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Louis F Stumpers
Number of games in database: 47
Years covered: 1932 to 1969
Overall record: +13 -27 =7 (35.1%)*
   * Overall winning percentage = (wins+draws/2) / total games
      Based on games in the database; may be incomplete.

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Most played openings
D94 Grunfeld (3 games)
E60 King's Indian Defense (2 games)
B59 Sicilian, Boleslavsky Variation, 7.Nb3 (2 games)

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LOUIS F STUMPERS
(born Aug-30-1911, died Sep-27-2003, 92 years old) Netherlands

[what is this?]

 page 1 of 2; games 1-25 of 47  PGN Download
Game  ResultMoves Year Event/LocaleOpening
1. L Stumpers vs J Lehr  1-019 1932 EindhovenD18 Queen's Gambit Declined Slav, Dutch
2. E Sapira vs L Stumpers 0-125 1938 NBSB - FlandersD94 Grunfeld
3. L Stumpers vs E Spanjaard  1-055 1938 Dutch Ch prelimE02 Catalan, Open, 5.Qa4
4. H van Steenis vs L Stumpers  1-025 1939 NED-ch11B02 Alekhine's Defense
5. J van den Bosch vs L Stumpers  ½-½58 1939 NED-ch11A48 King's Indian
6. L Stumpers vs S Landau  0-141 1939 NED-ch11D33 Queen's Gambit Declined, Tarrasch
7. A J van den Hoek vs L Stumpers  1-027 1941 BondswedstrijdenB10 Caro-Kann
8. T van Scheltinga vs L Stumpers 1-035 1942 NED-ch12D94 Grunfeld
9. L Stumpers vs J H Marwitz  1-040 1946 NED-ch prelim ID31 Queen's Gambit Declined
10. L Stumpers vs Euwe 0-130 1946 NED-ch prelim IE60 King's Indian Defense
11. W Wolthuis vs L Stumpers  ½-½52 1946 NED-ch prelim IC58 Two Knights
12. C B van den Berg vs L Stumpers  1-058 1946 NED-ch prelim ID19 Queen's Gambit Declined Slav, Dutch
13. L Stumpers vs H van Steenis 0-124 1946 NED-ch prelim ID28 Queen's Gambit Accepted, Classical
14. G Fontein vs L Stumpers  ½-½26 1946 NED-ch prelim ID94 Grunfeld
15. L Stumpers vs Cortlever  ½-½50 1946 NED-ch prelim IE60 King's Indian Defense
16. V Soultanbeieff vs L Stumpers  ½-½46 1947 Int BD96 Grunfeld, Russian Variation
17. Tartakower vs L Stumpers 1-024 1947 Int BD74 Neo-Grunfeld, 6.cd Nxd5, 7.O-O
18. L Stumpers vs H van Steenis  0-133 1947 Int BD23 Queen's Gambit Accepted
19. L Stumpers vs Grob 1-060 1947 Int BA55 Old Indian, Main line
20. L Stumpers vs F Henneberke 1-043 1948 NED-ch14C92 Ruy Lopez, Closed
21. J T Barendregt vs L Stumpers  0-126 1948 NED-ch14C86 Ruy Lopez, Worrall Attack
22. L Stumpers vs A Vinken  0-133 1948 NED-ch14E21 Nimzo-Indian, Three Knights
23. L Stumpers vs C Vlagsma  0-145 1948 NED-ch14C65 Ruy Lopez, Berlin Defense
24. L Stumpers vs T van Scheltinga  1-047 1948 NED-ch14C97 Ruy Lopez, Closed, Chigorin
25. L Stumpers vs H Kramer  0-140 1948 NED-ch14B92 Sicilian, Najdorf, Opocensky Variation
 page 1 of 2; games 1-25 of 47  PGN Download
  REFINE SEARCH:   White wins (1-0) | Black wins (0-1) | Draws (1/2-1/2) | Stumpers wins | Stumpers loses  

Kibitzer's Corner
< Earlier Kibitzing  · PAGE 251 OF 251 ·  Later Kibitzing>
Jan-05-15  schweigzwang: <Best Before July/2015.

How does the soup know to go bad by July?>

Another question: Best for *whom*?

Jan-05-15
Premium Chessgames Member
  al wazir: <Sneaky>: The trouble with my version of the card analogy is that instead of a 3:1 bias in favor of turning up a K or Q of the same color as the near observer's A, the distant observer finds a 3:1 bias in favor of the *other* color. But I think I've found a way to repair it. Better get another cup of coffee; this is really, really complicated.

I'll describe the preparation process in six steps, only one of which is new.

1. Use the same hybrid cards and same envelope that I described previously. Arrange all cards face up aligned north and south. Stack the aces so that the upper one shows the spade half pointing north and the heart half pointing south and the lower shows the heart half pointing north and the spade half pointing south. Do the same with the pair of kings and the pair of queens. You now have three stacks, each containing two identical but oppositely aligned cards of the same rank. Turn them face down.

2. Without changing the orientation of the cards, randomize the order of the three stacks so that it's impossible to tell which stack contains the cards of any rank.

3. Then spin them through some random number of 180-degree rotations, so that it's impossible to tell which way (north or south) each half of any card is oriented. The stacks should be rotated *together*, so that the alignments of the aces, kings, and queens continue to be correlated.

4. Now reverse the orientation of ONE of the three stacks, that is, give it an extra 180-degree rotation. (At this point you don't know whether it's the A, K, or Q stack.) This is the novelty.

5. Keeping the cards face down, place the upper card from each of the stacks in the corresponding compartment of one envelope, north end first.

6. Then shuffle the three remaining cards (without changing their orientation) so that the three are in some random order and place them in the three compartments of the other envelope, again north end first. Close the flaps of both envelopes. Transport one envelope to, say, Mars. The other stays where it was.

Now you turn the stay-at-home envelope over (so that the cards inside are facing up) and open *one* section of the flap, revealing half of *one* card and hence its identity. Let us say it is the ace, and the half you can see is the black (spade) half. You now know that when your Martian counterpart looks at the compartment holding the ace in his envelope, he will see the red (heart) half of the card.

Suppose now that he pulls out a K or Q. First suppose it is the K. There are three equally likely possibilities.

(1) The stack that got the extra 180-degree rotation was the Q stack. In that case the A stack and the K stack are still correlated, so he will see the same color as if he had drawn an A, i.e., a red K.

(2) The stack that got the extra 180-degree rotation was the K stack. In that case the A and the K are *anticorrelated*, so he will see a *black* K.

(3) The stack that got the extra 180-degree rotation was the A stack. In that case the A and the K are again anticorrelated, so he will again see a *black* K.

Thus the odds are 2:1 that he will see a black K, and by exactly the same reasoning, if he draws a Q there is a 2:1 chance that it will show the same color as the first observer's A.

So now we have a bias in the right direction. I will leave it as an exercise for the interested reader to find a modification of this scheme (using an idea from the previous, unsuccessful one) that delivers the desired 3:1 bias in favor of the same color as the first observer's A.

Jan-06-15
Premium Chessgames Member
  Sneaky: <al wazir> I am very interested but again I plea lack of coffee. Let me say that I am skeptical that it can be done at all, so I am going to read your effort with a critical eye; no offense.

In the meantime WannaBe's poser seems more achievable.

<Yes, for the sake of argument, let's say there are ∞ of names on the ballot.> OK, I get the problem them. Crooked voters wanting to push through nominees. I guess they get paid under the table for each one who gets through.

I'll get back to both of you.

Jan-06-15
Premium Chessgames Member
  al wazir: <Sneaky>: Don't waste your time. I think the subject of quantum cards has been worked to death.

But I'd like to know what I'm missing in the raft problem. I have a suspicion that I'll call foul when you tell me. It must be a trick.

Jan-08-15
Premium Chessgames Member
  Sneaky: The raft problem in a nutshell (as my grandfather taught it to me) —

Which travels faster downstream: a heavily laden raft or a lightly laden raft?

The counterintuitive answer is that the heavily laden raft travels faster, because there is more of it underwater for the stream to push against.

Jan-08-15
Premium Chessgames Member
  Sneaky: <WannaBe> I'm starting to think that your answer of 13 1/3 is right (in other words, 13, since there will be one player without a third the required votes).

The logic is: we have j judges with 10 votes each, and each player needs 3j/4 votes to be inducted, then we know the judges have only 10j votes among them so 10j / (3j/4) = 13 1/3.

Let's test the theory with a simple case: 4 judges, 10 votes each, and 13 players we'll call ABCDEFGHIJKLM. I can get all 13 players in by voting like this:

judge 1: ABCEFGIJKM
judge 2: ABDEFHIJLM
judge 3: ACDEGHIKLM
judge 4: BCDFGHJKLX

(I threw that "X" at the very end to represent the one player who got a third the required votes.)

I'm not sure if that's all there is to say about this. If there isn't, it wasn't particularly interesting, but there's always pleasure in the journey.

Jan-08-15
Premium Chessgames Member
  WannaBe: <Sneaky> Thanks you for your time spent and interest!

The reason for that (not-so-stumper) is because of Baseball HoF voting, and some voters complaint that too few get in (75% of vote).

Then you get someone like Randy Johnson that got over 90% (Plenty of votes to spare) & someone like Mike Piazza who missed out by 24 votes.

(There were over 500 voters this year for MLB HoF)

Jan-08-15
Premium Chessgames Member
  al wazir: <Sneaky: The counterintuitive answer is that the heavily laden raft travels faster, because there is more of it underwater for the stream to push against.> That makes no sense to me. Imagine that you transform to a rest frame attached to the moving water. (Your statement of the problem specified that the current is steady.) What forces act in that frame on an object floating in the water? Only gravity and buoyancy, which are balanced. There might be an additional force due to dynamic air pressure, but that could be in any direction. The wind could act either to retard the object with the greater freeboard or to assist it. That's why I ruled out wind forces.
Jan-08-15
Premium Chessgames Member
  WannaBe: Really don't know which forum is the best, so I pick here. =))

(One version of) Poker is nearly solved:

http://news.yahoo.com/self-taught-c...

Jan-08-15
Premium Chessgames Member
  al wazir: <Sneaky: I can't say I've ever actually done the experiment—it would be a good one for Mythbusters to settle.> I suspect the experiment has already been done: http://en.wikipedia.org/wiki/Poohst...
Jan-10-15
Premium Chessgames Member
  Sneaky: <That makes no sense to me.> Like I said, I am taking this just on faith, and I realize it sounds like "urban folklore science." So to some degree I share your skepticism. However it makes some sense to me. The idea is that it's like a sailboat, where the portion *under* the water is like the sail. So the one with the bigger sail moves faster.
Jan-10-15
Premium Chessgames Member
  Sneaky: Consider an extreme case. Take two air-balloons and slip a heavy rock in one of them (smooth, so not to puncture) and then blow them both up.

Putting them in the water, the one with the rock is 80% under the surface. The other one is as you'd imagine, 99% above the surface.

Here, clearly because of wind factors alone, the submersed ballon would win the race. The air resistance met by the balloon that's 99% of the the water would be enough to make it lose.

But I didn't think wind resistance had anything to do with it. The real question is the gedanken experiment: what would happen if we could do the same thing in a vacuum? Would they tie the race, or would one balloon have some advantage? I still feel like the one mostly underwater would get the biggest thrust and get to the finish line first but I confess I'm just going by gut feelings.

Jan-11-15
Premium Chessgames Member
  al wazir: <Sneaky: The idea is that it's like a sailboat, where the portion *under* the water is like the sail. So the one with the bigger sail moves faster.>

An ordinary sailboat has to overcome the resistance (friction) of the water. Other things being equal, the deeper the keel, the greater this resistance will be. The analog of that resistance for your rafts is the action of the *air*. It will exert a greater force on whichever raft has the higher profile above the surface of the water. In the terms of your analogy, the rafts are "sailing" through the air medium, propelled by the water "winds" acting on their underwater "sails." I have assumed that the rafts have no air friction to overcome (that is, I have assumed that we can ignore the force exerted on the rafts by the ambient air). That's not a very realistic assumption. But my justification for assuming it was that the wind could blow either upstream or downstream, and so could favor either one of the rafts.

If you haven't yet done so, check out the link I posted.

Jan-11-15
Premium Chessgames Member
  dalbertz: <al wazir: <Sneaky: The idea is that it's like a sailboat, where the portion *under* the water is like the sail. So the one with the bigger sail moves faster.>>

It sounds to me like a confusion about speed and acceleration. I think your intuition that both rafts would move at the same speed in the frame of reference defined by the water's current is correct, <once the two rafts reach the equilibrium speed of the current>. However, when they are first put in the water, it is the force of the current on the raft that causes it to accelerate and therefore the one presenting more surface area to the current would experience the greater force and so it would accelerate faster.

Jan-11-15
Premium Chessgames Member
  al wazir: <dalbertz: However, when they are first put in the water, it is the force of the current on the raft that causes it to accelerate and therefore the one presenting more surface area to the current would experience the greater force and so it would accelerate faster.> Good point.
Jan-11-15
Premium Chessgames Member
  Sneaky: I did see that link, about the Winnie the Pooh contest of stick-racing. It's fascinating on many levels but didn't really answer this question.

(I suppose if the heavily laden raft canard is true the best way to win the contest would be to find some really heavy piece of mahogany that would barely float.)

<I have assumed that we can ignore the force exerted on the rafts by the ambient air. That's not a very realistic assumption. > Yes, I know that. I've been trying to ignore it as well, because the interesting question is the one which ignores air resistance. We can always stipulate that our river is on a planet that has no atmosphere—it just makes it hard to actually carry out the experiment.

<dalbertz> Good point indeed. It may be that the heavy raft gets off to a faster start, but eventually everything reaches the same speed (again, ignoring air.)

Jan-22-15
Premium Chessgames Member
  al wazir: What is the longest English word in which no letter appears more than once? Offhand (and without recourse to Google) the best I can do is nine letters: "frightens" and "traducing." But there must be some with at least eleven or twelve.
Jan-22-15  diceman: 15 letters.
Jan-22-15  disasterion: "Facetiously" is eleven and has all the vowels in alphabetical order, for added smugness; but there must be something longer.
Jan-22-15  tbentley: The usual answer is 15, although there is an argument for 16 or 17.
Jan-22-15
Premium Chessgames Member
  al wazir: <disasterion: "Facetiously" is eleven and has all the vowels in alphabetical order>. Of course. I *knew* that. I had the feeling when I posted the question that I was doing it mistakenly (10 letters).
Jan-22-15
Premium Chessgames Member
  OhioChessFan: abstemiously is another with the vowels in order. For smugness value, I named facetiously and abstemiously off the top of my head when someone asked for words that fit that pattern. Never checked to see if there were more.
Jan-22-15
Premium Chessgames Member
  al wazir: <OhioChessFan>: Yes, but this time the question isn't about vowels.
Jan-23-15
Premium Chessgames Member
  al wazir: <OhioChessFan>: I spent a quarter of an hour with an unabridged dictionary and found "areligious," denoting someone without religion. Presumably the adverb "areligiously" is a real word too.

As a problem purist (and possibly for other reasons), you will of course object that the letter 'i' occurs twice. But at least they're consecutive, so the vowels appear in the correct order.

Jan-25-15
Premium Chessgames Member
  kellmano: Here's a nice Maths puzzle I saw recently: What is the sum of 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 .....

One of those annoying cases where I thought for a little bit, had an idea and checked the solution before spending enough time on it. I was wrong, but the solution in the version I saw was very elegant and I wish I'd thought harder. I showed it to someone who is very good at Maths and he had an unfortunately ugly solution, which I would consider a cook. He declared it a bad problem as his solution was inelegant, but I think it's just his solution that was bad (though correct).

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