Errors in puzzles?

[mith]

When I get back, I am going to work on going through all the old puzzles and correcting errors, formatting, etc. Maybe adding some graphics if anyone has anything fitting for the more recent puzzles. So, if anyone knows of any problems (spelling, bad answers, etc.), post them here. Include puzzle number, please.

[Bicho the Inhaler]

I think the answer to puzzle #6 is oversimplified, if not completely wrong. Couldn't I just as easily do this (invisible):

Separate the set of positive integers into the set A of numbers containing no 3, and the set B consisting of numbers with at least one three. We can make a rearrangement of the sequence "1, 2, 3, 4, 5, ..." by interlacing the two sets like this:

1, 3, 2, 13, 4, 23, 5, 30, 6, 31, ...

alternating between the sets but counting in order within each set. What percentage of the first 10 numbers have 3's? 50%. The first 1,000? 50%. The limiting ratio is clearly 1/2. We could actually make the limiting ratio be any real between 0 and 1 by finding some rearrangement. I'm surprised this isn't mentioned in the solution, since it's a pretty well-known idea.

[Bicho the Inhaler]

In puzzle 94, should "heresy" read "hearsay" in the second paragraph?

Originally posted by Kevin (emphasis mine):

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It also may be relevant that the women are notorious gossips. If a woman discovers a man to be untrue, either on her own or through heresy, this fact will quickly be known throughout the village...

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[Gomez]

In the Spy on the Submarine puzzle, could you please change the first sentence to read:

During the Cold War, a submarine communications officer received the following message via a fax which read:

Originally the bolded part read in morse code which I remember confused some people in the original thread.

Secondly, in the Statues in the Park puzzle, could you please delete the capitalsed inscription MAGICAL MYSTERY TOUR, since it wasn't released on the Apple record lable.

Thanks.

[DropOfaHat]

Regarding the Statues in the Park problem...

Unfortunately, removing the "Magical Mystery Tour" inscription won't quite do the trick, because it's still a reference to "I Am The Walrus", which is on MMT, which wasn't released on the apple label. I mean, you'd really have to change the statue itself to resolve the issue (maybe make it a reference to a beatles song on an album that WAS released under the Apple label - Maybe an octpus in a garden?)

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[morrisonlucas]

Bicho - Rearranging the terms like that (puzzle 6) is a well-known trick, but it is not allowed, for exactly the reasons you point out.

There are rules to infinite series. Generally you evaluate them in order. You are allowed to rearrange a little bit using rules that I'm having a hard time expressing. The rules ensure that the value stays the same. (I believe that you are allowed to clump terms and rearrange within a clump, but not to rearrange to entire infinite series.)

I can't find a reference in my textbooks right now, maybe someone else reading this can.

[Bicho the Inhaler]

morrisonlucas, yes, that is precisely correct. Given a sequence, it is generally not safe to rearrange the terms when studying it. As another example, there is a well known theorem that says for any series that converges but does not converge absolutely (i.e., the series Sa_n converges but S|a_n| diverges), you can rearrange its terms to approach any sum, or even diverge. This is the Riemann Series Theorem.

However, notice that the original problem reads:

quote:

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What percentage of all integers contains at least one instance of the digit three?

For example, 13, 31, 33 and 103 all contain the digit "three" at least once.

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There is no mention of any sequence; clearly, the integers are to be interpreted as a set, not a sequence. Agree? The question as worded has no well-defined solution. If we approach the problem by taking limiting ratios of some sequence containing every integer, then the answer depends on the sequence we choose. The sequence 1, 2, 3, 4, 5, 6, ... is not special here; its standing is equal to that of any other sequence, including the one I proposed in reply 1. The fact that each term is greater than all preceding terms is a nice property that this particular sequence has, but it doesn't give us a reason to choose this sequence over any other.

[CrystyB]

ow yeah and #126 should be " 'Fattest Stack' Puzzle Revenge" . Maybe with a link to it (it is #65).

Answer to #16 mentions Millikan Oil Drop Experiment. Would it be possible to make that a link to some site explaining the experiment?

while i agree with the puzzle being oversimplified i do not agree with you choosing another sequence. Limits are defined through n-->inf, and even if the function has humps, it still approaches 1. ((i think))


[This message has been edited by CrystyB (edited 08-08-2002 01:19 PM).]

[Bicho the Inhaler]

Crysty: Sorry, I don't really understand your last sentence...if you mean that the limiting ratio of numbers with "3" in them in the sequence I chose is actually 1, not 1/2, then I think you are mistaken.

[morrisonlucas]

Bicho-
I see your point that this is a set, not a sequence or series. However I still agree with the answer as posted.

Since it is an infinite set the number cannot be calculated exactly. (I'm pretty sure that thinking too hard about infinite sets has driven smarter people than me insane.) One strategy would be to calculate the expectation of a random number having a three in it (=1-0.9^(numDigits)) but it's meaningless to have a uniform random distribution over this set (ie what is the average numDigits).

Generally infinite sets are dealt with as limits. I was going to say that any reasonable limit would produce an answer of '1', but your example is a good counterexample.

Maybe

Originally posted by Puzzle 6 Solution:

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...illustrates one of the many "problems" associated with trying to apply concepts (like percentages) used for regular sets on the infinite...

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is more apt than we thought and the question is meaningless.

What if the question itself were rephrased as "What percent of the series of integers contains ..."?

[Bicho the Inhaler]

That phrasing does somewhat hint at the intention of treating the set of naturals as the sequence "1, 2, 3, ...", but I think it does only that. To arrive at the stated solution, we would still need to make the assumptions that (1) we're being asked to treat the naturals as a sequence and take a limit, and (2) we're supposed to take the limit of a certain property of the sequence "1, 2, 3, ..." and not some other sequence that contains every integer exactly once. I don't think mentioning the word "sequence" or "series" in the question is sufficient to fix the problem in a satisfying way, and moreover, speaking of "percentages" of sequences is imprecise mathematical language. To arrive at that solution, I think it would need to be phrased something like this: "What is the limiting percentage of all numbers less than some number k as k becomes arbitrarily large?" But that destroys the beauty of the puzzle.

Okay, so here's my humble opinion. We need to glean the essence of the puzzle, and I think Kevin hit the nail on the head with this:

Originally posted by Kevin Lin:

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This seeming paradox illustrates one of the many "problems" associated with trying to apply concepts (like percentages) used for regular sets on the infinite.

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(Except "problems" shouldn't be in quotes, since it is a genuine problem.) I think the question is fine as is, and the answer mostly is too, except that it is wrong to claim that the ratio is 100%. I think we would lose very little (actually, may gain some) "coolness" in the solution by keeping what's there (which at first glance looks reasonable enough), but then noting that we just need to shift the set around a bit, which shouldn't make any difference at all since sets don't have order, to make the ratio completely different. We could easily make the ratio any rational, and with a little extra effort, any real number from 0 to 1 inclusive. This would really drive home the difficulty of generalizing everyday concepts of finite sets to infinite sets.

[SplitCrystal]

Speaking of puzzle #6...
Part of the answer reads:
"What percentage of the thousand numbers contains at least one three? Twenty-seven point one (27.1) percent."

Isn't it actually 25.2%??

[Bicho the Inhaler]

Suppose we choose a random number of up to 3 digits (from 0 to 999, inclusive, I mean) by choosing 3 random digits (between 0 and 9, inclusive). What's the probability that it doesn't have a 3 anywhere? 0.9 * 0.9 * 0.9 because each digit has a 0.9 probability of not being 3, and the 3 digits are completely independent. So the probability is 0.729, which makes the probability that it does have a 3 1 - 0.729 = 0.271 = 27.1%?

[SplitCrystal]

Bicho,
I'm sorry. I must be dense but why would you do it that way?

[Bicho the Inhaler]

It seems correct to me; is there a flaw somewhere? You still haven't shown how you got 25.2%...

[cha]

Bicho - To expand on your earlier reply: what if we chose a random number of up to 30 digits by choosing 30 random digits (between 0 and 9, inclusive). What's the probability that it doesn't have a 3 anywhere? 0.9 * 0.9 * 0.9... because each digit has
a 0.9 probability of not being 3, and the 3 digits are completely independent. So the probability is about 0.042, which makes the probability that it does have a 3 1 - 0.042 = .958 = 95.8%. The larger the number of digits we choose, the closer we get to 100% certainty that it contains a three.

[Chuck]

Numbers with threes, 271 of them:

3 13 23 30 31 32 33 34 35 36 37 38 39 43 53 63 73 83 93 103 113 123 130 131 132 133 134 135 136 137 138 139 143 153 163 173 183 193 203 213 223 230 231 232 233 234 235 236 237 238 239 243 253 263 273 283 293 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 403 413 423 430 431 432 433 434 435 436 437 438 439 443 453 463 473 483 493 503 513 523 530 531 532 533 534 535 536 537 538 539 543 553 563 573 583 593 603 613 623 630 631 632 633 634 635 636 637 638 639 643 653 663 673 683 693 703 713 723 730 731 732 733 734 735 736 737 738 739 743 753 763 773 783 793 803 813 823 830 831 832 833 834 835 836 837 838 839 843 853 863 873 883 893 903 913 923 930 931 932 933 934 935 936 937 938 939 943 953 963 973 983 993

[Chuck]

For thousand digit numbers it's about 99.9999999999999999999999999999999999999999999825212874827734839034%

[Bicho the Inhaler]

cha, I agree, and that's essentially what it says in the solution to puzzle #6. My point in the first reply was that if you choose to look at something other than the size of the number, you might get a totally different limit.

[celticsrule]

for puzzle #10 (The Poker Game):

I don't know if this answer was established already, but there's nothing posted there so here it goes...

If I pick all four kings and a ten (any suit) I will win. Assuming he picks four aces and a ten (as the problem states), I can trade in three of the kings for the 9, J and Q of the suit of the 10 I have giving me a straight flush to the K. The highest hand my opponent can now get is the four aces he already has because all four kings have already been used so a straight flush is out of the question. Thus, the four aces and a ten does not work.

[This message has been edited by celticsrule (edited 08-15-2002 11:20 AM).]

[Chuck]

He could choose four aces and the jack that's the same suit as your ten.

[celticsrule]

you're right, I misread and assumed the other player was always going to choose the four aces and ten. Now, we can simply change it to four tens and a jack to win no matter what. It's impossible for the opposing player to break up all potential straight flushes for me and the highest straight flush he can create is a 9-high (in which case I will be able to create an ace high)

[Nauplius]

celticsrule are you choosing that as your starting hand?

If so, I choose three aces, the queen and 9 of the same suit as your ten and jack. This prevents you from getting a straight flush at all, and I can still draw a low straight flush A2345, or four aces.

[JToomey]

Re: Mr. Plow:

I think it's reasonable, given no contrary evidence, for the solver to assume in this puzzle that the snow falls at a constant rate, as stated in the solution. However, it should be clearer in the proposition that the plow's speed is inversely proportional to the snow depth.

I'd bet someone else's hat that plows' speeds are not related to snow depth linearly -- after all, the plow isn't just pushing area, it's pushing volume, and the volume increases with the square of the speed of snowfall. Rather than the v = c/rt expressed in the solution, I'd estimate something more like v = c/t(r^2).

If I were editing, I'd make this assumption of the rate of change of the plow's speed up front, perhaps by sticking something like "Given that a plow's speed is inversely proportional to the snow's depth..." in there.

Normally I'm not one to pick too many nits, but, hey, what the Hell -- you asked.

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The Big, Stupid Puzzle:
http://www.yark.org/puzzle

[mith]

Re: Bicho's posts.

I understand your point, but I'd like to see what you think about the following sequence:

1, 3, 2, 5, 7, 4, 9, 11, 6, ...

Would you argue that there are not actually 50% even numbers?

[Bicho the Inhaler]

mith, yes, I would argue that there aren't actually 50% even numbers. I think the question of "what percentage of elements of set A are in subset B" can only be answered meaningfully if B is finite or A\B is finite.

[mith]

*~shrug~* ok.

[Snatch]

Here is another twist on that. You start with 0-9 which is 1/10 then 10-99 which is 18/90 then 100-999 which is 252/900. In the end they each are scaled by a multiplier according to the probability that a random number would be that length. 1 digit = 1/infiniti 2 digits = 2/infiniti infiniti digits = infiniti/infiniti....Then you sum the scaled numbers to get the percent. Basically we assume a random number length then each digit of that number is random. I see 99.999999999999999999999999999999999999999% but not 100 . How to randomly generate a number between 0-9 simple. Between 0-infiniti I have no idea wont happen on a computer at least hehe.

Snatch

[Santa Claws]

Actually, if you consider the numbers from zero to infinity, you will find that 100% of you are faggots and should just stop.

[Satan Lives!]

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ANSWER TO THE POKER GAME
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Draw all four 10's, the fifth card doesn't matter. This prevents your opponent from drawing a royal flush, while still allowing you that possibility at the moment. Therefore, your opponent must prevent you from assembling a royal flush or he will lose. The only way he can do this is by drawing at least four cards that are higher than 10. Then you draw a straight flush with 10 high. Now your opponent can only draw the lowest straight flush, 5432A, (if he drew an A the first time), or otherwise no straight flush at all. Therefore you win.

[Santa Claws]

That doesn't work, you faggot!

[Chuck]

If you don't hold an ace you can't draw more than three cards.

[CrystyB]

Which begs for an appendum to that poker game... Could you please include some rules (maybe through a link, maybe not) of what i am allowed to change, how many, which cards are used, etc?

Also Bicho, don't you think morrisonlucas' "Rearranging the terms like that (puzzle 6) is a well-known trick, but it is not allowed, for exactly the reasons you point out. - There are rules to infinite series. Generally you evaluate them in order." could work? I mean forget about the arbitrarily 9*10^n blocks, just do the limit. Wouldn't it be 1? ((i think it would, but i keep forgetting to actually compute it ))

[This message has been edited by CrystyB (edited 08-23-2002 10:41 AM).]

[Chuck]

The Poker Puzzle

In the How Many Threes puzzle,choosing to put the integers in numerical order just to prove your point seems no less arbitrary than anyone else use any other order.

[Beartalon]

How Many Threes?

I think the puzzle is simply an interesting paradox. What's the possibility that a randomly chosen 4 digit number is not the number 3333? We know that there is only one 3333 in the enumeratable range 0000 - 9999, so the probability is 0.0001 (0.01%) that it is 3333, and 99.99% that it isn't.

However, by the theory used above, the possibility that 3 is not the first number is 0.9.
That 3 is not the second = 0.9, that 3 is not the third = 0.9, and that 3 is not the last = 0.9. All numbers are independently chosen, so the possibility that the number is not 3333 is 0.9*0.9*0.9*0.9 = 65.61%? That can't be right. What am I missing?

[Chuck]

That's the probability that no digit is a 3, not that none of them are.

[mith]

(That's the probability that no digit is 3, rather than the probability that not all the digits are 3. I think that's what Chuck's trying to say. )

[CrystyB]

Actually you can't define a discrete distribution, and the continuum one is of almost no use since both the sets are À₀. HEY! But... that's nice - it could be sustained that ANY infinite subset of N (the primes, the cubes, etc) should be awarded 100% of N.

Also, Chuck, thanks! *feels silly*
But i disagree on the ordering. That's not arbitrary, that's how they were created. The very definition of the N creates them starting at zero and going to its successor and so on.

[This message has been edited by CrystyB (edited 08-23-2002 02:56 PM).]

[morrisonlucas]

-How many threes-

Consider the following sets, each a subset of the positive integers represented in decimal:
{no threes} the set of all integers which contain no threes (e.g. 1245, 4, 287)
{one three} the set of all integers which contain exactly one three in position one (e.g. 3, 31, 3576108, 34
{some threes} the set of all integers which contain some threes (e.g. 3, 23, 33535)
{only ones} the set of all integers which contian only ones (e.g. 1, 11, 11111)

Defining the following symbols: '=', '<', '>' as refering the number of elements between two sets, we have:

{no threes} =¹ {one three} <² {some threes}
{some threes} <³ {only ones} <² {no threes}

Reasons:
1 - each element in one set can be mapped uniquely to an element in the second by adding/removing the leading three
2 - each element in the first set is an element in the second
3 - each element in the first set can be mapped to a uniqe element of the second, that is, the number containing as many ones as the first number represents. (i.e. 3->111, 13->1111111111111). The second set is larger since 11 is not mapped this way.

Therefore {some threes}>{no threes} and {some threes}<{no threes}.

-to CrystyB- the ordering is indeed not arbitrary. But it's irrelevant.

[CrystyB]

1. i can prove that indeed the ratio is 1. So, mith, now you can safely insert the "This can be considered over-simplistic by math wizards out there, but they can rest assure there is a thourough demonstration of this 100%. We're just not likely to include it... "
2. {no threes}<{one three} (you can have the single three on a position that isn't first)
3. Yes they are all of power À₀. So they are all 'equal' in power, and 'equal' to N. That's the beauty of infinity and infinite sets.
4. Ordering is NOT irrelevant. Read post #1.

[This message has been edited by CrystyB (edited 08-23-2002 03:12 PM).]