chessgames.com
Members · Prefs · Laboratory · Collections · Openings · Endgames · Sacrifices · History · Search Kibitzing · Kibitzer's Café · Chessforums · Tournament Index · Players · Kibitzing

Louis Stumpers
L Stumpers 
 
Number of games in database: 56
Years covered: 1932 to 1969
Overall record: +13 -32 =11 (33.0%)*
   * Overall winning percentage = (wins+draws/2) / total games.

Repertoire Explorer
Most played openings
D94 Grunfeld (3 games)
D45 Queen's Gambit Declined Semi-Slav (2 games)
D31 Queen's Gambit Declined (2 games)
E60 King's Indian Defense (2 games)
B59 Sicilian, Boleslavsky Variation, 7.Nb3 (2 games)
C65 Ruy Lopez, Berlin Defense (2 games)


Search Sacrifice Explorer for Louis Stumpers
Search Google for Louis Stumpers


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

[what is this?]

Frans Louis Henri Marie Stumpers was born in Eindhoven, Netherlands, on 30 August 1911. (1) He was champion of the Eindhoven Chess Club in 1938, 1939, 1946, 1947, 1948, 1949, 1951, 1952, 1953, 1955, 1957, 1958, 1961 and 1963, (2) and champion of the North Brabant Chess Federation (Noord Brabantse Schaak Bond, NBSB) in 1934, 1935, 1936, 1937, 1938, 1939, 1940, 1941, 1942, 1943, 1944, 1946, 1948, 1949, 1950, 1951, 1952, 1953, 1954, 1955, 1959, 1961, 1962, 1963, 1964, 1965, 1966 and 1967. (3) He participated in five Dutch Chess Championships, with a 4th place in 1948, (4) and represented his country at the 1st European Team Championship in Vienna in 1957 (two games, vs Josef Platt and Max Dorn). (5) From 1945 and until about 1956, he was first Secretary and then Chairman of the NBSB. (3)

Stumpers was a physicist, and worked for the Philips company as an assistant from 1928. During 1934-1937, he studied at the University of Utrecht, where he took the master's degree. (6) In 1938 he was again employed at Philips, (6) and at a tournament in 1942, he supplied the hungry chess players with food from his employer. (3) After the war, he made a career in physics, with patents and awards on information ("radio") technology. He received degrees from several universities and colleges, including in Poland and Japan. (1, 3, 6) He retired from Philips in 1972, but continued teaching, (6) partly as professor at the University of Utrecht (1977-1981). (7) He was also Vice President (1975-1981) and Honorary President (1990-2003) of URSI, the International Union of Radio Science. (8)

Louis Stumpers married Mieke Driessen in 1954. They had five children, three girls and two boys. (6)

1) Online Familieberichten 1.0 (2016), http://www.online-familieberichten...., Digitaal Tijdschrift, 5 (255), http://www.geneaservice.nl/ar/2003/...
2) Eindhovense Schaakvereniging (2016), http://www.eindhovenseschaakverenig...
3) Noord Brabantse Schaak Bond (2016), http://www.nbsb.nl/pkalgemeen/pk-er... Their main page: http://www.nbsb.nl.
4) Schaaksite.nl (2016), http://www.schaaksite.nl/2016/01/01...
5) Olimpbase, http://www.olimpbase.org/1957eq/195...
6) K. Teer, Levensbericht F. L. H. M. Stumpers, in: Levensberichten en herdenkingen, 2004, Amsterdam, pp. 90-97, http://www.dwc.knaw.nl/DL/levensber... Also available at http://www.hagenbeuk.nl/wp-content/...
7) Catalogus Professorum Academię Rheno-Traiectinę, https://profs.library.uu.nl/index.p...
8) URSI websites (2016), http://www.ursi.org/en/ursi_structu... and http://www.ursi.org/en/ursi_structu...

Suggested reading: Eindhovense Schaakvereniging 100 jaar 1915-2015, by Jules Welling. Stumpers' doctoral thesis Eenige onderzoekingen over trillingen met frequentiemodulatie (Studies on Vibration with Frequency Modulation) is found at http://repository.tudelft.nl/island...

Last updated: 2018-01-16 05:40:42

 page 1 of 3; games 1-25 of 56  PGN Download
Game  ResultMoves YearEvent/LocaleOpening
1. L Stumpers vs J Lehr  1-0191932EindhovenD18 Queen's Gambit Declined Slav, Dutch
2. Prins vs L Stumpers  1-0391936NED-ch prelimB20 Sicilian
3. E Sapira vs L Stumpers 0-1251938NBSB-FlandersD94 Grunfeld
4. L Stumpers vs E Spanjaard  1-0551938NED-ch prelimE02 Catalan, Open, 5.Qa4
5. A J Wijnans vs L Stumpers  1-0361939NED-chB05 Alekhine's Defense, Modern
6. J van den Bosch vs L Stumpers  ½-½581939NED-ch11A48 King's Indian
7. L Stumpers vs S Landau 0-1411939NED-ch11D33 Queen's Gambit Declined, Tarrasch
8. H van Steenis vs L Stumpers  1-0251939NED-chB02 Alekhine's Defense
9. L Stumpers vs S Landau  ½-½341940HilversumD31 Queen's Gambit Declined
10. L Stumpers vs H Kramer  0-1361940HilversumE25 Nimzo-Indian, Samisch
11. A J van den Hoek vs L Stumpers  1-0271941BondswedstrijdenB10 Caro-Kann
12. T van Scheltinga vs L Stumpers 1-0351942NED-ch12D94 Grunfeld
13. W Wolthuis vs L Stumpers  ½-½521946NED-ch prelim IC58 Two Knights
14. L Stumpers vs J H Marwitz  1-0401946NED-ch prelim ID31 Queen's Gambit Declined
15. G Fontein vs L Stumpers  ½-½261946NED-ch prelim ID94 Grunfeld
16. L Stumpers vs H van Steenis 0-1241946NED-ch prelim ID28 Queen's Gambit Accepted, Classical
17. C B van den Berg vs L Stumpers  1-0581946NED-ch prelim ID19 Queen's Gambit Declined Slav, Dutch
18. L Stumpers vs Euwe 0-1301946NED-ch prelim IE60 King's Indian Defense
19. L Stumpers vs Cortlever  ½-½501946NED-ch prelim IE60 King's Indian Defense
20. L Stumpers vs Grob 1-0601947Int BA55 Old Indian, Main line
21. L Stumpers vs H van Steenis  0-1331947Int BD23 Queen's Gambit Accepted
22. Tartakower vs L Stumpers 1-0241947Int BD74 Neo-Grunfeld, 6.cd Nxd5, 7.O-O
23. V Soultanbeieff vs L Stumpers  ½-½461947Int BD96 Grunfeld, Russian Variation
24. L Stumpers vs C Vlagsma  0-1451948NED-ch14C65 Ruy Lopez, Berlin Defense
25. L Stumpers vs A Vinken  0-1331948NED-ch14E21 Nimzo-Indian, Three Knights
 page 1 of 3; games 1-25 of 56  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 42 OF 42 ·  Later Kibitzing>
May-18-18
Premium Chessgames Member
  Marmot PFL: I have heard of viruses being used to treat patients with bacterial infections that resist antibiotics, but it doesn't seem to be approved in the USA.
May-21-18
Premium Chessgames Member
  al wazir: <johnlspouge>: I can't help continuing to think about the expression you derived for the probability that the plane is still aloft after 2N - 1 time units, which is the same as the probability that the gambler in the coin-tossing game I described has not been wiped out: P( 2N - 1 ) = 2^(-2N + 1) * c( 2N - 1, N - 1 ). I will rewrite this as P(2N-1) = 2^(1-2N)(2N-1)!/(N!(N-1)!, using the notation I prefer. If you check this formula for the case of N = 2, you find that of the eight possible sequences of three heads and tails

HHH
HHT
HTH
HTT
THH
TTH
THT
TTT

three are consistent with staying alive. Three out of eight is 0.375, and this agrees with P(3) = 2^(-3) x 3!/1!2! = 0.375.

However, it cannot have escaped your notice that 2^(1-2N)(2N-1)!/(N!(N-1)! is identical with the probability of getting N heads and N-1 tails in 2N-1 tosses. For N = 2, three of the eight possible sequences of heads and tails do indeed have two heads and one tail (namely, HHT, HTH, and THH).

But there is something strange about this: they are *not* the same as the three sequences that correspond to survival.

Similarly, for N = 3, where the survival probability is P(5) = 10/32 = 0.3125, ten of the 32 possible sequences have three H's and two T's, but they are not the same as the ten survival sequences.

Very well. P(2N-1) is the probability of getting N occurrences and N-1 non-occurrences of *something*, both with probability 0.5, out of 2N - 1 tries. What is this "something"? If I could find the answer, that would allow me to derive the formula in one line.

I had another thought. There must be other ways to derive your formula, e.g., by induction.

P(1) = 0.5, which agrees trivially with the result of a single toss. For any N > 1,

P(2N+1) = 2^(-1-2N)(2N+1)!/(N!(N+1)!

= 2^(1-2N)(2N-1)!/(N!(N-1)! x 2N(2N+1)/4N(N+1)

= P(2N-1) x (2N+1)/4N(N+1) = P(2N-1) - (1/4) x [2/(N+1)] x P(2N-1).

This has a simple interpretation. The probability of staying in the game is the same after 2N+1 tosses as after 2N-1, *except* when the gambler holds precisely $2K after 2N-1 tosses. In that case there is a 1/4 chance of getting TT in the next two tosses and losing. The last term describes that possibility, where the probability of ending up with $2K after 2N-1 tosses is [2/(N+1)] x P(2N-1).

Now all I need is an independent way of showing that.

May-21-18
Premium Chessgames Member
  johnlspouge: @<al wazir>: You are restlessly inquisitive. I tremble at the thought of what you could have accomplished if your inquisitiveness had been focused ;>)

In general, the kind of simple interpretation you seek indicates a 1-1 mapping between configurations ("decompositions" in the jargon of combinatorics).

Let me use the equivalent (and possibly simpler) expression

P( 2N-1 ) = 2^(-2N) * (2N)! / N! / N!

Represent a climbing step by "+" and a drop by "-". (The following omits the apostrophes.) Given a configuration of + and - that does not cause the plane to crash in 2N - 1 (or equivalently 2N) time units, it is an exercise for the reader to show that the following is a 1-1 mapping. Look back in time from the earliest (leftmost) - in the configuration and move successively to each later -, finding for each - the last (compensating) +, and then padding the set of compensating +s with the rightmost +s remaining. Under this selection of +s, each configuration of + and - not causing the plane to crash corresponds to exactly one choice of N +s, and vice versa.

Here is the inverse. Let * being one of the N selected +s; 0, one of the other N time units. The inverse mapping starts at the left * and puts a corresponding - at the first 0, and iterates, moving to the right, until running out of positions to place a -. It then puts + everywhere remaining.

As a worked example, consider N = 2.

There are

6 = 4!/2!/2!

allowed configurations where every - is preceded by a +, namely

00**
++++

0*0*
++-+

0**0
+++-

*00*
+-++

*0*0
+-+-

**00
++--

Sometimes, the 1-1 mapping indicates interesting ways to generalize the problem by showing which generalizations are "insightful", i.e., which generalizations maintain some version of the mapping.

May-21-18
Premium Chessgames Member
  john barleycorn: let's take a simple view.

1) The plane is at 1000ft. The probability going up to 2000 ft is .5 and so is the probabilty for crashing.

2) being at 2000ft the probability of going up to 4000 ft in the next two minutes is .25, the probability of being back to 2000ft after 2 minutes is .5, and the probability for crashing is .25. Total probability to crash after 3 minutes: .5 + .5*.25 = .625

3) being at 4000 ft after 3 minutes means one crash happens in the next four minutes with a probability of 1/16. And being at 2000 ft after 3 minutes means probability of a crash within the next four minutes is .25 + .75*.25.

Which gives the probabilitity of a crash after 7 minutes of

.5 + .5*(.25 +.75*.25) + .25*.25*.5 = .75

I cannot get this number number out of the formulae given so far. Please advise.

May-21-18
Premium Chessgames Member
  al wazir: <john barleycorn>: Using the notation of my analogous gambling problem, there is one way of crashing after one minute:

T, probability = 0.5;

there is one new way of crashing after three minutes:

HTT, probability = 0.125;

there are two new ways of crashing after five minutes:

HHTTT, probability = 1/32 = 0.03125
HTHTT, probability = 1/32 = 0.03125;

there are five new ways of crashing after seven minutes:

HHHTTTT, probability = 1/128 = 0.0078125
HHTHTTT, probability = 1/128 = 0.0078125
HHTTHTT, probability = 1/128 = 0.0078125
HTHHTTT, probability = 1/128 = 0.0078125
HTHTHTT, probability = 1/128 = 0.0078125.

Total: 1/2 + 1/8 + 1/16 + 5/128 = 0.5 + 0.125 + 0.0625 + 0.0390625 = 0.7265625.

P(7) = (7!/4!3!)/2^7 = 35/128 = 0.2734375 = 1.0 - 0.7265625.

May-21-18
Premium Chessgames Member
  Marmot PFL: <P(7) = (7!/4!3!)/2^7 = 35/128 = 0.2734375 = 1.0 - 0.7265625.>

Going back to work I saved from 5/15 that's the number I also got.

May-21-18
Premium Chessgames Member
  johnlspouge: Turn the probability problem into a counting problem by pretending the plane can continue to fly underground, so all its up- or down-one paths are equally likely. Then count paths that occur in N time units, and paths corresponding to the subset where the plane does not crash. Division then gives the probability you seek.
May-21-18
Premium Chessgames Member
  johnlspouge: My answer is the same as <al>‘s, without his numbers.
May-21-18
Premium Chessgames Member
  johnlspouge: Essentially...
May-21-18
Premium Chessgames Member
  john barleycorn: < <al wazir: ...

P(1) = 0.5, which agrees trivially with the result of a single toss. For any N > 1,

P(2N+1) = 2^(-1-2N)(2N+1)!/(N!(N+1)!

= 2^(1-2N)(2N-1)!/(N!(N-1)! x 2N(2N+1)/4N(N+1)

= P(2N-1) x (2N+1)/4N(N+1) = P(2N-1) - (1/4) x [2/(N+1)] x P(2N-1). ...>

Well, these are a dangerous just by looking at them. some well-placed braces could structure them.

May-21-18
Premium Chessgames Member
  john barleycorn: < johnlspouge: Turn the probability problem into a counting problem by pretending the plane can continue to fly underground, so all its up- or down-one paths are equally likely....>

I looked at it that the plane cannot go underground. After a crash the plane can neither crash again nor go up again. (it looks so reasonable but unsymmetric :-))

that is why I broke up the case of having reached 2000ft after 3 minutes into the case of being crashing after 2 minutes (.25) and the probability of being still in the air after 2 minutes and then crahing in the remaining 2 minutes (.75*.25)

In fact, as some poster mentioned you cannot evade the Catalan numbers in this problem. but I do not see them in any solution presented here.

May-21-18
Premium Chessgames Member
  johnlspouge: Catalan numbers count binary trees of various kinds. It should not be hard to find them if you are looking ;>)
May-21-18
Premium Chessgames Member
  Marmot PFL: Turn the probability problem into a counting problem by pretending the plane can continue to fly underground, so all its up- or down-one paths are equally likely....>

Maybe Malaysia Airlines Flight 370 is still flying, somewhere beneath the Indian Ocean.

May-21-18
Premium Chessgames Member
  al wazir: <Marmot PFL: Malaysia Airlines Flight 370 is still flying, somewhere beneath the Indian Ocean> with probability 1.0.
May-21-18
Premium Chessgames Member
  john barleycorn: < johnlspouge: Catalan numbers count binary trees of various kinds. It should not be hard to find them if you are looking ;>)>

haha yes they are ubiquitous in branching processes. but you forgot them here :-)

May-21-18
Premium Chessgames Member
  john barleycorn: < al wazir: <john barleycorn>: Using the notation of my analogous gambling problem, there is one way of crashing after one minute: ...

P(7) = (7!/4!3!)/2^7 = 35/128 = 0.2734375 = 1.0 - 0.7265625.>

Well, not even in my dreams.

1. there are 128 outcomes possible.

2. to win the player has to start with a head which cuts it down to 64 events.

3. out of that 64 events the outcomes of having 6H or 6T, or 5H, 1T, as well as having 4H,2T or 3H, 3T are symmetrical.

4. all in all the Player loses in 1+6+15+10+64= 96 times.

5. and 96/128= .75

May-21-18  nok: Those still pondering can have a look at the wiki's Fig. 1 where the ground is drawn as a red diagonal with the "sky" below it. They count the paths that crash at 2n+1 without crashing earlier.

https://en.wikipedia.org/wiki/Catal...

May-21-18
Premium Chessgames Member
  john barleycorn: < nok: Those still pondering can have a look at the wiki's Fig. 1 where the ground is drawn as a red diagonal with the "sky" below it. They count the paths that crash at 2n+1 without crashing earlier.>

Thank you for the link. the problem is here you have to sum up the earlier crashes as well. the best way to do that is that after the plane has reached 2000ft height after 1 minute to go on with 2 minute time intervalls. If you take D for going down 1000ft and U for going up 1000ft then it comes to handling properly (D+U)^n with n even.

May-21-18
Premium Chessgames Member
  WannaBe: Y'all lost me at <
May-21-18
Premium Chessgames Member
  john barleycorn: <WannaBe: Y'all lost me at > that is why Stumpers page was invented :-)
May-21-18
Premium Chessgames Member
  WannaBe: No, I mean I am lost, I was looking for the Kenneth Rogoff page.

Knew I shoudda made a left turn at Albuquerque.

May-22-18
Premium Chessgames Member
  john barleycorn: Yeah, same feelings here. I have laid out two baits for <al wazir> but he is not responding with a post of at least 5 inches long. Where has the world come to when <al wazir> skips his routine?
May-22-18
Premium Chessgames Member
  al wazir: <john barleycorn: I have laid out two baits for <al wazir> but he is not responding>. You mean, you're waiting for me to tell you why you got 96 instead of 93?

Nah. I'll let you find your mistake yourself.

May-23-18
Premium Chessgames Member
  john barleycorn: <al wazir: ...

Nah. I'll let you find your mistake yourself.>

haha, no need for me to find them myself as I built them in. Most obvious, 6!/3!/3! = 20 not 10. and the claim about the symmetry is another story.

Anyway, the correct number of losses in a 7 game play of Head/Tails is 64+29 and so probability is 93/128.

May-23-18
Premium Chessgames Member
  john barleycorn: Read 6!/3!/3! as (6!/3!)/3!
Jump to page #    (enter # from 1 to 42)
search thread:   
< Earlier Kibitzing  · PAGE 42 OF 42 ·  Later Kibitzing>
NOTE: You need to pick a username and password to post a reply. Getting your account takes less than a minute, totally anonymous, and 100% free--plus, it entitles you to features otherwise unavailable. Pick your username now and join the chessgames community!
If you already have an account, you should login now.
Please observe our posting guidelines:
  1. No obscene, racist, sexist, or profane language.
  2. No spamming, advertising, or duplicating posts.
  3. No personal attacks against other members.
  4. Nothing in violation of United States law.
  5. No posting personal information of members.
Blow the Whistle See something that violates our rules? Blow the whistle and inform an administrator.


NOTE: Keep all discussion on the topic of this page. This forum is for this specific player and nothing else. If you want to discuss chess in general, or this site, you might try the Kibitzer's Café.
Messages posted by Chessgames members do not necessarily represent the views of Chessgames.com, its employees, or sponsors.
Spot an error? Please suggest your correction and help us eliminate database mistakes!


home | about | login | logout | F.A.Q. | your profile | preferences | Premium Membership | Kibitzer's Café | Biographer's Bistro | new kibitzing | chessforums | Tournament Index | Player Directory | Notable Games | World Chess Championships | Opening Explorer | Guess the Move | Game Collections | ChessBookie Game | Chessgames Challenge | Store | privacy notice | contact us
Copyright 2001-2018, Chessgames Services LLC