Excellent Math Riddle

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I am pretty sure that there is no way to get 998 and that you have a fatal flaw in your analysis.

My guess is that you want the first wizard to subtract the smaller hat number from the larger hat number and give that as his choice. Then, the second wizard would see three numbers missing, figure out which two one would subtract to get the choice of the first wizard, and select the third as his own. And so on.

If that is so, then it doesn't work except in specific hat distributions.

For example, what if the first wizard doesn't see hats number 500 and 510. So he selects 10 as his choice. The second wizard would then look to see which hats were missing and determine that the two seen by the first wizard was 500 and 510. If hat number 750 wasn't seen, then he would choose 750 and get it right.

Sounds good so far.

But what if he sees that hats 500, 510, and 520 are missing? All he can know is that the first wizard either sees 500 and 510 as missing or 510 and 520 as missing. He is left with a guess between hat number 500 and hat number 520.

Even if the second wizard's hat was 750, when you get to a wizard with either hat 490 or hat 520, he is going to be left with a 50% chance of getting it right.

With a method like this, you really can't guarantee that any wizard will be faced with anything but a 50-50 guess and the total number of winners could easily be significantly less than 500.

Remember that for you to guarantee that 998 wizards win, the method must work for every possible distribution of hats. If there is even just one distribution that will not guarantee a win for 998 wizards, then you don't have a solution that guarantees wins for 998 wizards.

I really think that the best one can do is a guarantee that 500 wizards win like Somberlain suggested in which the odd numbered wizards give clues specifically for the even numbered wizards. His suggestion of doing a kind of round robin where the guess can wrap is brilliant. I wish I had thought of that.

I actually tried to play it out on a piece of paper- but with only ten wizards and eleven hats.

Labeling the unused hat N11, and labeling the number picked by the first wizard N1, the second N2, and so on...up to N10.

And picking random numbers between 1 and 11 for each,AND following the rules.

I assumed the ploy of the first guy trying to tip off the rest by giving the difference between the two missing numbers (N1minus N11, or the opposite, whichever is bigger) instead of a real guess.

The values happened to two and seven - so each wizard knew the difference was five.

And went through each wizards options. Very complicated.

But usually each wizard would have three options - the number of unseen numbers- minus the numbers that had already been picked by previous wizards. Knowing that the point spread had to be five reduced that number of choices- not usually to one- but to two typically. So it did increase many wizards odds somewhat.

As it has already been said a few times, and assuming all wizards coöperate and don’t needlessly ruin each other’s chances to get promoted, each one has two possible hat numbers after discarding those he sees in front of him and those previously guessed.

Each wizard’s correct option will be, obviously, his own hat number, and his incorrect option will be the option the previous wizard discarded, no matter whether it would have been correct for him or not. That is, each wizard will say one of his two options and hand the other to the next wizard as the incorrect option. The first wizard’s incorrect option is the missing hat number.

Therefore, for any given wizard to be certainly promoted, the previous wizards have to use their guesses in some way to convey him the one bit of information he needs to determine which one of his two options is correct, according to the strategy they agreed on before the trial.

If we just want to make sure 500 wizards are promoted, it can be done by having each wizard at an odd-numbered place in the row use his guess to send that information to the next wizard. A given odd-numbered wizard will know his two options and the next wizard’s hat number. Let’s arrange these three numbers in decreasing order, calling them a, b and c, with a > b > c. The next wizard will know what these numbers are—his own two options and the number the previous wizard “guessed”—and therefore can also order them and know which is a, which is b and which is c. Therefore, Somberlain’s strategy can be used: if the even-numbered wizard’s hat number is a, the previous, odd-numbered wizard says b; if it’s b, he says c, and if it’s c, he says a. Now the even-numbered wizard can tell which of his two options is correct from the number the previous wizard “guessed”.

In order to make sure 999 wizards are promoted, there’d have to be a way for every wizard other than the first to deduce which of his options is the correct one from the previous guesses and the knowledge that they’re all right, with the possible exception of the first. The first wizard can never be guaranteed to get his guess right, so all the remaining wizards have to be. I haven’t managed to work this out.

We need a new strategy, because, if the first and the second wizards follow Somberlain’s, there are cases in which the third has no way to deduce which of his two options is correct. Namely, if the second wizard’s guess is less than the first’s, he’ll know his own incorrect option is between those two numbers, and otherwise that it’s outside that range. But if his real hat number also happens to meet the same condition, he won’t be able to tell it apart from the wrong option.

Considering all wizards except the first have to be right, only the first can actually use his “guess” to send information, even if the other wizards have to make use of other guesses to decode it. He’ll also know that the option he discards will reach all the other wizards as their own incorrect option.

b) The 998 solution:

The first wizard tells the average of the two unknown. There will always be 3 unknowns for the guessing wizard, and from 3 unknowns, the arithmetic mean of the 2 will always be unique, eliminating confusion. The wizard will use rounding in case of encountering averages with decimals.

Edit: Oh, that was stupid. Whatever.

The 999 solution is not possible, because the first wizard should send the data of the two unknowns in one number. The boundary ''over 1001 is not allowed'' forces the first wizard to say another wizards number, ruining the 999 solution.

And 998... Well, it is only possible if we can manage to point two unknowns with a single number within boundaries.

One more thought in case it comes in handy: if a wizard “guesses” a number which isn’t one of his two options, which necessarily is the hat number of a wizard in front of him, any wizards in the middle will have an extra option (they’ll see in front of them a number already said by a wizard behind them, so the chance to eliminate another number will be lost), and the wizard whose name was “guessed” will have one less option again, because he won’t have the correct one. In other words, when a wizard says an invalid option, he’ll relay his two plausible ones to the next wizard, rather than just one, and this excess will be purged when the wizard whose hat number was said fails to have a correct option.

This can be nested, so each time a wizard “guesses” a number other than his plausible options, the number of options is increased by one, and it’s decreased by one when the turn comes for a wizard whose hat number has already been said to make his guess.

Of course, this can’t be used if you aim to promote 999 wizards with certainty.

If they are true wizards they can easily cast a magic spell to see what number hat is on their head without having to perform any calculations. If they can't do at least that, then what is the benefit of being a wizard?

I guess that’s against the rules and they’d get caught

Ditto if being a mathematician means prioritizing the consistency of the story over an interesting mathematical riddle

The strategy of having the tirst wizard 'tip-off' the rest of the wizards by NOT doing the expected thing of picking a guess from one of the two numbers that he doesnt see on the heads of the wizards sitting in front of him- but by shouting the DIFFERENCE between the two missing numbers- probably would not work.

The problem is that if he did not use that strategem but did it the normal expected way- he would leave each of the subsequent wizards with a choice of two to guess from- giving each a fifty-fifty chance.

This is because he knows N2 through N1000, and only N1001(the unworn hat), and N1 (his own hat)are hidden from him. So he has a fifty-fify choice.

If he makes a real guess, he will pick either N1, or N1001.

So when the next wizard gets his turn that wizard will be forbiddened to guess either N1, or N1001 (whichever his predecessor picks). But he will not be able to see N2 (because it is sitting on his own head). So even though either N1, or N1001 have been eliminated- N2 is added. So like his predecessor he has a 50-50 choice of two.

And so it goes down the line.

But if WIzard one makes a fake guess- and gives out the point spread between N1, and N1001, but does not give out either of those numbers-this creates two problems- it guarantees two losers (himself- and the wizard who really has the number of the point spread number on his hat).

But it also means that he fails to eliminate one of the two mystery numbers because he didnt say either N1, or N1001. He said a third number. So wizard two cant see N1, N2, nor N1001, and he cannot discount ANY of them because none were guessed by wizard one. So he has -not two- but THREE mystery hats to pick from. So his odds have ostensibly been REDUCED from one/two to one/three!


But he does know the point spread between N1 and N1001- so that may or not partially compensate. Probably- in the long run for all he wizards most ofthe time- it would cancel out. So the subtracting strategy probably would not yield a net increase in winners because - though it gives with one hand (by giving the clue of the point spread) it takes away with the other (by forcing each wizard to have three -rather than just two- choices). And it also guarantees two losers.

This could be wrong but I revised my strategy and I think it solves for 998.

I initially generalized to 10 wizards and 11 hats and my original strategy for this was a range of sum totals the first wizard sees with correlative values in two separate ranges of 1-11 and it covered the range of sum totals but it can't be generalized to 1000 wizards and 1001 hats and I realized it wouldn't work.

The first wizard needs to express the two missing values and somehow express it with 1-1001.

So the first wizard expresses the sum of the two missing values. I constructed a range of possible sums for the two missing values and you get something like this: 3-2001.

Take this range of sums and overlay two ranges of 1-1001 for expressions. If the sum of the two missing values correlates to the first expression between 1-1001 he guesses in a high tone and if it falls in the 2nd then he guesses in a low tone. One of the ranges is expressed in a low tone and covers the first half of the sum totals range and a high tone for the second part of the range. Using the process of elimination, the second wizard knows the two missing values and while this robs one subsequent wizard of a correct guess, that wizard simply states one of the 2 missing values expressed by the first wizard.

Is this flawed?