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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.

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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)
C65 Ruy Lopez, Berlin Defense (2 games)
D45 Queen's Gambit Declined Semi-Slav (2 games)
D31 Queen's Gambit Declined (2 games)

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

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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),, Digitaal Tijdschrift, 5 (255),
2) Eindhovense Schaakvereniging (2016), http://www.eindhovenseschaakverenig...
3) Noord Brabantse Schaak Bond (2016), Their main page:
4) (2016),
5) Olimpbase,
6) K. Teer, Levensbericht F. L. H. M. Stumpers, in: Levensberichten en herdenkingen, 2004, Amsterdam, pp. 90-97, Also available at
7) Catalogus Professorum Academiæ Rheno-Traiectinæ,
8) URSI websites (2016), and

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

This text by User: Tabanus. The photo was taken from

Last updated: 2018-08-17 13:29:49

 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 H Kramer  0-1361940HilversumE25 Nimzo-Indian, Samisch
10. L Stumpers vs S Landau  ½-½341940HilversumD31 Queen's Gambit Declined
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, Nxd5, 7.O-O
23. V Soultanbeieff vs L Stumpers  ½-½461947Int BD96 Grunfeld, Russian Variation
24. L Stumpers vs T van Scheltinga  1-0471948NED-ch14C97 Ruy Lopez, Closed, Chigorin
25. Prins vs L Stumpers  ½-½301948NED-chD02 Queen's Pawn Game
 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
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Premium Chessgames Member
  moronovich: Convoluted - Lol!

Sounds like we are on the Stampers page.

Premium Chessgames Member
  al wazir: <Count Wedgemore: In other words, there's 50% chance of gaining x, and 50% chance of losing x.> You can say that in a simpler way.

If you don't look inside the envelopes, there's a 50% chance of getting the big payoff and a 50% chance if getting the little one, no matter how many times you shuffle the envelopes.

Which is why it's trivial. It's not paradoxical at all.

But the version of the problem where you open up one envelope before deciding whether to switch is *nontrivial*.

Think about the way I formulated the problem: the donor repeats the transaction over and over with many subjects -- dozens of them. You observe them, so you know whether or not they switched and what resulted.

When recipients open the first envelope and find $x in it, some of them decide to switch and some don't. (The amount x varies unpredictably from one trial to the next, say, between $10 and $100.) Those who chose to take the other envelope find $2x in it 50% of the time and $x/2 in it 50% of the time. Of course not everyone decides to switch; those who stand pat wind up with whatever is in the first envelope and never find out what is in the second one.

Now it's your turn. You open an envelope you pick at random and find $50 in it. What do you do: switch or stand pat?

This problem is an example of what are called *choice paradoxes*. Perhaps the most famous is Newcomb's Paradox. My take on it is decidedly heterodox: Louis Stumpers (kibitz #6565).

Premium Chessgames Member
  beatgiant: <al wazir>
If you know the minimum is $10 and the maximum is $100 as in your example, you definitely do switch if your envelope contains less than $20 and definitely don't switch if it contains more than $50. Your dilemma occurs only when it's between $20 and $50.
Premium Chessgames Member
  al wazir: <beatgiant: If you know the minimum is $10 and the maximum is $100 as in your example you definitely do switch if your envelope contains less than $20 and definitely don't switch if it contains more than $50.> Why? Because that would have been a good strategy in the past? You don't know that $10 and $100 will always be the limits.

Haven't you ever heard the adage "past performance is not indicative of future results"?

Premium Chessgames Member
  Marmot PFL: Didn't see that Newcomb problem before. If you know nothing abut the "predictor" (maybe just some guy flipping a coin) go materialist and take A+B. If "predictor" is right more often than not go with B only but as you say "past performance is not indicative of future results" (I would say not <always> indicative).

A shame that Sneaky is gone.

Premium Chessgames Member
  beatgiant: <al wazir>
Still, if we are dealing with the real world, there will be some known limits. The currency has a minimum unit of account (e.g. 1 cent, if it's dollars), and Forbes puts the net worth of the wealthiest person at 100 billion dollars.

In those conditions, I don't think it's possible that the probability of doubling is always the same as the probability of halving.

In fact, if the amount in the envelope is an odd number of minimum units, you switch. If the amount in the envelope is over 50 billion dollars, you don't switch.

Premium Chessgames Member
  al wazir: <beatgiant: In fact, if the amount in the envelope is an odd number of minimum units, you switch.> Granted. But it never is. (All our examples have been even.)

<If the amount in the envelope is over 50 billion dollars, you don't switch.> That's you. I probably wouldn't switch at any amount. I'm betting-averse.

Oct-12-18  john barleycorn: < al wazir: ...

<If the amount in the envelope is over 50 billion dollars, you don't switch.> That's you. I probably wouldn't switch at any amount. I'm betting-averse.>

Nah, probably you are easy to please. Winning 1 dollar may give satisfaction to one guy, while winning a million dollars may not mean anything to another guy. everybody has to decide for him/herself when and with what amount s/he feels "happy". If I remember right one of the Bernoulli clan invented the "utility" function to describe the phenomenon

Premium Chessgames Member
  beatgiant: <al wazir>
<All our examples have been even.>

If it is known that the amount in the envelope is always even, that's equivalent to using minimum unit of account of 2, instead of 1. Now if the amount is not a multiple of 4, you switch.

<<If the amount in the envelope is over 50 billion dollars, you don't switch.> That's you. I probably wouldn't switch at any amount. I'm betting-averse.>

I'm not sure if you're making a joke or missing the point. We don't switch because <we know the maximum is 100 billion> so any amount greater than 50 billion must be the 2x and not the x. It's no gamble.

Premium Chessgames Member
  al wazir: <beatgiant: I'm not sure if you're making a joke or missing the point.>

This problem was beaten to death the first time it came up. (<Marmot PFL> posted it that time too: Louis Stumpers (kibitz #3133).) If I keep posting I'm afraid I'll contradict what I said back then.

Premium Chessgames Member
  beatgiant: <al wazir>
I don't think the earlier kibitzing addressed these issues of limits at the bottom or the top. But that's because we all know the whole point is to show how infinity breaks our reasoning.
Premium Chessgames Member
  Marmot PFL: 2012, I had totally forgotten the topic except vaguely remembering seeing it somewhere. Shows how badly my memory is declining I guess. I find games on this site sometimes that I commented on years ago and have no recollection of having seen the game before.
Oct-12-18  john barleycorn: <Marmot PFL: ... Shows how badly my memory is declining I guess. ...>

Now, a real stumper :

<Are we of the same age?>

Premium Chessgames Member
  beatgiant: <Are we of the same age?> I am of the anthropocene, by some definition of the term.
Oct-12-18  john barleycorn: <anthropocene>? never heard that. was it before the earth became a red fiery ball?
Premium Chessgames Member
Oct-12-18  john barleycorn: <beatgiant> thanks. I did not know you are 12000 years older than me. :-)
Oct-12-18  nok: PC talk for <capitalism>

Premium Chessgames Member
  al wazir: I want to return to the airline passenger problem because I'm unhappy with where we left off.

Yes, it's easy to show that the probability that the last of n passengers to board has a 50% chance of getting his assigned seat for the special cases n = 2, 3, 4, 5, . . . (I say it's easy because it *is* easy, even though I messed up the first time.) But that doesn't *prove* that the probability is 0.5 for n = 100, and I wouldn't want to try to carry out the calculations needed to check it by hand .

Here's an illustration of why a “proof” by examples isn't good enough. Consider the formula p = n^2 + n + 41. For n= 0, 1, 2, 3, 4, . . ., and as far as you probably care to check, p turns out to be prime. Mirabile dictu! A formula that automatically generates prime numbers!

But for n = 41 it transparently fails.

The airline passenger problem looks as if it *ought* to be provable by mathematical induction, yet there is no obvious way to do it. But maybe there's a clue (not a proof, just a clue) lurking in those examples.

Let the passengers be designated by capital letters and the corresponding assigned seats by small letters. (As is well known, airline seats are always smaller than the passengers who sit in them.) For n = 2, there are just two possibilities: either A, the first passenger, sits in a, his assigned seat, or he doesn't, with equal probabilities of 1/2. So a table of the results looks like this:

AB [0.5]
BA [0.5]

With three passengers there are four possibilities. If the first passenger sits in the right seat (probability 1/3 = 0.3333...), then everyone does. If not, there are two other seats available, b and c. If he sits in c (again with probability 1/3), then b is available for B, and C is left with a. But if A sits in b, there are two possibilities. Both of these last two cases have the same probability, 1/3 x 1/2 = 1/6 = 0.1666....

ABC [0.3333...]
CBA [0.3333...]
BAC [0.1666...]
CAB [0.1666...]

The four probabilities add up to 1.0, as they should, and the probabilities of the two cases where C gets to sit in c add up to 0.5.

Continuing in this manner, for n = 4 we find eight possible arrangements and their probabilities:

ABCD [0.25]
BACD [0.08333...]
CABD [0.04166...]
DABC [0.04166...]
DACB [0.08333...]
CBAD [0.125]
DBAC [0.125]
DBCA [0.25]

Again the probabilities of the possible arrangements add up to 1.0, and those of the subset with D sitting in d add up to 0.5.

And for n = 5, the results are

ABCDE [0.2]
BACDE [0.05]
CABDE [0.01666...]
DABCE [0.008333...]
EABCD [0.008333...]
EABDC [0.01666...]
DACBE [0.025]
EACBD [0.025]
EACDB [0.05]
CBADE [0.06666...]
DBACE [0.03333...]
EBACD [0.03333...]
EBADC [0.06666...]
DBCAE [0.1]
EBCAD [0.1]
EBCDA [0.2]

These 16 probabilities add up to 1.0, and those of the eight cases with E sitting in e add up to 0.5.

Now some patterns can be discerned. First, in each example half of the possible arrangements resulted in the last passenger getting his assigned seat. Second, the number of possible arrangements where the last passenger sits in his assigned seat doubled each time, from 1 to 2 to 4 to 8, and the total number of possible arrangements also doubled, from 2 to 4 to 8 to 16. (So if we decided to work out the probability by hand for 100 passengers, the number of cases we would have to examine would be 2^99 ≈ 6 x 10^29 — nearly a quadrillion quadrillion!)

A third pattern is less obvious. The two cases found for n = 3 that have the last passenger sitting in the last seat, ABC and BAC, have the *other two* passengers (A and B) arranged in *all* of the ways that were possible for n = 2. The four cases found for n = 4 that have the last passenger sitting in the last seat, ABCD, BACD, CABD, and CBAD, have the *other three* passengers (A, B, and C) arranged in *all* the ways that were possible for n = 3. A similar relationship holds between n = 4 and n = 5. The explanation is straightforward: the permissible arrangements in which the first n – 1 passengers occupy the first n – 1 seats are precisely the ones that leave the nth seat available for the nth passenger.

Premium Chessgames Member
  al wazir: (cont'd)

This doesn't take us very far towards explaining why the probability that the last passenger occupies the last seat is exactly 0.5. There is one more clue, however, which does help.

Look at the n = 5 example. Each of the arrangements that has 'E' in the last place has a probability equal to that of another arrangement that is exactly the same, except that the passengers in the first and last seats are interchanged:

ABCDE and EBCDA both have probability 0.2;

DBCAE and EBCAD both have probability 0.1;

CBADE and EBADC both have probability 0.06666...;

BACDE and EACDB both have probability 0.05;

DBACE and EBACD both have probability 0.03333...;

DACBE and EACBD both have probability 0.025;

CABDE and EABDC both have probability 0.01666...;

DABCE and EABCD both have probability 0.008333....

The first member of each pair ends in 'E' and the second doesn't, so it follows immediately that the probability of E sitting in e is 0.5. A similar pairing occurs for n = 4 and for n = 3. For n = 2 it's trivial.

Why does this happen?

Consider any arrangement in which E is *not* sitting in the last seat. Then some other passenger (A, B, C, or D) must be. Swapping them *must* yield a permissible arrangement. This is a tricky point, but it's crucial to the proof. At the time when that passenger sat in e, his own assigned seat must have been unavailable, or else he would have sat in it; and the seat that E eventually sat in (because it was the only one left) must also have been available. So that passenger could instead have sat in the seat eventually occupied by E, and what's more, since both seats were then available and the seating of the other three passengers is otherwise unchanged in the two arrangements, they must have the *same probability*.

Got that?

But what holds for n = 5 holds for any other value of n, including 100, by exactly the same reasoning.


Premium Chessgames Member
  Marmot PFL: As the 1st passenger (lost ticket) enters he has an equal chance (1/100) of getting the correct seat #1, thereby "winning" the game) or of getting the seat of #100 (in which case the game is "lost").

In 98/100 cases neither will happen. He sits in another seat, lets say #10. Now #2-#9 sit where assigned, and #10 has an equal chance (1/90) of taking seat #1 ("winning" as all other passengers now take the correct seats), or of taking seat #100 ("losing").

In 88/90 cases he takes another seat and the "game" continues, always with increasing chances of seats #1 or #100 being taken (the rest are irrelevant to the outcome) but the chances of either seat being taken by any other passenger are always the same.

If by some miracle passenger #99 enters and the only seats left are #1 and #100 the chance remains 50-50.

Premium Chessgames Member
  al wazir: <Marmot PFL>: I don't understand that proof. Yes, each time a passenger finds his assigned seat occupied, the probability of taking seat #100 is the same as the probability of taking any other available seat, including the one he actually winds up choosing.

But that doesn't tell us the probability that, over the whole process of boarding, no one out of the 99 will take seat #100, or equivalently, the probability that *someone* will. It could happen with the first passenger, or the second, or the third, and so on. Each has a certain probability of taking #100, contingent on what has happened previously, and the total probability is the sum of all of those contingent probabilities.

Your proof is probably right; I just don't understand it.

But if there is a 50% probability that passenger #100 winds up sitting in his own seat, then it doesn't take a "miracle" for it to remain unoccupied after the first 98 passengers board. And I find the idea of "winning" and "losing," of this being a game, jarringly dissonant.

Premium Chessgames Member
  al wazir: <Marmot PFL>: Okay, now I understand your proof. It's a restatement of the one I posted, but it improves on my version. It's not only more concise, it shows explicitly why the two arrangements with the same probability differ only in swapping the *first* and last passengers.

Sorry I was so obtuse.

But, although I see how it could be made into a game, I still don't like to think of it in terms of "winning" and "losing."

Premium Chessgames Member
  beatgiant: <al wazir>
Proofs in terms of a game are common in certain fields (cryptology comes to mind).
Premium Chessgames Member
  Marmot PFL: By winning I just meant the last passenger gets his assigned seat and losing that he has to take whatever is left, but the terms aren't too accurate here.
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