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Louis F Stumpers
Number of games in database: 46
Years covered: 1932 to 1969
Overall record: +12 -27 =7 (33.7%)*
   * Overall winning percentage = (wins+draws/2) / total games
      Based on games in the database; may be incomplete.

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B59 Sicilian, Boleslavsky Variation, 7.Nb3 (2 games)
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LOUIS F STUMPERS
(born Aug-30-1911, died Sep-27-2003) Netherlands

[what is this?]

 page 1 of 2; games 1-25 of 46  PGN Download
Game  ResultMoves Year Event/LocaleOpening
1. L Stumpers vs J Lehr  1-019 1932 EindhovenD18 Queen's Gambit Declined Slav, Dutch
2. L Stumpers vs E Spanjaard  1-055 1938 Dutch Ch prelimE02 Catalan, Open, 5.Qa4
3. H van Steenis vs L Stumpers  1-025 1939 NED-ch11B02 Alekhine's Defense
4. J van den Bosch vs L Stumpers  ½-½58 1939 NED-ch11A48 King's Indian
5. L Stumpers vs S Landau  0-141 1939 NED-ch11D33 Queen's Gambit Declined, Tarrasch
6. A J van den Hoek vs L Stumpers  1-027 1941 BondswedstrijdenB10 Caro-Kann
7. T van Scheltinga vs L Stumpers 1-035 1942 NED-ch12D94 Grunfeld
8. G Fontein vs L Stumpers  ½-½26 1946 NED-ch prelim ID94 Grunfeld
9. L Stumpers vs Cortlever  ½-½50 1946 NED-ch prelim IE60 King's Indian Defense
10. L Stumpers vs J H Marwitz  1-040 1946 NED-ch prelim ID31 Queen's Gambit Declined
11. L Stumpers vs Euwe 0-130 1946 NED-ch prelim IE60 King's Indian Defense
12. W Wolthuis vs L Stumpers  ½-½52 1946 NED-ch prelim IC58 Two Knights
13. C B van den Berg vs L Stumpers  1-058 1946 NED-ch prelim ID19 Queen's Gambit Declined Slav, Dutch
14. L Stumpers vs H van Steenis 0-124 1946 NED-ch prelim ID28 Queen's Gambit Accepted, Classical
15. V Soultanbeieff vs L Stumpers  ½-½46 1947 Int BD96 Grunfeld, Russian Variation
16. Tartakower vs L Stumpers 1-024 1947 Int BD74 Neo-Grunfeld, 6.cd Nxd5, 7.O-O
17. L Stumpers vs H van Steenis  0-133 1947 Int BD23 Queen's Gambit Accepted
18. L Stumpers vs Grob 1-060 1947 Int BA55 Old Indian, Main line
19. J Baay vs L Stumpers  1-040 1948 NED-ch14E37 Nimzo-Indian, Classical
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 46  PGN Download
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Aug-18-14
Premium Chessgames Member
  Sneaky: <1. I once heard a soccer commentator say that if one side in a match scores the first two goals, it's more likely than not that the other team will score the next goal. Is this plausible?>

I can't imagine how, unless the sports announce simultaneously regarded soccer as a game of luck, and also believed in the gambler's fallacy.

I would imagine that if you did a statistical analysis of thousands of soccer games you would find that if a team scores the first two goals, they are likely to score the third goal.

<I have frequently heard sports commentators say something like this: "The team that scores first wins 60% of the time." (I'm not sure of the exact percentage, but it's more than half.) Is this plausible?>

Of course it is! Some teams are better than others. The team that's better is more likely to score first, and they better they are the more likely it becomes. I don't even see what the controversy is here.

The one about Femalaysia and Womania is interesting. At first I thought it was trivial but the more I think about it the less clear it becomes. Well, the correct answer may be trivial, but there's a lot to say about it.

Aug-19-14
Premium Chessgames Member
  Sneaky: <Femalaysia and Womania>

I liked this one because it's very simple but not obvious, at least not to me. I reasoned this through by example, and then I saw what was really going on.

It is not necessary to assume that the odds are exactly 50/50, but for the sake of discussion let's assume for a moment they are. Start with the case of four families on each island, and considered what the likely outcomes would be:

<Femalaysia: B, B, GB, GGB (3 girls, 4 girls)>

<Womania: G, G, BG, BBG (4 girls, 3 boys)>

It seems at first that Womania is winning. But then I thought about it for 16 families:

<Femalaysia: B, B, B, B, B, B, B, B, GB, GB, GB, GB, GGB, GGB, GGGB, GGGGB (15 girls, 16 boys)>

<Womania: G, G, G, G, G, G, G, G, BG, BG, BG, BG, BBG, BBG, BBBG, BBBBG (16 girls, 15 boys)>

Now it's obvious to me that neither island is winning: as the number of families approaches infinity the ratio turns into a nice neat 1:1 for both islands. They all end up with a 50/50 split between girls and boys in spite of their family planning.

As I said, it doesn't have to be 50/50. Suppose there was a biological disposition to males (I think there actually is) of 51/49. Then we'd find both islands seeing that very distribution, in spite of their efforts to combat it.

Finally I saw what's really going on here. Having a child, in hope of a certain sex, is exactly like a gambling game: in one case you put your hopes on "girl" and hope you win, in another case you put your money on "red" and hope you win.

So let's suppose that there really is a strategy child of birthing to ensure a predominance of one sex over the other, whether it's one of the strategies suggested in the stumper, or something entirely different. You could then take that very strategy, walk into any casino, and clean the house out. Simply pretend that 'winning at roulette' is the same thing as 'having a girl', repeat the strategy over and over, and finally you'll walk out a millionaire.

The strategy of Femalaysia would equate to "Keep betting on red as until you lose, then end the session." The strategy of Womania would be "Keep betting on red until you win, then end the session."

Would either of those strategies make you a roulette winner? We all know that's ridiculous, and so must be the strategies of Femalaysia and Womania, or any other idea intended to produce the same result.

Aug-19-14
Premium Chessgames Member
  kellmano: <sneaky> yes the fact is that whatever national policy is in place when any particular child is born the odds are 50-50. In your examples I would go for the 16 families maxing out at 4 children (or any other number for that matter) to give:

<Femalaysia: B, B, B, B, B, B, B, B, GB, GB, GB, GB, GGB, GGB, GGGB, GGGG (15 girls, 15 boys)>

Aug-19-14
Premium Chessgames Member
  kellmano: Actually, imagine if 100% of couples could have one child and 50% could have 2 children for biological reasons, and no-one could have more than 2.

Then, 8 families gives B,B,B,B,G,G,GB,GG.

Of course it still doesn't make any difference for the reason I gave above, but for a moment I though it did. OK I have thought about this problem too long now and forgotten the original intuition. It's that boys would increase I think. Perhaps that is because you think of half of all families having a single male child.

Aug-19-14
Premium Chessgames Member
  kellmano: Apologies for the monologue but the betting point is a good one. In the UK something called betfair is very popular, in which you can lay bets (take them) meaning you can but up a big stake at very short odds for something likely, eg. Chelsea beating Burnley at football. You then tell everyone that you have got a way to make easy money and brag about it on social media.

Really all your doing is saying I'll bet one pound on red, then 2 if I lose, then 4, then 8, then 16. Sure you win almost all the time, but when you lose it's massive. I've got no problem with gambling for leisure but it's less dangerous to put small stakes on long odds.

Aug-20-14
Premium Chessgames Member
  al wazir: Here is my solution to #1.

The obvious answer is no. You would expect that the team that scored the first two goals did so because it's stronger than its opponent. However, there's an element of luck in soccer. I've heard that 50% of everything that happens in a game is chance, whatever that means. If it were 100%, then the probability of the opposing team scoring the next goal would be 50%.

But it's not all luck. Ability plays a role, and so do tactics. A team with a two-goal lead might be inclined to start playing conservatively. If it played very conservatively – if it went into a complete shell – its opponent would have little chance to score. But teams that play totally defensively, with no attack at all, have no chance whatever to score goals for themselves.

My feeling is that the only sure way to answer this question is to look at a representative sample of matches that started 2:0 and find out who scored next. The top European leagues each have 20 teams playing one another in a double round robin during the regular season, making a total of 380 matches. In matches between two sides of comparable strength roughly half should start off 2:0 or 0:2, and at least one additional goal should be scored in roughly half of those, since on average around three goals are scored per match. Hence about a quarter of all matches, roughly 100 per league, would go into the sample. There must be searchable databases somewhere that record the order of scoring. If I find them, that will be a project for some of my idle hours.

Aug-20-14
Premium Chessgames Member
  al wazir: Here is my solution to #2.

Again, you would expect that the team that scored first did so because it's stronger than its opponent. That certainly seems plausible for low-scoring sports like soccer and hockey, and even for those with somewhat more scoring, like baseball and American football. But what about basketball? How significant can it be that team A makes the first basket before team B when each team's final score is on the order of a hundred points? Well, quite significant, because the average *margin* of victory is only ten or fifteen points.

Sports commentators, by and large, are not the world's deepest thinkers, and their statistics can be pretty meaningless. Take the old standby that “the team leading at halftime,” or after three innings, or whatever, wins x percent of the time. Sure, and the team leading after three quarters does even better, and one that's ahead at the two-minute mark does better still. Then there's one that goes something like this: the Assassins “have never lost a game” in which their batters hit five or more home runs.” Well, I should think not.

And consider those statoids based on ridiculously small samples. Baseball lends itself to the worst abuses, because it has so many different things that can be counted: batter Smith “has hit safely four times in six at-bats over his career” against pitcher Jones; and he “leads the league in walk-off base hits” (he has a grand total of three); but Jones “hasn't blown a save so far this season” (in five appearances).

I no longer listen to the commentary when I watch televised sports.

Aug-20-14
Premium Chessgames Member
  al wazir: Here is my solution to #3.

The simple, straightforward answer is that in both countries girls will be born at the natural rate, because when conception takes place the chromosomes do their thing and nothing any potentate says is going to change the outcome. (Selective abortion, guided by ultrasound screening, is another matter: http://en.wikipedia.org/wiki/Sex-se....) Of course it's implicit in this problem that the natural rates of birth are the same in both countries, but that's a realistic assumption. Male and female babies are born in almost equal numbers, though males outnumber females by a few percent everywhere in the world.

Not convinced yet? All right, then I have to resort to algebra. Suppose that the probability of a male birth is m and the probability of a female birth is f, where m + f = 1. I'm assuming that m is the same in both countries (so f must be too), but I'm not assuming that m = f. I'm also assuming that couples can keep on having children indefinitely, so there's no upper limit to the size of families. (That's not so realistic.) I'll disregard multiple births (twins, triplets, etc.) for simplicity. They don't change the result.

In Femalaysia the possible birth sequences can be: boy; a girl, then a boy; two girls, then a boy; three girls, then a boy, and so on. Their respective probabilities are m; f*m; f^2*m; f^3*m; ... . The numbers of girls corresponding to these possibilities are 0, 1, 2, 3, …, so the *expected number* of girls in Femalaysian families is the total of

0*m + 1*f*m + 2*f^2*m + 3*f^3*m + … = f*m*(1 + 2f + 3f^2 + …) = f*m/(1 – f)^2 = f/m.

[The identity 1 + 2x + 3x^2 + … = (1 – x)^2 follows from differentiating the familiar expression for the sum of a geometric series, 1 + x + x^2 + … = 1/(1 – x).]

The proportion of girls in the population is thus f/m divided by 1 + f/m, which is equal to f. In Womania the possible birth sequences are girl; boy, then girl; two boys, then a girl; three boys, then a girl; and so on. By going through exactly the same steps, but interchanging f and m and “girl” and “boy” (or cutting to the chase and just interchanging them in the final result), you find that the expected number of *boys* in Womanian families is m/f. Hence the proportion of females there is 1 divided by 1 + m/f, which is again f. QED.

Aug-20-14
Premium Chessgames Member
  al wazir: Here is my solution to #4.

Bicyclists take less time going downhill than up, so the probability of my seeing one who is descending is smaller. It doesn't matter whether *I* am going up or down, since I drive at about the same speed either way. But suppose I were biking up and down that hill too. (Not likely.) Then I guess it would be a wash.

Aug-20-14
Premium Chessgames Member
  al wazir: Here is my solution to #5.

If there is only one elevator and I am the only one in the building, it will come in response to my signal from wherever it was resting. That would probably be my own floor. (But sometimes elevators are programmed to return to the ground floor when not in use.) If other people are in the building along with me, however, it will probably come from one of the lower floors, since there are more of those than there are floors above me. In that case it makes just as much sense to say that it is stopping on the way down as on the way up. The same reasoning applies if there are multiple elevators. When I press the “down” signal button on the eighth floor at some random time, I don't know where they are. Each one might be anywhere, but since they spend most of their time *below* me, the next one that stops is more likely to be coming from below than from above.

But the situation is a little different at the start of the workday, when workers are arriving in large numbers. The elevators are in the service of everyone, not just me, and everyone except me wants to go up. The elevators are all likely to be in continual use. Each one fills up on the ground floor and ascends to one of the top floors, stopping several times on the way to let people out. The highest floor it reaches might be the eighth, but more likely not. If it stops at the eighth floor it probably does so not for me but in order to let someone out and then continue up. (I can get on now or wait for it to stop again on its way back down.) The next one that stops there on its endless cyclic journey from the bottom to the top of the building and back is probably stopping to let someone out, not to take me on a bespoke ride to the second floor. So it is more likely than not to be on its way up, not down.

Aug-20-14
Premium Chessgames Member
  al wazir: Here is my solution to #6.

I'm afraid it's really a trick problem. The clue is that those were *day trips*: Zeke always left early in the day and returned late. When you drive on a highway you pass many more cars going the other way than in the same direction you're going, so most of the license plates you see are on cars headed in the opposite direction. (This would still be true regardless of whether you go much slower than most drivers or drive like an Andretti.) If it was dark on his return trip Zeke wouldn't have seen any going in the other direction because of the headlights, but that still leaves the ones he saw in the morning.

Most people with Nevada license plates live in Nevada. When they go somewhere they usually start out from home and when they return they head towards home. So when traveling between Vegas and L.A., they're mostly going toward L.A. early in the day and toward Vegas late in the day. (This would be true even if they didn't live in or near Las Vegas, though as it happens most Nevadans do.) Zeke was going *with* most of the flow of Nevada cars when he lived in Vegas and *against* it when he lived in L.A., so he saw more of them when he started from and returned to L.A. then when he started from and returned to Vegas.

Aug-20-14
Premium Chessgames Member
  Sneaky: <al wazir: Here is my solution to #4.

Bicyclists take less time going downhill than up, so the probability of my seeing one who is descending is smaller. It doesn't matter whether *I* am going up or down, since I drive at about the same speed either way.>

Brilliant. That is the definitive answer (although creative explanations would work as well.) I was thinking about something like that but I couldn't quite put my finger on it.

Aug-21-14
Premium Chessgames Member
  kellmano: <The top European leagues each have 20 teams playing one another in a double round robin during the regular season, making a total of 380 matches. In matches between two sides of comparable strength roughly half should start off 2:0 or 0:2> not sure what percentage of games have two or more goals, but I'd guess around 70.
Aug-21-14
Premium Chessgames Member
  kellmano: Regarding number 1, I used to get annoyed to see teams in soccer 'sitting back' after taking a lead. Often it would be the underdog who played very well until they took the lead and then just let the team with the better reputation pass the ball about with leisure. I saw an interview with a manager once who acknowledged this as a bad strategy but said that 'sitting back' was not a reasoned response but a psychological habit of the players.

Not specifically relevant to here, but does relate to some potentially interesting maths. Say team A and team B have probability of scoring in a ten minute period of 0.2 and 0.15 respectively. With 20 minutes remaining and with a 2-0 lead Team A can elect to adopt a defensive strategy to change it so that both teams have a probability of 0.1 of scoring in the ten minute period. Is it a rational decision to do so?

Assume that the result is all that matters and the margin of victory is unimportant. Assume that if a goal is scored team A will get to make the same decision again.

Aug-21-14
Premium Chessgames Member
  kellmano: In relation to the above whether the decision *looks* rational depends on whether team B scores quickly.

Relates to <al wazir>'s point about the lack of deep-thinking in commentators. If you watch a game of football over here players are consistently criticised for making a poor decision by shooting from 30 yards away. If it flies in to the top corner, no-one criticises the decision. Happens in other sports too actually - cricket springs to mind and the attitude adopted towards a batsman who plays big shots.

Aug-21-14
Premium Chessgames Member
  al wazir: <kellmano: not sure what percentage of games have two or more goals, but I'd guess around 70.> That was just a horseback guess. During the WC I heard (from a deep-thinking commentator?) that the goals-per-match average during the group stage was 2.76, if I recall correctly. It was lower during the knockouts, but I have the ungrounded impression that it's higher during league play.

<I saw an interview with a manager once who acknowledged this as a bad strategy but said that 'sitting back' was not a reasoned response but a psychological habit of the players.> For the life of me I've never been able to figure out what managers do beyond selection, or why they get all the credit for a winning season and all the blame for a losing one. If Ferguson was such a genius, why did he anoint Moyes? How is Van Gaal going to turn Man U around? I'll believe it when I see it. I have always thought that great managers are great because they have great players.

<Is it a rational decision to do so?> I really do think that 50% of what happens in football is down to chance. I think that close to 0% of success is down to what sportswriters constantly harp on: the team's "courage," "attitude," "spirit," and like intangible emotional factors. And as in war, no game plan survives first contact with the adversary.

Aug-21-14  nok: For #3 you can say first-borns are split naturally, second-borns too etc. no matter who gets to play the next round.
Aug-21-14
Premium Chessgames Member
  al wazir: <kellmano: Say team A and team B have probability of scoring in a ten minute period of 0.2 and 0.15 respectively. With 20 minutes remaining and with a 2-0 lead Team A can elect to adopt a defensive strategy to change it so that both teams have a probability of 0.1 of scoring in the ten minute period. Is it a rational decision to do so?>

Back to your problem:

Is team A playing to win, or just not to lose? That is, is a draw acceptable? Likewise for team B. That's four different scenarios. And there are more than just those four possibilities, depending on the two clubs' standings in the table. Maybe team A can use 1 point, but 3 points are better; whereas team B absolutely needs 3 to have a hope of winning the league title, and with a draw there is only a faint chance for them even to qualify for the Champions League.

A 0.2 probability of scoring once in ten minutes translates to 0.02 goals per minute. A proper treatment requires working with a continuous distribution, since there is nothing in principle to prevent a team from scoring twice in a minute or even faster. (Nothing but those stupid minute-long made-for-TV celebrations, that is.) But to keep it simple I'll say that it's equivalent to a 0.4 probability of scoring once in twenty minutes, and likewise that team B has a 0.3 probability of scoring. The answer should be close to what would be found using a Poisson distribution.

Assume that a draw is worthless; both teams need to win. With strategy #1, team A has 0.4 chance of scoring once, (0.4^2 chance of scoring twice, (0.4)^3 chance of scoring three times, etc. Expected no. of goals: 0.4 + 2*(0.4)^2 + 3*(0.4)^3 + ... = 1.11. Team B's expectation is 0.061 goals. Looks pretty safe. But what are the chances of not winning? If team B scores twice, team A must score at least once; if team B scores four times, team A must score three or more times, etc. This is getting complicated. I'll just sketch the steps.

The probability that team A fails to score is 1.0 - [0.4 + (0.4)^2 + (0.4)^3 + ... ] = 1/3. The probability that team B scores twice or more is (0.3)^2 + (0.3)^3 + (0.3)^4 + ... = .1286. Their product is 0.0439. The probability that team A scores once and team B scores three times or more is 0.4*[(0.3)^3 + (0.3)^4 + (0.3)^5 + ...] = .0096. The probability that team A scores twice and team B scores four times or more is (0.4)^2*[(0.3)^4 + (0.3)^5 + (0.3)^6 + ...] = .0002, and we can stop there; the rest of the terms are negligible. Adding them up we get 0.0537. So with this strategy team A has a 0.946 chance of winning.

With strategy #2, both teams have a 0.2 chance of scoring in the remaining 20 minutes. The probability that team A (or B) fails to score is 1.0 - [0.2 + (0.2)^2 + (0.2)^3 + ... ] = 0.75. The probability that team B scores twice or more is (0.2)^2 + (0.2)^3 + (0.2)^4 + ... = .032. Their product is 0.024. The probability that team A scores once and team B scores three times or more is 0.2*[(0.2)^3 + (0.2)^4 + (0.2)^5 + ...] = 0.002. Everything beyond that is negligible. The probability that team A wins is thus 1.0 - (0.024 + 0.002) = 0.974.

Hence #2, the defensive strategy is preferable for team A.

Aug-21-14
Premium Chessgames Member
  kellmano: Thanks <al wazir> so it turns out the numbers I gave did make it moderately close.

You're obviously a statistician, but I think the statement <A 0.2 probability of scoring once in ten minutes translates to 0.02 goals per minute> is wrong as it will be higher than that. I'd be happy for the model to allow two goals to occur instantaneously, but to say the probabilities can be divided like this is inaccurate.

If I have a 0.02 probability of scoring in 1 minute, the probability of scoring at least once in twenty minutes is 0.02 + (1-0.02)*0.02) + ((0.02 + (1-0.02)*0.02))*0.02 ........

Perhaps given the values your version is considered a good enough approximation, I genuinely don't know.

Aug-21-14
Premium Chessgames Member
  Marmot PFL: <But what about basketball? How significant can it be that team A makes the first basket before team B when each team's final score is on the order of a hundred points? Well, quite significant, because the average *margin* of victory is only ten or fifteen points. >

Not really, because 10 points out of 180 (both teams) is a much smaller percentage than 1 goal out of 4-6. And one basket (2 points) is even more negligible.

Say a baseball team loses 7 of its first 10 games. Not good, but there are still 152 games to go, and plenty of teams have started worse than that and made the postseason.

In football a 3-7 team has almost no chance of turning it around in time for the playoffs. So the context of each sport does make a difference.

Aug-21-14
Premium Chessgames Member
  al wazir: <Marmot PFL: Not really, because 10 points out of 180 (both teams) is a much smaller percentage than 1 goal out of 4-6. And one basket (2 points) is even more negligible.> You are making the same mistake most sportscasters and writers -- and nearly all fans -- make. The "game-winning" final basket counts exactly the same as the first one, no more and no less.

<Say a baseball team loses 7 of its first 10 games. Not good, but there are still 152 games to go, and plenty of teams have started worse than that and made the postseason.> Same thing. Every one counts the same. It's true, however, that the multi-team format punishes excellence by giving mediocre teams a shot at winning the cup or Super Bowl or World Series.

Typically, first place in at least one, and usually a couple, of the six divisions in MLB and the NBA comes down to the last game of the season. True, all they're competing for is a more favorable match-up in the playoffs. But by the same token, one or two teams miss out entirely from those playoffs by a single game -- a game that they lost back in the first month of the season because so-and-so committed a brainless technical foul or what's-his-name was day-dreaming and got picked off first.

There are some subtleties, however. Every big-league team has a "closer," the relief pitcher the manager thinks is the best on the staff on getting outs over an inning or less. He only comes in to protect a late-inning lead, almost never in a game where the team is behind. Why wear out your best pitcher when the chances of winning are slim? Likewise, if the playoffs are out of reach at the end of the season, back-up players and minor leaguers get to play. Basketball coaches empty the bench during "garbage time," and football coaches bring on the second- or third-string running back when the game is safely in hand.

Aug-21-14
Premium Chessgames Member
  al wazir: The average margin of victory in the first 60 NBA games played in 2014 was 11.5 points. The statistical error in a sample that size is pretty big -- 13% -- but this means that the true value is probably between 10 and 13 points. Some of those 60 games were blowouts, but some were close. Two of them were won by one point and three by two. That's *not* statistically significant, but I expect that it's in the ballpark.

Players don't know in the first quarter of a game if it's going to be a blowout, so they should play as if it's going to be close, because there's a good chance that it will be. They're pros, after all. And they're certainly paid enough.

Aug-22-14  beatgiant: <al wazir>
#2: Since it actually happens in the real world, it's plausible, so I stand corrected.

#3. It seems everyone has the assumption that the birth rate per gender is uniform across all couples. I don't think that's known to be true in the real world

What if some couples are more likely to have girls, and other couples are more likely to have boys? Then the policy in Femalaysia (stop after the first boy) will disproportionately affect the boy-prone couples, while the policy in Womania (stop after the first girl) will disproportionately affect the girl-prone ones.

For example, consider the extreme case where some couples can have only boys, and other couples can have only girls. If those couples live in Femalaysia, the boys-only couples will be limited to one child, while the girls-only couples can have as many as they want.

Aug-22-14
Premium Chessgames Member
  al wazir: <beatgiant: What if some couples are more likely to have girls, and other couples are more likely to have boys?> This does happen. I don't know much about it, but I think it's a very small percentage. Gender dependence in death rates plays a much bigger role. In developed countries women live longer, whereas men do in some undeveloped countries. There are also gender differences in child mortality for some diseases.

I think we're getting down to the fine print in the footnotes here.

Aug-22-14  beatgiant: <al wazir>
A source about the phenomenon:
http://tinyurl.com/mjmwfgc

Besides the issue of sex-selective abortions (for example, in Femalaysia, a couple wanting more than one child might abort when the ultrasound shows a boy), some other factors mentioned were age of the couple, maternal stress, and frequency of intercourse.

I haven't read the full cited studies, but the abstracts claim the effects are significant, and those factors sound like they could affect a lot of people. And all such factors tend toward Femalaysia having a higher rate of girls than Womania.

The article also reports that more girls than boys are born during wartime. So if Femalaysia and Womania are at war with each other, all bets are off....

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