The National Coin Flipping League Championship Series
No tech today, but a little basic math.
In baseball, a sport I know little about, apparently the Boston Red Sox have recently come back from a three game deficit to win a best-of-seven series against their traditional rival team, the New York Yankees.
Baseball is a game which attracts statisticians, and many have noted that this is the first time in major league baseball history that a team has won a best of seven series after being down three games to none.
However, it has happened twice in hockey.
I have a modest proposal. Suppose once a year, the National Hockey League and Major League Baseball decide all their various championships without going to all the trouble and expense of playing the game. Rather, they could simply hold a best-of-seven coin-flipping championship. (Call it the Numismatic Hockey League if you'd like.)
Suppose Boston calls heads. The odds of Boston flipping T T T and then coming back to win with H H H H are one in 128.
Therefore, there should be one such occurrence on average every 128 series. There are four such series a year: the American and National League finals, one "world" series (for which only North American teams are eligible, strangely enough), and one Stanley Cup. You'd expect to wait 128 / 4 = 32 years on average between occurrences.
We've been playing pro baseball and hockey, what, about a hundred years in North America?
Three such series, in about a hundred years -- or, roughly one every 32 years. It seems like the math works out rather nicely. Maybe they have been deciding the games via coin flipping and just not telling anyone. Hmm...
Is Boston's victory really that impressive? I mean, the last time I played Risk I rolled three sixes on three dice and England crushed Iceland -- odds of that are 1/216, almost twice as long as Boston coming back from a three tail deficit in the National League Coin Flipping Championship. That's because my blue plastic army guys really worked together as a team and gave 110%!
And yet it didn't make headlines in even the local paper.
In related news, if Houston wins their championship, and it ends up being Texas vs. Massachussets in both baseball AND the presidential election, that's going to be freaky weird. What are the odds of that?
Comments
- Anonymous
October 21, 2004
Except that baseball has only had a seven game league championship series since 1985, and if you're going to factor in the league championship series, why not factor in the hockey conference finals? - Anonymous
October 21, 2004
...or hockey quarterfinals? They are 7 games as well. - Anonymous
October 21, 2004
What about HHH TTTT? That makes it 1/64 for evenly matched teams. - Anonymous
October 21, 2004
And reading "The Red Sox were the 26th team in Major League history to fall behind 3-0 in a seven-game series." from MLB.com will make it seem like even less impressive. - Anonymous
October 21, 2004
It troubles me greatly when people who admittedly know next to nothing about sports apply the kind of logic they ARE familiar with to sound reasonable. Don't try to sound smart by making something else sound dumb...especially something that's so extremely important to a great many people.
By the way, baseball attracts statisticians because of its complexities, not its simplicities.
Oh yeah, and GO CARDS! - Anonymous
October 21, 2004
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October 21, 2004
You do realize <a href="http://sports.espn.go.com/mlb/news/story?id=1778397">Boston</a> has the second highest payroll in baseball, no? - Anonymous
October 21, 2004
Alright, HTML doesn't exactly work.
The link: http://sports.espn.go.com/mlb/news/story?id=1778397 - Anonymous
October 21, 2004
OK, suppose the odds are 60-40 in favour of the highest paid team vs the second highest paid team.
Then the odds of a T T T H H H H series are... 1 / 181 instead of 1 / 128.
Big whoop. My victory over Iceland is still WAAAY more impressive at 1/216. - Anonymous
October 21, 2004
Oops I meant to say good luck with those Mariners, not Marines. Although I guess Marines apply to your little fantasy plastic toy guys too.
And Tad: You tell us, what is the difference in payroll for the Yankees and Red Sox?? Hmmm? It's about as big as some teams entire payroll!!!! - Anonymous
October 21, 2004
Eric you are not factoring in "The Curse"!!! - Anonymous
October 21, 2004
I'm going to have to go with about 57 million, which isn't that large considering Boston also has a 24 million dollar advantage on the third place team. And Boston is still going to have well over double the team payroll of whoever it plays in the WS. - Anonymous
October 21, 2004
You are not factoring in the curse either!! - Anonymous
October 21, 2004
Well we need to get this right. Exactly how much is the curse worth? 68 Million? - Anonymous
October 21, 2004
You're totally disregarding the psychology part of it. You'd be surprised at how many teams even get to game 5 after being down 3-0. - Anonymous
October 21, 2004
And assuming that a dollar spent equals a dollar in talent, the Yankees would only have 54-46 advantage. Anyway, time to go home. - Anonymous
October 21, 2004
Besides, we always herald the first, and this is a first in MLB. I'm sure there's been plenty of cases of someone rolling 3 6's to win as an underdog (heck, it probably happens every game). - Anonymous
October 21, 2004
Really? I'd be surprised? OK, I'll bite. What percentage of teams that are down 3-0 end up down 4-0?
If it were coin flipping, I'd expect it to be 50%. Given that being down 3-0 is evidence of being the worse team, in reality I'd expect it to be slightly lower.
I'd be surprised if it were, say, 25%. - Anonymous
October 21, 2004
Whether this was your intention or not, it would seem that your analysis is a classic proof by contradiction that Baseball is much more than coin flipping. - Anonymous
October 21, 2004
If I'm reading this correctly..
"The Red Sox were the 26th team in Major League history to fall behind 3-0 in a seven-game series. They are just the sixth to avoid a sweep, and now just the third to win two straight after dropping the first three."
I'd say that 20-26 teams in baseball lost 4-0 after going down 3-0. - Anonymous
October 21, 2004
So it's about 20%. I am surprised!
:) - Anonymous
October 21, 2004
Yep, I think he was feeling feisty and just wanted to post a major troll. The "modest proposal" should have given it away.
Score : -1, Flamebait - Anonymous
October 21, 2004
I'm shocked, shocked that you'd think I'd do such a thing. Shocked! - Anonymous
October 21, 2004
:D - Anonymous
October 21, 2004
The reason why no team had been down 0-3 and then won 4 straight games is because the team that won the first 3 games is a lot better, so of course they will almost always win 1 of the remaining 4 games. You need to make your coin weighted on one side to have the analogy hold. ;) - Anonymous
October 21, 2004
If only 25 people playing ball were as reliable as flipping coins... - Anonymous
October 21, 2004
If you take into consideration W vs. L over the course of the season, the Yankees are approximately 2 percent better than the Red Sox (ultimately W vs. L is a better gauge of a team's prowess than salary). So if the teams play any seven games together in a closed environment (not counting previous years because the teams were different back then) it should be split 3-4 one way or another.
That being said, numbers lie and Damon is my boy. - Anonymous
October 21, 2004
Well, sure, but that's not the issue -- given a 3-4 split, there are 35 equally likely ways to distribute the wins and losses.
Which raises an interesting point. Winning a series with HTHHTTH is every bit as unlikely as TTTHHHH, but no one ever makes a big deal out of it.
People make big deals out of things that seem particularly dramatic, not particularly rare. - Anonymous
October 21, 2004
you should check out a wonderful novel by robert coover titled "universal baseball association, j. henry waugh, prop." it's a hilarious exploration of almost this exact idea (and provides nice parallels to the job of the fiction writer). - Anonymous
October 21, 2004
> The odds of Boston flipping T T T and then coming back to win with H H H H are one in 128.
This is not true. Since the series ends when one team reaches 4 wins, there are not 2^7=128 valid combinations of wins and losses. In fact, there are only 50 valid combinations -- 1 way to win a series in 4 games, 4 ways to win in 5 games, 10 ways in 6 games, and 35 ways in 7 games. So, you'd expect 1 occurrance every 50/4 = 13 years. It is surprising then that we don't see it more often. - Anonymous
October 21, 2004
The comment has been removed - Anonymous
October 21, 2004
Ah, yes. Not all outcomes are equally likely. - Anonymous
October 21, 2004
Right. Chance of HHHH is 1/16, chance of HHHTH, HHTHH, HTHHH, THHHH is 4/32 = 1/8, so you can't count the four-game series and the five-game series as equally likely. - Anonymous
October 21, 2004
Eric,
I see noone has commented on you playing Risk.
You really need to acquaint yourself with http://www.boardgamegeek.com ( start with http://www.boardgamegeek.com/top10.htm ) and play some real board games.
;-) - Anonymous
October 21, 2004
<i>Which raises an interesting point. Winning a series with HTHHTTH is every bit as unlikely as TTTHHHH, but no one ever makes a big deal out of it.</i>
That's because people are perhaps unconciously trying to be statisticians.
This is another aspect of the difference between coin-flipping and baseball. It's already been mentioned that the odds of a Yankees win is any one game is not necessarily 50-50. But more than that, the odds of a Yankees win in any one game is largely unknown. (Certainly not well-defined, at any rate.)
So imagine you're a baseball fan, and you've just watched three games of the Yankees-Red Sox series. Since you're human, you want to be able to predict future events. So you try to guess who's going to win the series, and most likely you try to do so by unconciously estimating the odds that, say, the Yankees will win game four based on what you've seen in games one through three.
The moment you first try to estimate the probability of an event based on past trials, you've entered the realm of statistics. And given the fact that the Yankees have won the first three games, it <i>is</i> statistically valid to infer that the probability of the Yankees winning an individual game is likely higher than 50%. Not much higher...three data points make a poor statistical universe...but likely higher nonetheless. It would be much less valid to make the same inference if the Yankees only win one or two of the first three games.
So if the series ends up the way it did, it will be more surprising to the average fan than if it ends up HTHHTTH. But with good reason: the final outcome was not in agreement with the statistical expectation formed by the first three games. We don't see this effect with coin flips because the probability of heads vs. tails in a coin flip is known from the outset and thus statistics (as opposed to plain old probability) never comes into play. - Anonymous
October 21, 2004
No formatting tags, gotcha. - Anonymous
October 21, 2004
Basketball also has 7-games series in its playoffs and no team has come back from an 0-3 deficit. IIRC from the TV broadcast, there have been something like 320 cases where a team has gone 0-3 and only 3 have won the next 4 games. - Anonymous
October 21, 2004
We were eating lunch, when the restaurant owner, a gambling fan, told us about his latest trip to the casino. He laughed when he told us about a "chump" who, playing roulette, bet on red again after red had already come up three times: "stupid guy, what are the chances of red coming up four times in a row?" Wanting to avoid future food poisoning, we nodded in agreement ... - Anonymous
October 21, 2004
Ahh yes, the gamblers fallacy. - Anonymous
October 21, 2004
Actually there are 15 7-game series per year in hockey -- every playoff series is a best of 7, and they start with 16 teams.
15 hockey and 3 baseball; I don't know the basketball playoffs (whether there is more than 1 7-game series), so assume 1 for that.
128 / (15 + 3 + 1) = ~6.75.
Since there hasn't been such an upset every 7 years, we can safely assume that they're not flipping coins. - Anonymous
October 21, 2004
Basketball has 7 7-game series (the first round, like the first round in baseball is 5 games, but basketball has a 16 team 1st round while baseball has a 8 team 1st round)
so...
64 (HHHTTTT + TTTHHHH) / (15 + 3 + 7) = 2.37
'nuff said - Anonymous
October 21, 2004
What would REALLY be interesting is if someone won 7 games emerging from a 6 game Deficit.
Think about THAT. - Anonymous
October 21, 2004
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October 22, 2004
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October 22, 2004
The comment has been removed - Anonymous
October 22, 2004
This topic should make some interesting contributions to "Riddle Me This, Google: Part Three". - Anonymous
October 22, 2004
One certainly hopes so.
Between May and August of this year I was averaging 200-300 Google hits a day.
It was over 600 yesterday, and some of those have got to be gems. - Anonymous
October 22, 2004
I sense a great disturbance in the force... as if billions of Statistics professors cried out at once, and then fell silent. Your attempts to analyze a sports event using this statistical model is flawed at its core:
1. In statistics, we assume that a coin toss has exactly a 50% chance of each outcome. This is not true in a sports game.
2. In statistics, a series of coin tosses is "memory-less". The coin doesn't remember what happened the last time it was tossed, so it has the same 50% chance of each outcome on every toss. Humans do have memory.
Those are the two mathematical errors in your model. There are also errors in your analysis, such as the fact that there are far more 7-game series than you had mentioned, but that's not a mistake in the model; only in its validation.
In summary: never trust the words of a man who writes in a purple font... - Anonymous
October 22, 2004
Flawed at its VERY CORE, eh?
Your first complaint is the same one I addressed forty comments ago. Don't like the odds? Pick different odds. Weight the coin differently. The chances of getting a TTTWWWW series change, but not by much.
Your second point is also easily dispensed with. Suppose for instance a team has a .5 chance of winning the first game, and then a .75 chance of winning the next game if they won the previous game. In such a system, the chances of a TTTHHHH series go UP, not DOWN! Pick any conditional probability model you want, it's only going to make such events more common.
No, the real flaw is in the validation. The model is insufficiently complex to explain on probabalistic grounds why there have been so few TTTWWWW series given that there are more seven game series than I stated.
We could come up with a new model in which all these concerns were met. Divide all pairings of teams into "thoroughly outclassed" and "about even".
In the vast majority of "thoroughly outclassed" matchups, one team has such a higher probability of winning that four-game series are practically inevitable.
But in a few "about even" matchups, we should expect to see a proper coin-flipping distribution. About 1/8 of them should be sweeps, about 1/64 of them should be three losses followed by four wins, etc.
As argued in the comments above, clearly Boston and the Yankees were in the "about even" category. - Anonymous
October 24, 2004
The comment has been removed - Anonymous
October 24, 2004
In the spirit of this thread, here's an interesting logical/statistical conundrum: Suppose I offered you to choose one of two identical sealed envelopes, telling you that both contain an unspecified amount of money, but that one contains double the amount of the other. Obviously the choice is random.
Now suppose that before you open the envelope I give you the chance to exchange it. If we assume your envelope contains X $, then the other envelope contains either half of X or double X. So, you’ve got a 50% chance of loosing 1/2 X, and a 50% chance of gaining X. So it would seem that contrary to reason, it makes sense to exchange the unopened envelopes.
And say I offered you the chance again, and again, and again. It would seem that we would stand there forever, exchanging the unopened envelopes. What’s wrong here? - Anonymous
October 25, 2004
This is a fairly well-known paradox. Google "Newcomb's Paradox" for another.
The resolution of this paradox lies in the fact that you're using "X" to mean two different things in one sentence. In one half of the sentence, X represents the larger amount, and in the second half it represents the smaller amount, so of course they're different.
Stop using "X" and start using dollar amounts and the paradox goes away. - Anonymous
October 25, 2004
But what if instead of picking heads or tails, Boston and New York each have to pick a sequence of three outcomes (e.g. HHT), and New York has to pick first, and whoever's sequence appears first wins ...
:) - Anonymous
October 25, 2004
Is this what you C&O people do for fun? - Anonymous
October 25, 2004
The Envelope Paradox gets into knots because of the probabilities. So let's state it without probabilities.
Statement 1: Let the amount in your envelope be X. You either gain X by switching, or lose 1/2 X by switching. Since it's obviously better to gain X than lose 1/2 X, you should switch.
Statement 2: Let the amount in the smaller envelope be Y. You either have Y or 2Y. If you switch, you might gain Y or you might lose Y. There's no benefit to switching.
Clearly at least one of these statements is wrong because they contradict each other. But now the misdirection is obvious:
"You either gain X (=Y) by switching, or lose 1/2 X (=Y) by switching."
X and X/2 can't both be Y in the same sentence! - Anonymous
October 25, 2004
Eric, one of the better answers that I got. Good for you! - Anonymous
October 26, 2004
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October 27, 2004
I don't think Mr. Lippert explained the envelope problem.
> "You either gain X (=Y) by switching, or
> lose 1/2 X (=Y) by switching."
> X and X/2 can't both be Y in the same
> sentence!
True but so what? The "quoted" sentence has this meaning:
"Either X = Y or X/2 = Y. If you open the envelope then you know the value of X but you don't know the value of Y until later.[] You either gain X (and later find out that X = Y) by switching, or lose X/2 (and later find out that X/2 = Y) by switching."
The paradox is not in that explanation.
The actual explanation was determined by some participants in rec.puzzles a few years ago. I almost remember enough math to understand it. There is no such thing as a uniform distribution over the set of integers.
I can add a bit to the explanation. Pretend that the assumed probability distribution (uniform over the set of integers) were possible. Your expected gain is infinity divided by infinity, or infinity minus infinity, however you want to express it.
[ Yes, part of the apparent paradox is that you don't have to open the envelopes in order to decide that you want to switch back and forth an infinite number of times. Nonetheless it is true that IF you open your envelope THEN you will know X while still not knowing Y. This still explains why the possibility of either X matching Y or X/2 matching Y is not a paradox.] - Anonymous
October 31, 2004
I would like to expand a bit on that. As already mentioned, the explanation that was determined by participants in rec.puzzles was: There is no such thing as a uniform distribution over the set of integers. An explanation of the explanation can be explained a bit better than I did.
Based on the fallacious assumption that there is a uniform distribution over the set of integers, one can compute that one's expected gain from switching is infinite. Similarly one can compute that one's expected gain from switching back to the original envelope is also infinite. Also the expected amount of money in both envelopes is infinite, and the expected gain is infinity minus infinity.
Actually the game can be simplified (though I haven't seen anyone do it). There is just one envelope. A probably malicious hacker persuades you that the amount of money was chosen at random from a uniform distribution over the set of all integers. You compute that the expected amount of money in the envelope is infinite. Watch out, the Sultan of Brunei and someone else might be tailing you.
This is not much different from fallacies involving division by zero. 10 == 20 therefore 1 == 2. Well, it was true that 10 == 20. - Anonymous
April 07, 2006
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