Red/blue ball challenge - Math or probability

If the OP had stated “Suppose you’re in a job interview and your prospective boss asks you the following question…” that would have been a completely different context.

FWIW, I reject the idea that a employers would rather hire someone who plunges ahead with imperfect information rather than asking for clarification. For example, if a customer brings a key into my locksmith shop and says “This key doesn’t work; make me another one.”… I’d be rather disappointed if one of my employees just went and duplicated the key without asking for more clarification. I can think of at least 4 different scenarios, all of which might be described by the customer with the words “This key doesn’t work.”

#1 Last week, the customer brought in a working key to be duplicated, we made a mistake and copied it onto the wrong key blank, and the duplicate won’t go into the lock at all.
#2 Last week, the customer brought in a working key to be duplicated, we didn’t clamp it into the vise properly, and the duplicate key goes into the lock but only turns if you wiggle it.
#3 Last week, the customer brought in a worn key which only works when you wiggle it, asking us to duplicate the key, our key machine was slightly miscalibrated, and the duplicate goes in but it doesn’t work at all, even if you wiggle it.
#4 The customer just bought a new car and it came with just one key (not made by us) and this key only works if you wiggle it.

Those four scenarios would be resolved in four very different ways, using various machines and different techniques. And I’m sure there are more than just the four I thought of, off the top of my head.

I would want my employee to ask for clarification about exactly how the key “doesn’t work” and where the key came from, before deciding which course of action to take to try to fix the problem. How can you fix a problem if you don’t even know what the problem is?

Hmmm…

I didn’t realize until I got into the thread that there was a reservoir. I interpreted the question as “pull two balls out at random. If X then look in the bag and pull the necessary coloured third ball.”

Yes, that thought crossed my mind as well. Suppose I remove two red balls. The rules say I’m supposed to “add in one extra blue ball”. Clearly, I can’t take a blue ball from my hand and put it into the bag because there aren’t any blue balls in my hand at all, just two red ones. The problem expressly told me that I only have 17 blue balls and now I need an 18th blue ball. So I thought maybe I read it backwards, that I was supposed to take a blue ball FROM the bag and add it to my hand.

I found a thread on reddit discussing this puzzle and david55555 made a similar observation: “There is an unexplained edge condition. What happens if you pull out two red balls. Since they are the same color you are supposed to add a blue, but from where?

One of my pet peeves is smart puzzles written by dumb (or sloppy) people. They are oblivious to all the glaring ambiguities and unstated arbitrary assumptions in their crappy writing.

I think the worst are the so-called IQ tests written by people below the level the test aims at.

It’s not just the sloppiness per se, but when you see a question written like that you don’t know if it’s an interesting problem. I don’t want to waste half an hour thinking about somebody’s idea of a “lateral thinking” problem when it turns out that the solution is to twist the meaning of the question by facile wordplay.

When I sent this puzzle to my kids, I worded it as follows:

You have an urn with 17 red balls and 17 blue balls, plus a reservoir of red and blue balls. You pull them out 2 at a time and…until there is only one left. What color is it? Simple and clear. You might want to vary it with m red balls and n blue balls.

As for the Monty Hall problem, the point is that Monty always opens a door that has a goat (or whatever) behind it. This forces you to assume that he knows the correct door. Incidentally, the best explanation I have for the solution is to redefine your problem as initially trying to choose a goat, for which you obviously have two chances in three. If you are successful and play the strategy of always switch, you win. So playing that stategy gives you a win 2/3 of the time.

In general, when such questions are asked what is wanted is how you go about it, not the answer. How do you think on your feet. For some jobs this might be important. For others not so much. For program developers, thinking deeply is more important than thinking swiftly, IMHO.

Agreed.

Though I’m a lot more inclined to attribute issues to incompetence rather than malice. Which also *really *torques me off when I do see a crappy problem statement that turns out to have been maliciousness all along. As so well described here:

Where’s my big knife when I need it?

Misrepresentation of facile wordplay as a “lateral thinking” problem is a strict liability crime. Even without intent, they still get the knife.

Malice and incompetence are not mutually exclusive. I once saw a puzzle that said to look for a mistake, followed by a bunch of numbers. There’s no mistake in the number section; the solution is that there’s a misspelled word in the instructions, and that’s the “mistake” you’re supposed to find. But then some Goober corrected the misspelling and forwarded the puzzle on. So the new version of the puzzle instructs you to look for mistakes and there aren’t any.

I think the Goober was just too deep for you. The mistake was that there was no mistake. Is your mind blown?

Well, if he didn’t know, then it wouldn’t be a “Monty Hall” problem, would it?

But this was apposite to the discussion of stating problems clearly. This critical aspect of the Monty Hall problem is often glossed over or not stated at all; perhaps originally because there was an assumption of familiarity with the show?

It’s striking that even the framing of Monty Hall on Wikipedia is ambiguous!

Although Wiki states correctly that the host knows what’s behind the doors, it doesn’t state clearly the host’s method for choosing which door to open. It is critical to know that the host looks behind the doors, then opens a door that has a goat because it has a goat. The situation is different if the host picked at random, and just happened to reveal a goat by chance on this occasion.

Incidentally, I find the easiest intuitive resolution of Monty Hall is this:

Suppose slightly different rules:
You choose a door.
Host offers you the better of the two things behind the other two doors, if they are different.
Of course you switch to the two doors, 2/3 is better than 1/3.
Now, the host reveals which of your two doors is better and which is worse (if there is a difference), and you take home the better one.

Now, you can see that the above process is functionally identical to the actual game. The host is effectively offering you the better of the other two doors (if there is a difference) if you switch. It’s just that he shows you which is better and which is worse (if there is a difference) a little earlier.

This simply isn’t true. In both the original presentation of the problem by Steve Selvin in American Statistician and Marilyn vos Savant’s restatement of it in the infamous 1990 Parade magazine column it was clearly indicated that the host was aware of which door (box in the original problem) contains the prize and which ones have goats (empty in the original problem). Furthermore, the question is rendered after the second door/box is opened and determined to not contain the prize, so whether the host knowingly or incidentally exposed the non-prize choice of the remaining doors/boxes is irrelevant; the only thing that matters is that the remaining choice is between keeping the original selection (with 1/3 chance of containing the prize) or the changing selection to the remaining box (which, because the host has opened one, essentially means that you’re getting to look behind both, giving a 2/3 chance). The line contention of whether the host knew and the psychology behind whether he would have opened a door if the contestant’s original selection was correct is outside the scope of the problem, which is an exercise in interpreting conditional probability.

The statement of the red/blue ball problem is perfectly clear, if not absolutely explicit. It assumes, quite reasonably, that the reader will understand that the original set of each 17 red and 17 blue balls are in a reservoir from which they are removed (in pairs without replacement) or added (one at a time) per the stated logic. The only real ambiguity worthy of nitpick is the final statement (the color of the final ball removed) which is admittedly unclear as it is stated that balls are only removed in pairs, so the question should be the color of the remaining ball. I understood this is in a quick reading of the problem. The statement is certainly not complete in the sense of a formal proposition in a proof, but it should be clear to people whose first language is English, which contains many accepted uses of grammar and implicit possession which are far more ambiguous, requiring context and an understanding of local vernacular to interpret.

Stranger

No, if this were so, why would the host need to know what’s behind the doors at all?

Suppose you initially choose door 1.
Now simulate:

If host chooses a goat deliberately, we have:
1/3 of the time the car is behind door 1, Monty opens door 2 or 3
1/3 of the time the car is behind door 2, Monty opens door 3
1/3 of the time the car is behind door 3, Monty opens door 2
Switching wins 2/3 of the time.

If host chooses randomly, we have:
1/6 of the time the car is behind door 1, Monty opens door 2
1/6 of the time the car is behind door 1, Monty opens door 3
1/6 of the time the car is behind door 2, Monty opens door 2 - CAR SHOWN
1/6 of the time the car is behind door 2, Monty opens door 3
1/6 of the time the car is behind door 3, Monty opens door 2
1/6 of the time the car is behind door 3, Monty opens door 3 - CAR SHOWN
It’s not clear in this simulation what you are allowed to do if Monty has revealed a car. But there are 4 equally likely cases where Monty has not shown a car, and in 50% of them it’s correct to switch.

In a nutshell, if Monty is picking randomly, then if your initial choice was a car, it is subsequently twice as likely that Monty will randomly show a goat. Conditional probabilties are calculated accordingly.

Stranger, I just looked at your two links - thanks for that, I had never seen the original publications that caused so much controversy.

In Steve Selvin’s original work, he says very explicitly that Monte is deliberately choosing a goat [actually an empty box in that version]:

But it’s striking that Marilyn vos Savant does not make this explicit, this is her entire framing:

It seems to me that a lot of the controversy from Marilyn’s column probably arose from this ambiguity. And clearly, people still haven’t got the point, because the ambiguity is retained in Wikipedia! Okay, one could argue that there’s no reason to say that Monte knows what’s behind the doors if he’s not going to use that information, so we should infer his deliberate choice of a goat. But still, I think it should be made explicit given that the result is so couniterintuitive.

Again, as I’ve laid out above, if Monte is picking randomly between the two remaining doors, and just by chance reveals a got, it is indeed an even chance to switch. Only when Monty is deliberately picking a goat because it is a goat is the couinterintuitive “classic” Monte Hall solution “2/3 chance if you switch” correct.

So all those PhDs and Nobel Prizes who supposedly got it wrong? I’m not convinced, maybe the whole thing was based on poor framing.

After Hall has opened one of the unchosen doors and revealed a goat (or empty box), it doesn’t matter whether he had foreknowledge or not. The fact is, the contestant now has some new knowledge about the two existing groups; his original selection (1/3 chance) and the group of two unchosen doors (2/3 chance). By switching after one of the unchosen doors has been opened he effectively gets to open two doors instead of one. Arguing that Hall would only open the door if he knew the contestant had chosen correctly to begin with is totally outside the scope of the problem and is in no way implied by the statement in either case.

Having a doctorate or Nobel Prize does not make one right in every instance. I work with and interact with a number of Ph.D.s and they’re frequently wrong about things outside of their area of expertise, and sometimes even within it. The reflexive answers–that the probability doesn’t change, or that it becomes 1:2 for each of the remaining doors–show that the respondent hasn’t thought through the conditionality of a prior and *a posteriori * states, which is the point of the problem. Making some tangential argument about how the host might try to second guess a knowledgeable contestant isn’t removing ambiguity; it is adding additional conditions to the problem that are to be found nowhere in the problem statement.

Stranger

Stranger, it’s not often that I’d expect to catch you out, but you’re wrong about this.

Game where Monte picks a goat deliberately:
If I pick a car initially (p=1/3), then Monte picks a goat with p=1
If I pick a goat initially (p=2/3), then Monte picks a goat with p=1

Game where Monte picks randomly:
If I pick a car initially (p=1/3), then Monte picks a goat with p=1
If I pick a goat initially (p=2/3), then Monte picks a goat with p=0.5

Now work through the conditional probabilities given that we know “Monte picked a goat” in each version.

Sorry, I have to go with Stranger on this one. Go read post 26. Define your task as finding a goat (or nothing, but let’s stick with goats here). You will succeed 2/3 of the time. Now if Monte opens a door finds a goat, switch. If you were successful on your first choice, you will automatically win a car and that is 2/3 of the time. If you never switch you will win those trials only 1/3 of the time. Now consider the 1/3 of the time Monte opens door and finds a car. Then you have lost already. Thus playing the always switch strategy, you will win 4/9 of the time (2/3 of the time Monte finds a goat times 2/3 of those times you win). Still better than the 1/3 if you play never switch.

If you don’t believe me, simulate.
Simulation of game where Monte chooses a goat deliberately.
Assume I initially choose door #1.
(1a) 1/6 of the time the car is behind #1, Monty opens #2 - reveals goat
(1b) 1/6 of the time the car is behind #1, Monty opens #3 - reveals goat
(2a) 1/6 of the time the car is behind #2, Monty opens #3 - reveals goat
(2b) 1/6 of the time the car is behind #2, Monty opens #3 - reveals goat
(3a) 1/6 of the time the car is behind #3, Monty opens #2 - reveals goat
(3b) 1/6 of the time the car is behind #3, Monty opens #2 - reveals goat
Note that here (2a) = (2b) and that (3a) = (3b) because of the forced “choice”. There was never a prior possibility that Monte could reveal a car.
Here, all 6 equal likely prior possibilities are included in the reckoning, because the condition “Monte revealed a goat” does not eliminate any of them. This gives the classic Monte Hall result, always switch for a 2/3 chance to get the car.
Simulation of game where Monte chooses randomly.
Assume I initially choose door #1.
(1a) 1/6 of the time the car is behind #1, Monty opens #2 - reveals goat
(1b) 1/6 of the time the car is behind #1, Monty opens #3 - reveals goat
(2a) 1/6 of the time the car is behind #2, Monty opens #2 - REVEALS CAR
(2b) 1/6 of the time the car is behind #2, Monty opens #3 - reveals goat
(3a) 1/6 of the time the car is behind #3, Monty opens #2 - reveals goat
(3b) 1/6 of the time the car is behind #3, Monty opens #3 - REVEALS CAR
Here, two of the prior 6 possibilities must be discarded from the reckoning, because of the condition “we know that Monte revealed a goat”. That’s what makes the difference. This conditioning under these rules makes the switch 50/50 between the remaining 4 possibilities.