Random number

It seems one cannot pick a random element uniformly from any "infinitely large" set (as in a metric space), since with relation to a fixed point O, the probability of picking a second number closer to O than the first random number should be 0, despite symmetry.

Wraith King Ulti

REINCARNATION!!!

Why does playing whimsically make you suffer so much...

Exact equal rating...

FORMAT  WXF
BLACK   me
RESULT  0-1
DATE    2014-05-30 17:54:21
EVENT   KGP Game ; 10m+0s
START{
 1. C8.5 c2.5   2. H8+7 h2+3   3. R9.8 r1+1
 4. C2.4 r1.6   5. A4+5 p3+1   6. R8+4 p9+1
 7. H2+3 c8.7   8. P3+1 r9+1   9. H3+4 c7+3
10. H4+3 c5.7  11. R1.2 r6+2  12. H3-1 c+-1
13. P1+1 h8+9  14. R8.6 r6+1  15. P7+1 e7+5
16. P7+1 e5+3  17. R6+2 a4+5  18. H7+6 r6+1
19. C5+4 c-.5  20. C5-2 r6.9  21. H1+3 h9+7
22. R6.3 c7.5  23. C5+3 e+-5  24. R3.7 r+.4
25. R7+1 r4+1  26. R2+3 r9+8  27. R2.3 r4.1
28. C4-2 r9-4  29. C4+2 r9.4  30. E7+5 c5.2
31. C4+4 c2+5  32. E5-7 r4+3  33. C4.5 k5.4
34. C5.2 p1+1  35. C2-4 p1+1  36. C2.6 p1.2
37. R3+1 r1+2  38. R3.6 k4.5  39. A5+4 r4+1 }END

Do I wear specs?

Someone doesn't seem to know.

Chess960

People have complained about the advantage memorising variations in opening theory gives in chess, while adding little value to the gameplay, such moves are not found by yourself, but already worked out by others. This, naturally leads to chess variations where opening theory is not fixed, scoring another defeat for rote learning.

The rules are more or less the same as chess -- pieces move the same way (castling is a bit different), and the starting position is different: pieces are randomly arranged along the home rank for white subject to 2 rules, and the position is mirrored. The two rules are:

1) Bishops are of opposite colours.

2) The king is in between both rooks (to allow for castling both sides)

Castling brings the kings and rooks to the normal final positions (as in standard International Chess) without regard to where they started from. This can result in long-range king flights (0-0-0-0-0-0!, except its still 0-0 for "king-side" castling and 0-0-0 for "queen-side" castling), and is subjected to the normal castling rules, ie.

1) King must not have moved.

2) Rook must not have moved, even if it had not been a rook then.

3) King cannot castle into, out of, or through check.

And thus I bring an interesting game which I will eventually fill the analysis for.

[Event "Let's Play!"]

[Site "Chess.com"]

[Date "2014.01.21"]

[White "Cookiejar"]

[Black "meeeep"]

[Result "0-1"]

[WhiteElo "1400"]

[BlackElo "1465"]

[TimeControl "1 in 3 days"]

[Termination "meeeep won by resignation"]

[Variant "Chess 960"]

[SetUp "1"]

[FEN "nrqkbbnr/pppppppp/8/8/8/8/PPPPPPPP/NRQKBBNR w KQkq - 0 1"]

1.d4 Nb6 2.e4 d5 3.e5 f6 4.Nf3 Bh5 5.Qe3 Qd7 6.h3 Nc4 7.Bxc4 dxc4 8.O-O-O O-O-O 9.g4 Bf7 10.Bc3 Qc6

 11.Rhe1 Bd5 12.Nd2 e6 13.b3 b5 14.f4 b4 15.bxc4 bxc3 16.Qxc3 Kd7 17.Qa5 Bg2 18.Ndb3 Ke8 19.exf6 Nxf6 20.d5 Bxd5

 21.cxd5 Nxd5 22.Nd4 Qc3 23.Qa4+ Kf7 24.Nab3 Nb6 25.Qa5 Bb4 26.Qa6 Rxd4 0-1

All images shall be resized, cropped screenshots of the chess.com board.

Ok let's do this move by move.

0.

This isn't actually very abnormal. The rooks are slightly displaced, the king went over to visit the queen, and accidentally pushed her one square away. The bishop hurriedly circumvented the regal duo. We shall call the side closer to the h-file the king side, and the other side the queen side. What happens here is that the queenside is already under the influence of the bishop pair once the pawns are out of the way. King might be safer on the kingside.

But at least there's no Ba1/Bh1, where b3/g3 opens up the diagonal for the bishop to stare straight through the kingside instantly.

There is Na1 though, which could prove to be a problem to activate.

1.d4 Nb6
d4 opens diagonals for bishop and queen, fights for the center. Nb6 solves the problem of the knight on the corner, develops it and exerts a little control over the center. Nothing much here.

2. e4 d5
And he has gained Space.

Boggle

Lately, I have found myself deeply attracted (addicted?) to Wordtwist, which is basically Boggle with a weirder scoring system. Let's explore methods of scoring well in Wordtwist on a good board (one with many words available).

Firstly, an example board:

b x o p
y e t h
o n a s
k e s w

Let's look at the board and talk about it in general.

Sure, the game tells us that there are many words to be found (a pre-game accessible table tells us, with a bit of computation, that there are 316 words waiting to be found), but where do we start looking?

The 'x' near the top left corner is not badly placed. It contains obvious words (which are rewarded with bonus score thanks to the presence of the "weird" letter 'x') such as exo and oxy. In fact, there are 6 3-letter words containing an 'x' in this board: exo, oxy, hox, pox, yex and tex. It pays to know your 3-letter words (I didn't; oh well).

There are 6 other words containing 'x': poxy, next, nexts, oxen, onyx and ethoxy, which is admittedly hard to find.

The other "weird" letter of this board is 'k'. 'k' contributes to surprisingly many short words to those unfamiliar with Boggle-like games: kas, ska, ken, kent, kon, and whatever not. In fact, for this board, the 'k' words are: kon, kons, ken, kens, kea, keas, oke, okes, nek, hank, sank, keno, kent, kents, tank, wank, yoke, yokes, kente, sanko, shank, snoek (and edible marine fish, what the...), snoke, stank, swank, thank and bethank. Of these, kon, kons, ken, kens, kea, keas, oke, okes and nek are 3-letter words and s-plurals. If one recognises 'nk' as an ending stem, one could in theory locate the words hank, sank, tank, wank, shank, stank, swank and thank with some degree of consistency (a glance would give hank, sank, tank and wank surely, and the rest are not that hard to locate if you're looking for words ending with 'nk').

Next, we might take a look at 3-letter words (with some repetition):

yon yok yex yet yen yea wat was wan wae top toe tho tex ten tea taw tas tan tae sny sha sen sea saw sat san sae pox pot poh pht pho oye oxy opt ony ons one oke ohs oes nye nth noy net nek neb naw nat nas nah nae kon ken kea hox hot hop hoe haw hat has han hae exo eth eta ess eon ens ene eat eas ean bye bey bet ben ate ass ash any ant ane ahs

We theorize that most 3-letter words are of the form . Words that fit this bill are:

yon yok yex yet yen wat was wan top tex ten taw tas tan sen saw sat san pox pot poh noy net nek neb naw nat nas nah kon ken hox hot hop haw hat has han bey bet ben

This kind of breaks my hypothesis to bits, accounting for only half of the words. Yet these 3-letter words are the easiest to find. Relooking at the board:

b x o p
y e t h
o n a s
k e s w

One might have a deeply organised attack on these words which go:

 xop xot xoh pot poh tox top toh hot hop hox bex bet ben bey xeb xet xey xen yet yeb yex yen tex ten tey teb neb ney nex net yon yok noy nok kon koy nat tan tah tas taw hat han has haw nat nah nas naw sas sat sah san saw was wan wat wah

This covers all the 41 words in the reduced list above, and took me slightly above 1 minute to type out in full. Knowing the words well could likely reduce this to about 40 seconds, minus away some consistency due to flawed recognition of words. Also, doing it with a more stable presence of mind will probably help in locating the plurals, such as hats, nats, tans, hans, haws and wans more easily.

The remaining 3-letter words:

yea wae toe tho tea tae sny sha sea sae pht pho oye oxy opt ony ons one oke ohs oes nye nth nae kea hoe hae exo eth eta ess eon ens ene eat eas ean bye ate ass ash any ant ane ahs

6 are plurals of 2-letter words:

ons ohs oes ens eas ahs

The rest are weird; go figure:

yea wae toe tho tea tae sny sha sea sae pht pho oye oxy opt ony one oke nye nth nae kea hoe hae exo eth eta ess eon ene eat ean bye ate ass ash any ant ane

The remaining words according to classification where I can find them:

Seeming plurals/third person present tense of 3-letter words:

yens yeas wats wans waes tens teas taws tans taes seas saws sans pots opts ones nets nats hots haws hats haes eths etas eons enes eats eans bets bens ants anes

Others:

yont yean yeah ybet wate wast wash wase wany want wane toph toey toea thaw than thae tene tass tash tane swat swan stop stey sten staw snye sneb snaw shot shop shoe shaw shat shan sent sene sena seat sean sate sash sant sane pote poet phot phat onto noes ness neat hote hate hast hant haet haen eyne etna esne eath east ease bets beth beta bent bene beat bean atop ants ante aeon

5-letter words (words with apparent 4-letter stems are bolded):

yenta yeast yeans yeahs waste wants waney wanes tophs toeas thaws thans thane tense tenes tease tawse tasse swats swath swash swans stens steno stean staws stane snaws snath snash shote shaws shans sents sente senas seats sants santo sanes potae poets phots phase netop neats neath hoten hawse haste hants hanse haets etnas enate eaten beths betas bents bento benes beats beath beast beany beans beano aeons

6-letter words:

yentas thanes swaths steans steane stanes sneath snaths snaste senate potass potash potaes hasten ethane ensate beaths assent

7-letter words:

washpot waeness teashop steanes sneaths shantey hotness hastens ethanes beneath

Notably, waeness and hotness both have the -ness stem.

Random notes that didn't fit in anywhere else:

b x o p
y e t h
o n a s
k e s w

With the letters n, e (x2), a, t, s (x2), we can form (not exhaustively):

eta, etas, tea, teas, tae, eat, eats, ate, east, sate, sae, eas, tas, sat, ass, tass, neat, neats, tane, stane, stanes, stean, steans, steane, steanes, net, nets, ten, tens, sten, stens, sea, seas, sen, ens, nas, san, sans, sane, ane, ean, eans, tan, tans, nat, nats, sene, ene, tene, tenes, taes, sent, ant, ants, sant, sants,

all of which are rather common words with my 5 favourite letters, 's', 't', 'a', 'n' and 'e'. In fact, a square containing 'teas' has 22 words, which are:

tea, teas, tae, taes, eat, eats, eta, etas, ate, ates, sat, tas, set, tes, est, sea, sae, eas, seat, seta, east and sate.

For 'nats': nat, ant, tan, ans, nas, san, tas, sat, tans, nats, ants, sant
For 'nets': ten, net, sen, ens, set, tes, est, nets, tens, sten, nest, sent
For 'sane': san, nas, ans, sea, eas, sae, ens, sen, ane, ean, nae, sean, sane, anes, eans, sena
For 'tane': tan, nat, ant, tea, tae, eat, eta, ate, ten, net, ane, ean, nae, neat, tane, etna, ante

The Two-envelope Problem

  Well, this is one of the recurring problems we've probably all seen, time and time again, with various "obvious!"-style solutions as well as a nice wall of text explaining away the paradoxes by claiming that it was all a case of confusing phrasing -- with an inspiredly sneaky re-expression of the problem, the paradox dissolves! However, there is more or less no true escape from the true Mathematics of this problem. Wikipedia has a nice and technical exposition on the various "solutions" and solutions of the paradox, and equally sneaky reformulations of the problem to expose the holes in the "solutions". You should read it if you actually want a formal explanation, but I believe mine follows the Wikipedia article somewhat faithfully, up to at least somewhere in the middle.

  The first thing to state is the question in a text-wall format, so that we can demolish it later for clearer insight. Well, here I go: Two envelopes are on a table. One contains twice as much money as the other, and there is no way to tell the contents of the envelope until you open it. You choose one, taking extreme caution not to open it (and cleverly avoid revealing the contents of the other as well, you sneaky person). You think to yourself: If I open this envelope and find X dollars, then the other envelope could contain X/2 dollars or 2X dollars. Those two outcomes have equal chance, since I pick either the larger or the smaller (contentwise) envelope randomly. My expectation of money in the other envelope is hence 5X/4 dollars. This does not make sense, since the expectation of money in my envelope is now also 5Y/4 dollars, where Y is the amount contained in the other envelope. We get X=25/16*X. That's depressing, since it solves to 0. :( In my depression, I claim a paradox!

  Well well, not the most accurate formulation of the problem, but at least somewhat intelligible. Now I shall point out an interesting feature of this problem which is usually ignored in discussion for being somewhat impossible to prove -- Consider, in a train of thought very far removed from the original paragraph, the following paradox: "X is 1. X is 2. Therefore 1=2. Contradiction?!". This is clearly a matter for the logicians out there to name. I'd personally file it under the classification of "Lies" and trash it. But, the thing is... our problem seems self-consistent... up till the paradox. So now let's rephrase it without the envelopes:

  I pick a number (call it X), and map IDs #1 and #2 randomly to the two positive numbers X and 2X. Now, you pick either #1 or #2 and we define the number it maps to be Y. Now the other number could either be Y/2 or 2Y with equal probability, and the expectation is 5Y/4. The problem statement is clearly sound, since these are more or less mathematical formulations, and if a contradiction exists, the subject would fall apart and all I learnt in school would be totally useless!!!!! We therefore call it a confusing argument and try to explain it away.

  Clearer (albeit somewhat spoilt by the ending). So now for "solution" 1. We notice that Y can either be X or 2X. As such, Y is inconsistently defined. We therefore redefine it in terms of X and 2X and it all works out to an expectation of 3X/2. The world is saved!

  Clearly this violates the principle of the matter. It's like saying: "Why don't I get dough when I mix flour with kerosene?" and answering it with "Everything is well if you just buy the dough directly from the supermarket, you twit. After all, they've derived the dough from flour as well.". This resolves the paradox by more or less ignoring it, so let's just make the paradox more annoying and harder to sidestep:

  Both the number mapped from #1 and #2 have equal expectations by symmetry, and the smaller of which is further defined to be a random positive real number. You pick #1. It maps to X. Now, the expectation of the value of #2 is (X/2 + 2X)/2=5X/4, since the probabilities of the pairs [X/2,X] and [X,2X] are equally probable. The problem is: the expectation of the other number is always higher than your no matter what it is, so on average, knowing a number causes the other number's expectation to be 5X/4, which is 5/4 times its own expectation. Based on that, before you even know what #1 maps to (ok call it f(1), and similarly define f(2) -- this is getting tedious), you know that f(2) is on average 5/4*f(1). OMG contradiction!

  The resolution for this is a one-liner (Note: see Zagier's one sentence proof for an idea of how a mathematical line might look): X*5/4=X, where X is the cardinality of positive real numbers (commonly known as infinity, though not the only infinity). Well the thing is that since positive reals stretch indefinitely, the chance of choosing a finite number is infinitesimal. And for infinities, the statement f(1)=5/4*f(1) can make perfect sense.

  One last sneaky trick to throw in (and a solid text wall to help obfuscate matters just that little bit more):

  "We shall now be so nice, going out of the way to help you define just a specific case of the problem, and we'll even babysit you by stating precisely how we chose the numbers. More formally, we shall enlighten you as to the probability distribution of the amounts in the envelopes. We have an old friend here, called Mr. Fair Coin, who is a truly erratic fellow at the best of times and completely random at the worst of times. Now, we all know that Charles Dickens once famously said "It was the best of times, it was the worst of times", but you see, truly erratic and completely random are consistent with each other, so you're not getting a paradox for free. We first choose the smaller number by setting it as 1, and then tossing Mr. Fair Coin until he lands on his head (that clumsy fellow!). For each toss where he does not land on his head, we double the value of our numbers (the charitability in me!). Oh and I unconsciously neglected to mention, since it was so banally trivial, that since Mr Fair Coin is so fair, he does not have a preference to landing on heads, nor does he have a preference to landing on tails, and is most assuredly not inclined to land on sides, but lands on one of the three with equal probability. Then we double our number to generate the second number."

--Awkward seconds pass--

  "That's quite close to a random number generator, isn't it?"

  Disregarding the disconnect in common sense, he makes a fair point. Let's list the argument in point form:

P1.1: If the number you choose turns out to be 1, the other is 2, which is truthfully larger than 1.

P1.2: If the number you choose turns out not to be 1, but instead is X, the expectation is 3/5*X/2 + 2/5*2X = 11X/10. The other number is expected to be larger anyway!.

P1.conclusion: CE[f(2)]>f(1) for all possible values of f(1), where CE denotes conditional expectation.

P2: The chance of f(1) being infinite is limit (2/3)^n as n --> infinity = 0. Therefore f(1) is finite, and CE[f(2)] is finite as well. Therefore the previous argument regarding infinities does not hold.

  Final resolution (for this post): (highlight for spoilers below)

  The number of real numbers between 0 and 1 is a humongously huge infinity, infinitely (lol) greater than the number of positive integers. If we were to pick a random number from 0 to 1, therefore, the probability that it is any particular number is in fact 0. But this does not mean that it is impossible, i.e. P(A)=0 does not mean A is impossible, since the number randomly generated from the first try between 0 and 1 will have occurred, despite having 0 probability of occurring.