was behind one of those two other doors.

Ah, but there's the rub: There is a 2 out of 3 chance that one of the two other (non-chosen) doors will reveal a car, but if you open one to reveal a goat, you also remove 1/3 of the original 100% of possibilities (or half of the 2/3 probability that you agreed the two doors had). The remaining non-chosen door will now only have its original 1/3 chance of being "car-ful". Since that is equal to the 1/3 possibility (of the original 100%) for the door you have chosen, but you now know that there is a 100% probability that the car IS behind one of the two remaining non-opened doors, we must conclude that both choices now have a 1/2 chance of revealing a car. Therefore, even though revealing the goat DID increase your chances from 1/3 to 1/2, you cannot improve your chances by switching.

I would love to see an actual, statistically sound, physical experiment proving otherwise! Remember, in the actual experiment, there is also the possibility that BOTH non-chosen doors will have goats, and ALWAYS switching will abandon 1/3 of the probability of having the car behind the originally chosen door. ALWAYS switching is just as bad, or good, as NEVER switching.

~~~~~~~~~~~~~~~~~

Wait a minute. The above was what I was GOING to say. At least, until I concluded my own "thought experiment" as follows:

If instead of one car, we have 1000 cars to give away (make it 999, to avoid confusion) and they are equally distributed, i.e. for every set of of three doors (999 sets) and an equal number of games (trials), there will be 333 behind door A, 333 behind door B and 333 behind door C, with only one car per game. Yes, switching every time loses me 333 cars, but with 666 cars remaining, and the host ALWAYS revealing one of those two doors to have a goat, I will get ALL of the other 666 cars. Not a bad trade off. Or in other words my chances of getting a car by always switching is 2/3. So,....I guess....I was wrong. THIS TIME!

Man, this was a tough one. No wonder there was no consensus. Even knowing, and understanding the right answer, I find it hard to fault my original logic! I think it only goes to show you that probability is only meaningful when you have a relatively large sample size. For one instance, it does not matter much if you lose that you had only a 1% chance of winning or a 99% chance of winning, you still lost! Only 100% probability has a 100% guarantee! That's why chronic gamblers will always lose money in the long run. The sample size becomes quite large and the odds are ALWAYS in the House's favor!

Oh, and if you must have an apology for my inference that you were full of feces on the bb gun challenge, very well, I APOLOGIZE. In fact, you cannot be full of feces, since so much of it has already come out of you!

(Just kidding! Lighten up! )

Hi STL Engineer.

You probably have absolutely no conception how aggressively you come over. And not just when you write to me. I fear it is you who seem heavy handed, and your apology meaningless, as you have simply not modified your behaviour. Lighten up yourself - stop insulting your virtual colleagues and pretending it's in jest - it simply doesn't wash (pun intended).

On this topic, considering the single case works just fine - provided you junk the fallacy that you can change the relative probabilities of something being in a particular position without moving anything around (this is effectively what the posts are doing when they assume that the result of the host removing a goat is 50-% between the remaining positions).

Also, this particular experiment is quite simple to set up - put the car in any of the three positions, and go through your possible choices in turn. If you get the car first time, (1/3), you'll lose out by changing your mind. If you fail to choose the car initially, you'll definitely get it if you change your mind (2/3). Easy. The problem is the rule-based way that statistics is mistaught. Don't know anything about Vos Savant, however.

Fyz

Wow, Fyz, from the tone of your last post, I would guess that you are truly angry, and for that I am sorry. What you take as genuine insult was really, truly only meant in jest, and I am sorry that you took it otherwise.

I guess we have a different outlook on social interaction here in the US than you do. One of our favorite ways of teasing our friends involves telling them they are full of it in interesting and inventive ways. We often trade insults in jest, and often the only way to tell if it IS a jest or not is if there is some cue, i.e. a smile, a grin, or online the smiley icon or abbreviations like ROFL. The classic example is the Western gunslinger in the movies who thinks he might have just been insulted, pulls out his gun, and says, "Smile when you say that, boy!"

OK, you pulled out your gun. I am SMILING!

Back on topic: Marilyn Vos Savant was a weekly columnist for the Sunday national newspaper supplement, PARADE MAGAZINE. Her column was called "Ask Marilyn", and readers would send in challenging academic questions from any and all fields, usually logic-, philosophy-, or mathematics-based, and Marilyn would print the question, with an answer to be found on another page, upside-down and in fine print, just like a crossword puzzle answer. She would often go into a more elaborate explanation in the next week's column. Many of her answers were quite controversial, and she was not always 100% right, however her track record was extremely good. She was supposed to be some kind of certified genius, and I often wondered if Vos Savant was just a pen-name, playing on the word "savant". The column also ran her picture, and she definitely was a good looking woman, as if to say, "look, smart women can be pretty, too!". Or is it "pretty women can be smart, too!"?

According to one of the websites referenced, it was in a letter to Vos Savant that the "Monty's Dilemma" question first appeared, with her answer being given as you and others described. Too bad I missed that one! <grin>

Thanks. It's more likely local or generational than national - I worked in the USA for some time and never experienced this type of low-level insult. It was once quite common in parts of Eastern Europe, but seems to have vanished - except when people there actually mean it (so be careful what you say if you ever venture into these regions).

I looked up Marylin vos Savant (reportedly, she took her surname from her mother) on Wikipedia. It also talks about Monty Hall. She was born in St Louis, and managed to upset the American Society for the Deaf - perhaps I'm not alone in my reaction to St Louis humour.

Fyz

I guess we have a different outlook on social interaction here in the US than you do. One of our favorite ways of teasing our friends involves telling them they are full of it in interesting and inventive ways. We often trade insults in jest, and often the only way to tell if it IS a jest or not is if there is some cue, i.e. a smile, a grin, or online the smiley icon or abbreviations like ROFL.

You by no means speak for the entire US, STL. "Ranking out" friends (as we called it back in the 60's) is, by and large, confined to boys of middle school and high school age. Occasionally, it is seen in adults who have not had the opportunity to mature normally, and who are enduring the stresses of rough neighborhoods, crummy jobs, etc. I've spent time in both the US and England, (and in various other countries), and have not found hurling insults common anywhere among normal, polite, adults. Certainly, saying that someone is full of feces, (or more colorful language to the same effect) simply because they present a well-reasoned argument counter to your own can correctly be viewed as just plain insulting, not kidding. Throwing in an occasional emoticon does not change that.

I am not immune to being abrasive or overbearing, but I would certainly not say that all Americans are abrasive and overbearing. Better that I should accept my own flaws as my own, rather than blaming everyone else.

Nolo contendere.

You put it more bluntly than ever I would, but thanks for the compliment ("well reasoned")

DON'T GIVE IN STL!

you are being intimidated into wrong thinking. there is a simulator here that shows 50% prob for not switchnig, not 33%:

http://www.remote.org/frederik/projects/ziege/empirie.html

The simulation is programed to show 66% for switching -- but it makes the wrong assumption, just as the 50% proves! How can you have a combined prob of 116%!!???

So maybe others are not so smart after all!

Forget history. ten coin tosses, all heads. What is prob, now of tails. 50% NOT something greater than that!

I do research (which uses stats) all the time! I have a PhD in math and can tell you for sure that if you've got two doors the chance if 50-50. IT can't be any other way! Your gut feel was right.

I would like to agree with you, guest. I really would. However, on careful examination of the logic of both the intuitive approach and the mathematical proof given on the same website tp which you provided a link, I have to ignore my "gut", and go with the paradoxical answer espoused by many others, i.e. the contestant that ALWAYS switches will have a 2//3 probability of winning. Now, if the contestant flips a coin to decide whether or not to switch (50% odds of choosing one or the other), then his probability does go back down to 1/2.

And I do not base my answers on verbal intimidation, but on my analysis, faulty or not, of the facts presented to me.

I did not understand your statement, "The simulation is programed to show 66% for switching -- but it makes the wrong assumption, just as the 50% proves! How can you have a combined prob of 116%!!??? "

Can you elaborate?

Hi STL, guest here again.

here's a link that proves marilyn savant was wrong. The prob is 50%, just like you said.

http://www.wiskit.com/marilyn/gameshow.html

When given a puzzle to solve, one has to assume that all necessary information to solve the puzzle is contained within it or is common knowledge. Otherwise it is just a question with multiple interpretations and answers. For example, the original puzzle never stated that the host had ulterior motives. It also did not state that the doors were opaque. Marilyn Savant gave the most logical answer.

Unfortunately, that link doesn't 'prove' anything. I see both arguments presented.

Why don't you check out this link...

http://math.ucsd.edu/~crypto/Monty/montybg.html

It is an excellent link - but it most certainly doesn't show 50% - it shows in what respects the question was inadequately framed and what is the consequence of different assumptions. For the case of the 'original' game show, Monty Hall put it rather well in his two direct quotes: "You should switch"; and "You need to know what mood I'm in". It has also been suggested (Wikipedia) that Monty "played on the psychology of the contestants", a game two could play...

However, in order to convert it to a mathematical challenge, you simply need to state your assumptions. The scary thing is that not everyone has solutions that correspond to their apparent assumptions.

Fyz

You say it makes "the wrong assumption", but don't say what that was. Is that a way for a researcher with PhD to communicate? Under "fair" assumptions (host always opens one of the unchosen doors, and that door always has a goat behind it), the probability for switching is 66.6..%, for not switching is 33.3..%. Total, 100%.

What assumption do you think we should be making?

Fyz

Hi STL Engineer

I omitted to say - your ability openly to correct your mistakes wins my deep respect.

Fyz

Thanks Fyz,

And I am truly sorry for any grief my feeble attempt at humor caused. You were correct in saying, "You probably have absolutely no conception how aggressively you come over."

However, I must comment on a few of the things that Blink added to your own rebuttal:

"You by no means speak for the entire US, STL.", said Ken. I never said I did, Blink.

Blink then added,"'Ranking out' friends (as we called it back in the 60's) is, by and large, confined to boys of middle school and high school age. Occasionally, it is seen in adults who have not had the opportunity to mature normally, and who are enduring the stresses of rough neighborhoods, crummy jobs, etc."

Ranking out...being roasted...getting served....whatever you want to call it is hardly "confined to boys of middle school and high school age" or occasionally "seen in adults who have not had the opportunity to mature normally, and who are enduring the stresses of rough neighborhoods, crummy jobs, etc."

In fact, if you look at the stand-up comedy routines of most professional comics, you will probably find that insulting others, individuals or even groups, is probably one of the most popular forms of "shtick", right next to the self-deprecating form. Look at comedians like Don Rickles (What are you looking at, you hockey puck?"), Adam Sandler, Eddie Murphy, Jeff Foxworthy ("You may be a redneck...."), Jerry Seinfeld, Robin Williams, Chris Rock,and many others. And of course, amateur humorists tend to mimic what they see and hear on stage, radio, and TV, often with less than stellar results. And it is certainly possible to go over the line from humorous put-down to impolite insult, without even realizing it. Of that I may certainly be guilty. But, regardless of what you may think you know, you can in know way know what was in my mind at the time, and I find your comparison of me to juveniles and immature losers to be highly insulting. So practice what you preach, Blink!

I'm sorry to have so offended you. Actually, I was not putting you in the class of misfits and losers. On the contrary, I was considering you a professional engineer who would be above such things. I have nothing but tolerance for people who have found themselves in situations in which foul language and insults are the norm. In fact, I have worked in industrial plants in which the universal adjective (fxxxxxx) is used more than perhaps any other word. I would never admonish those people to clean up their acts.

You are not such a person. I have great respect for your thoughts and contributions. So when you say that Fyz is full of it, it seems out of place.

And sure, there are professional comedians who are both filthy and funny. You pay to see them, and you expect raunch. Everyone in the audience has paid, because they want raunch. Here, not everyone in the audience has paid.

We all like you. I hope that you can understand that if you say rude behaviour is common in America, (but perhaps not so common in the UK) that I, as an American, might take exception. You can probably understand how I would want to publicly say that you do not speak for all of us -- at least not for me.

I've always enjoyed your posts, and expect that I will continue to do so.

Hi STL

There is a place for everything, and stand-up comics work in a particular environment. Most of them don't behave similarly outside of that environment - in fact, some are painfully shy, and may adopt a stand-up persona in a situation where this is expected partly as a release. (Whoops, bovine feces psychology, wash your mouth out Fyz).

Personally, I try to avoid put-downs, unless things are becoming desperate (which is usually too late).

Regards

Also, the motives of the host are unknown. The only "known" element regarding the host is that he reveals one of the booby prizes.

If the contestant knew the host's motives, would that change the probabilities?

If the host opens the door with the car behind it pick that door, there's a car there.

The assumption must be that the host knows where the car is and doesn't pick it.

The answer is switch because your odds are improved to 66% as discussed above.

Chris S

I like Davo's comments (#1) on trying to out thinking the Host ... but if it were me, I'd stick with the original choice hoping that you odds will continue their upward movement from 33 1/3% to 50% then from 50% to 100%.

by the way English Rose .... I like your latest floral arrangement the best so far ....

Thank you - it's the crest of Rose Brothers Ltd; the firm I work for is located on one of their old sites. I have one of the crest plates at home - my Dad saved it when the Rose Packing machines he worked on at WD & HO Wills were scrapped.

Incidently, the efficacy and ingenuity of the packing machines designed by Rose's led to Cadbury's naming an assortment pack after the firm...can you guess which one!

E R....i thought that crest looked familiar...it my younger day i worked on many rose-forgrove wrapping machines ...did penny bubble gum twist wrap.....big heavy cast frame, wonderfull works....fun to take apart and work on....i think they were sold here in US by "package machinery" (now gone) as i remember some of the other name plates above the rose name....some of the units were from 1940-50's.....stilll ran well in the mid 70's....only problem was some threads were british wentworth, which was a issue in US and many machine ended up with a mixture of threads.....db

Some of those 1950's machines are still running (although in Glasgow rather than Bristol) packing Golden Virginia 1/2 oz packets. Some are in the Bristol Industrial Museum; Dad moved them to Hartcliffe in 1972 and then to the Museum & Glasgow in the late 80s. Apparently, no other firm has ever made a machine better at the task!

I don't know about the Cadbury's machines, but I'd have a small wager that they're still going! Anyone know?

English Rose

While I don't totally disagree with Uncle Red regarding your latest floral arrangement, it is interesting. But I liked the Blue rose. It was simple, elegant and distinctive.

With regards to the choice of doors, I don't believe there is anything logistically you can do to maximize your probability of winning. It's purely chance.

Joe

I agree with you Uncle Red on the "Upward Movement" of the odds.

The very fact that you are on the show (being asked about the doors) was an increase from 0% (like the rest of us watching at home) to 33 1/3%.

So stay the course my contenstant friend .... 0% to 33 1/3% .....33 1/3% to 50% .... 50% to 100% .....

also, if you did change and lost then you'd forever regret that you could have been the owner of a Gremlin....

Assuming that the host always follows the same procedure: you should switch. For your original choice there is a one in three chance that you're right and a two in three chance that you're wrong. Once the host has revealed a goat there's still a 1 in three chance that your original choice was correct, but a two in three chance that the car is behind the "third" door.

Unless of course you like goats!

How do you justify saying that there is a two in three chance that the car is behind the other door?

Before any door was opened all doors had a 1/3 chance. Now that 2 doors are left surely both doors have a 1/2 chance. When the door was opened to reveal the first goat, how did that door's probability leap onto one or other of the other doors?

From the moment you make your first decision: you have divided the doors into two groups the door you chose with a 1/3 chance of the car and the other two with a 2/3 chance of the car between them. Whether or not the host reveals a goat, he is effectively saying you can either stick with your original choice, or, you can pick the "other two" doors.

Suppose there are 1000 doors you choose 1; the host then reveals 998 goats. Do you stick with your original one in a thousand guess, or, switch to the one other door the host has left unopened.

I'm making assumptions here about the "format" of the end game, and, the host's knowledge of the car's location.

To make your comparison fair , the host would reveal 333 doors/goats and you'd have been able to select any half of the reminder - those 333 you initially picked or the 333 untouched.

Haven't we seen this question before. Randall is right. To try to make it clearer, there are only two choices:

1. Make one pick and then keep that pick or,
2. Make one pick and then change that pick.

• If you make the long odds on the first pick and get the car, 1 chance in 3, keeping you selection insures a win. So the odds are 1/3 x 1/1 = 1/3
• If you make the short odds the first pick and get a goat, the host then eliminates the other goat. By switching your choice on your second selection you get the car every time you miss on your first guess. So the odds are 2/3 x 1/1 = 2/3 you win by switching.

I still think there's a flaw in this logic. Try this scenario:

1. You have a bag containing 3 balls: 2 black and 1 white.

What is the probability of pulling out the white ball? 1/3.

2. You pull out a black ball. You now have 1 black and 1 white ball in the bag. What is the probability of pulling out a white ball now? 1/2

The two choices are independent. The probability "counter" is reset between the actions, just as there is always a 1/6 chance of rolling a 6 on a fair die, however many times you have or haven't already rolled a 6.

With the boxes, you have no way of knowing what was in the box you first picked. If it was the car, then whichever box the host opened would have a goat. If it was a goat, the host only had one choice about which box he would open. Once he's opened that box, you are now in part 2 of the scenario above with the bag of balls and the chances of the car being in either box is 1/2 as all you know for sure is that the car and one of the goats are still to be located. The previous "knowledge" is invalid in the the new situation.

OK: you've got a bag with 99 black balls and one white ball in it. You select one ball and without checking the colour you put it in a drawer. Your friend now takes the bag and removes 98 black balls, placing them on a tray in front of you. You want the white ball: do you go for the one in the drawer or the one still in the bag.

You might be really unlucky; the ball in the drawer may be the white one, but, I bet you'd choose the one still in the bag.

Your odds of having originally pulled the white ball increased from 1:100 to 1:2. Same analogy as the 2 goats/3 doors example, you're still down to a 50/50 chance of choosing the correct door.... keyword is CHANCE.

Hi Guest,

Another keyword is "filter". The original ball was randomly selected so it had a 1/100 chance of being white. The second ball was selected by filtering out the black balls so it has a 99/100 chance of being white.

The odds would even up if you put the two balls back in the bag and randomly selected one, but this is not the way the game show is run.

"The original ball was randomly selected so it had a 1/100 chance of being white."

Yes, that part is correct.

"The second ball was selected by filtering out the black balls so it has a 99/100 chance of being white."

GONG! Oh, sorry, you made an incorrect statement. But thanks for playing our game anyway! For your answer to be correct (99/100 chance of being white), you must ASSUME that one of the 99 is white, but what if you had the white ball all along? Then there is a 0/100 chance of the "filtered" ball being white!

Really, Davo. You should have stuck with your original answer, and not switched just because the opinions of some logic-impaired nincompoops sounded plausible.

When you are given the choice of switching or keeping your original, and you know that one has the goat and the other has a car, then what has happened before, even if "filtered" by the host, has no impact on the outcome. You still have a 1/2 chance of getting the car either way because you have the freedom to choose either one of two unknowns.

It's like Russian Roulette. If you have one bullet and six possible chambers, and spin the cylinder every time (neglecting gravity effects), you will ALWAYS have a 1/6 probability of firing the bullet. However, if you pull the trigger each time without spinning the cylinder, every time you pull the trigger the odds of firing the bullet on the next pull goes up, 1/5, 1/4, 1/3, etc. If you survived to where there were just two untried chambers, there would be a 1/2 chance that either would contain the bullet. Do you really think your odds of NOT getting the bullet would be any better by skipping ahead to the last chamber?

What Davo has stated is correct.

Says you!

Actually, the first time you pick the ball, chance of getting the white ball out of the bag is 1 out of 100, but for the second time, the chance of picking the white ball out of the bag is 1 in 2. But if you choose the bag in the drawer, you are still using the probability of 1:100. So, to increase you odd, you should pick the ball out of the bag because the ball isn't replaced inside the bag.

So, back to the challenge question, given the same logic above, the first time the door is pick, I have a 1:3 chance of getting the car. After the goat is revealed, the unpick door is now having a 1:2 chance of having the car, whereas if I stay with the original door, my chance is still 1:3, so I would change and pick the unpick door.

MidniteFighter

There is only one ball in the bag at that time.

1 ball in drawer with 1% chance of being white.

98 black balls on table with 0% chance of being white

1 ball in bag with 99% chance of being white.

--------------------------------------------------

100 balls accounted for and 100% chance that one of them is white.

My explanation does not require preknowledge for the second choice. In fact there is no real second choice.

You can either assume you correctly guessed the right door on your first pick or you can assume you guessed the wrong door. There are no more choices after this.

If you assume you guessed correctly, then there is one chance in three of you being right. The host removing one goat does not change this probability.

If you assume you guessed wrong, then changing your selection gives you 2 chances in 3 of winning.

As stated the host is not picking randomly, he knows were the goats are. If he was picking randomly then he could select the car and changing your selection would not help.

Hometwo

I think we have to consider that the host acts acording to certain rules, and in this case his rule is ALWAYS OPEN THE DOOR NOT SELECTED BY THE CONTESTANT, in the first trial he have two doors to open and he select it randomly, in the second trial he opens the door not selected by the contestant.

Under this consideration we can say this is a fair play, not a triky one and the logic to solve it is in the 185 replay.

Ciro

the odds do not change when the host reveals one of the goats, they are 1/3 for choosing any given door, and the remaining door has odds of 2/3 since he revealed one of them with a goat behind.