I believe that that is precisely why a bear is specified. If it were merely “an explorer”, there would be a countless number of starting points.
For example, just north of the south pole, you will a line of latitude of length 10 miles around, and south of that, one of length 5 miles around, and south of that, one of length 10/3 miles around; etc. Any point on a line of latitude 10 miles north of any of these lines of latitude will work.
Quoth Ellis Dee:
The question is incompletely specified, such that it is impossible to determine the answer without more information. To get the answer you spoilered, you need to specify that the game show host always acts this way, not that he just happened to do so this game. From the actual Monty Hall show, however, this was not a good assumption.
Indeed, you need that whenever you pick door number 1 and have it wrong, the host will open the sole other door with a goat. And, you also need, and this is less obvious, that conditioned on door number 1 being right, the host will open the other two doors with equal probability [at least, if you are able to distinguish between the host selecting door 2 and the host selecting door 3]; otherwise, his particular selection of door number 2 could be giving you information about whether door 1 was correct or not. You also need that the prize initially has an equal probability of being behind any of the three doors, even conditioned on your selection of door 1; this is implicit, of course, but not actually tautological. You basically need to know exactly what the a priori probability distribution on events is, and what information you learn is and isn’t relevant to this and how.
The problem with probability riddles is that so much turns on details of the relevant probability distribution generally left not explicitly specified, which can legitimately be reconstructed in a multitude of different ways.
The (insanely oblique) way I heard that one was If JASON is a DJ who plays FM and AM, what is his girlfriend’s name?
June
Abdullah the camel merchant has died. His three sons, Abdul, Ahmed and Ali, were having difficulty with the will. The will specified that Abdul was to inherit one half the herd of camels of which Abdullah had died possessed, Ahmed was to inherit one third of the herd, and Ali one ninth of the herd.
Owing to some unfortunate reverses in the camel market, Abdullah had died with a herd of exactly seventeen camels, The sons were in the middle of a heated argument about which camels to butcher when the kingdom’s vizier rode by.
After being apprised of the situation, he suggested a simple solution that allowed each son to inherit a whole number of camels. What was the solution?
The vizier loaned Abdullah’s estate the camel he was then riding. Abdul inherited nine camels (18/2) Ahmed six (18/3) and Ali two (18/9) leaving one remaining, which was used to repay the Vizier his loan.
One wonders, though, what Abdullah’s will wanted to be done with the last 1/18th of his herd…
I regret to say that the puzzle did not preserve that information.
That same vizier encountered another camel problem on his way home, where there was a huge traffic jam caused by a couple of slow-moving camels. Enquiring, the vizier learned that two men had been having an argument over who had the slower camel - and they were settling it by riding towards the city at a snail’s pace. The vizier pondered this for a moment, then made a suggestion. A minute later the road was clear and the two camels were heading citywards leaving a plume of dust behind them. What did he say?
“Ride each other’s camel”
You don’t need to know this, although without this assumption solving the problem requires more subtle mathematical ideas. You simply note that the problem statement gives you no reason to believe any conditional distribution over any other one. Then notice that no strategy outperforms the canonical one consistently (i.e., over all possible distributions). Therefore, unless you’re given additional information, you stick with the canonical strategy.
Again, you don’t need this assumption. We could note that the problem statement gives no reason to favor one door over another. Or, you could ensure that your probability of having picked the correct door is 1/3, by making your decision with a uniform distribution.
This is a fancy way of saying “Unless you’re given additional information, let’s mix every distribution together ‘uniformly’ into… one in which, conditioned on door number 1 being right, the host will open door 2 or door 3 with equal probability”. (Of course, that’s mixing the distributions with a particular weighting in mind to consider ‘uniform’, and that weighting already implicitly carries the assumption that door 2-opening and door 3-opening are equiprobable).
In other words, you’re just saying “If you don’t know anything else, you should assume the problem intends the ‘uniform’ marginal as your distribution”. Whatever. The way I see it is as “Without further information about the relevant probability distribution, the winning probabilities are undefined; however, with suitable assumptions about the relevant probability distribution, which are perhaps implicit but which one could ask to be made explicit, the relevant facts about the winning probabilities are provable”.
The same response to this as above; plus, there’s the assumption in here that your choice of door is independent of which door is correct. This is definitely implicit in the problem, and I have no problem with treating it that way, but I wanted to point out its assumption all the same. It needn’t apply to all probability distributions; one could imagine a distribution in which, as it happened, your choice of door was highly correlated with which door was correct. This might be useful for modelling, say, the possibility that you are psychic…
Hm, or maybe the response should be this:
No strategy outperforms “Always stick with door number 1” over all possible distributions either. (After all, one distribution has the prize always being behind door number 1). So this kind of reasoning doesn’t demonstrate why the canonical strategy should be considered superior to various other strategies. The only reason to consider the canonical strategy ideal is because one has reason to consider the canonical distribution the intended one.
An old favorite:
You possess a balance scale, and twelve coins. All of the coins are identical in every respect, but one. One coin is counterfeit, and it differs from the others by weight. You do not know whether the counterfeit coin is heavier or lighter, only that its weight is different from the other coins. You cannot find the counterfeit by touch.
In three weighings on the balance scale, find the counterfeit.
Solution:
[spoiler]First weighing: 1 2 3 4 against 5 6 7 8. Note which side is heavier and which is lighter.
Second weighing: 1 2 3 5 against 4 9 10 11. Again, note which side is heavier and lighter. This is where the gimmick lies: swapping one coin from each side of the scale to the other.
Now, let’s talk about the possibilities, of which there are only three. If the same side is heavier or lighter in the second weighing, then you know that the counterfeit must be 1, 2, or 3, and you know whether the counterfeit is heavy or light. Just weigh 1 against 2 for your final weighing to find the counterfeit.
If the opposite side is heavy or light for the second weighing, you know that the counterfeit must be either 4 or 5, because those coins switched sides, but you don’t know yet if the counterfeit is heavy or light. Weigh 4 against a known good coin to find the counterfeit. If they unbalance, it’s 4; if they balance, it’s 5.
And if the second weighing results in a balance, you know that the counterfeit is one of the coins you removed (6, 7, or 8), and you know whether the counterfeit is heavy or light. For the third weighing, weigh 6 against 7.
And of course there was an additional possibility at the first weighing that it would result in a balance. If weighing number one balances, there’s no need to swap coins; just make the second weighing 1 2 3 against 9 10 11, and go from there.
Note that it’s possible to find the counterfeit without determining whether it’s heavier or lighter in certain situations. However, that doesn’t matter; the only problem the puzzle poses is “find the counterfeit”, not determining whether it’s a heavy or light counterfeit. Simply weighing the coins as groups of four will not find the counterfeit in all situations, but the method of switching places with two coins for the second weighing works no matter where the counterfeit is.[/spoiler]
My answer was "If you don’t stop blocking the road I’ll have you both beheaded"Which does seem more in character for your standard vizier.
From Max Torque’s spoiler:
Actually, it is possible to determine whether it’s heavy or light in all cases.
Though I’ve always seen this problem listed with 12 coins, it also works for 13!
14 coins, however, and it can’t be guaranteed done in only 3 weighings.
No, that’s a standard Grand Vizier. This guy was still several rungs lower on the ladder of Vizierdom, and building a powerbase.
To continue with the coin problem, You can always, as Chronos points out, tell whether the counterfeit is heavier or lighter with 12 coins, but (and correct me if I’m wrong) with 13 coins it should be (as Max Torque states) you can always determine the counterfeit…but not always whether it is heavier or lighter.
Bonus!
I was just out walking the dogs with my wife and explaining/contemplating the coin puzzle. As we talked, I realized something I never had about this little problem.
So now I challenge you with an extension of this classic lateral thinking puzzle.
I posted a little while ago that the problem doesn’t work if you have to find the fake among 14 coins in three weighings. But I was wrong. It can be done provided you obtain one simple thing. Can you tell me what that one simple thing is (and it does not seem to be in violation of the spirit of the puzzle) and then tell me how you determine the fake coin?
Does this one simple thing allow you to find the fake among any number of coins or just 14 in three weighings? (Just trying to get a better feel for what sort of thing it is, though I suppose, if it’s not in violation of the spirit of the puzzle, the answer is almost certainly the latter…)