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?

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?

2. mike says:

…or hockey quarterfinals? They are 7 games as well.

3. Pedant says:

What about HHH TTTT? That makes it 1/64 for evenly matched teams.

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.

5. Fryguy says:

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!

6. G says:

Yeah, I agree with Fryguy and feel the need to make a major correction here.

Why are you assuming the chance of any team beating any other team is 50%? The Yankees have a payroll that far and away outstrips ANY other team in MLB. It’s not like the NFL where the salary cap means that parity rules. In baseball, there is only the luxury tax and Steinbrenner has no problem paying it out due to crazy revenues from TV contracts, merchandising, etc. etc.

Remember also that the Sox were down 3-0 and in the 9th inning of game 4 they were within 3 OUTS of being eliminated!!

So, what are the chances of a team loaded with misfits will come from down 3-0 and losing in the 9th inning, to win over a button-down team of the best that money can buy? Compute that one buddy.

Oh wait, your smug "isnt sports so silly" attitude only reveals your shallow attempt to snub something you know nothing about. Stick with Technology… and good luck with those Marines (Baaahahahaha)

You do realize <a href="http://sports.espn.go.com/mlb/news/story?id=1778397">Boston</a&gt; has the second highest payroll in baseball, no?

Alright, HTML doesn’t exactly work.

9. Eric Lippert says:

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.

10. G says:

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!!!!

11. G says:

Eric you are not factoring in "The Curse"!!!

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.

13. G says:

You are not factoring in the curse either!!

Well we need to get this right. Exactly how much is the curse worth? 68 Million?

15. domovoi says:

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.

And assuming that a dollar spent equals a dollar in talent, the Yankees would only have 54-46 advantage. Anyway, time to go home.

17. domovoi says:

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).

18. Eric Lippert says:

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%.

19. Anonymous says:

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.

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.

21. Eric Lippert says:

So it’s about 20%. I am surprised!

🙂

22. G says:

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

23. Eric Lippert says:

I’m shocked, shocked that you’d think I’d do such a thing. Shocked!

24. G says:

😀

25. Mike Dunn says:

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. 😉

26. If only 25 people playing ball were as reliable as flipping coins…

27. Zube says:

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.

28. Eric Lippert says:

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_.

29. jeffcal says:

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).

30. Drew says:

> 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.

31. Eric Lippert says:

Drew’s analysis is plausible, but completely incorrect. Anyone else want to take a stab at explaining why, or shall I?

32. Drew says:

Ah, yes. Not all outcomes are equally likely.

33. Eric Lippert says:

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.

34. Anon says:

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.

😉

35. Orbifold says:

<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.

36. Orbifold says:

No formatting tags, gotcha.

37. Larry says:

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.

38. Dan Shappir says:

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 …

Ahh yes, the gamblers fallacy.

40. Eric says:

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.

41. sims says:

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

42. Jorge says:

What would REALLY be interesting is if someone won 7 games emerging from a 6 game Deficit.

43. Mat Hall says:

"What would REALLY be interesting is if someone won 7 games emerging from a 6 game Deficit."

The most impressive thing about that would be playing 13 games in a seven game series… 🙂

Unfortunately by default our brains are hopeless at stistics. The UK national lottery is a 6-balls-from-a-bag type deal, and if the sequence 1-2-3-4-5-6 came up people would be amazed, astounded, and possibly shouting "FIX!", but it’s no more unlikely than 7-18-24-29-35-39 and no-one’s going to be up in arms if THAT turns up. (Apart from me — if those are the winning numbers tomorrow I’ll be REALLY upset.)

I quite enjoy making people angry with the Monty Hall dilemma (Google for it if you’re not familiar with the principle). Some people just refuse to accept the answer matter how many decision trees you draw or practical demonstrations you do as it flies against common sense…

44. Fryguy says:

Ok, flip-heads. Given I spent the past two nights at Busch Stadium watching my Cardinals send the Astros back to their B-hive (by the way, you’d only understand what I’m talking about if you gave a crap about the game) and that in celebrating those two fantastic victories, I tried to do my part in reducing the surplus of Bud Light here in the St. Louis area, by what margin did I defy the odds of making it into work on time this morning?

45. Eric Lippert says:

> I see noone has commented on you playing Risk.

You’ll note that I did not mention WHEN I last played Risk. It was probably eight or ten years ago.

How is it then that I so clearly remember my blue armies in England crushing Iceland? Because my Risk strategy is:

* always play blue, and

* get into a huge, endless fight over Europe, sapping my strength until eventually the guy who’s held Russia/Asia for the last ten turns cashes in 120 armies worth of cards (card limits are for wimps!) and sweeps across Europe in a bloody rampage.

So at some point England ALWAYS crushes Iceland with triple sixes.

It’s not an _effective_ winning strategy, but it’s fun. Particularly if you can make a last stand on Iceland with a hundred armies or so.

46. Brian Butler says:

This topic should make some interesting contributions to "Riddle Me This, Google: Part Three".

47. Eric Lippert says:

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.

48. Oren says:

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…

49. Eric Lippert says:

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.

50. Anon says:

Eric,

<quote>

You’ll note that I did not mention WHEN I last played Risk. It was probably eight or ten years ago.

How is it then that I so clearly remember my blue armies in England crushing Iceland? Because my Risk strategy is:

* always play blue, and

* get into a huge, endless fight over Europe, sapping my strength until eventually the guy who’s held Russia/Asia for the last ten turns cashes in 120 armies worth of cards (card limits are for wimps!) and sweeps across Europe in a bloody rampage.

So at some point England ALWAYS crushes Iceland with triple sixes.

It’s not an _effective_ winning strategy, but it’s fun. Particularly if you can make a last stand on Iceland with a hundred armies or so.

</quote>

ROFL…

51. Dan Shappir says:

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?

52. Eric Lippert says:

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.

53. Zac says:

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 …

🙂

54. Eric Lippert says:

Is this what you C&O people do for fun?

55. Eric Lippert says:

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!

56. Dan Shappir says:

Eric, one of the better answers that I got. Good for you!

57. KC Lemson says:

I helped an elderly gentleman win thousands of dollars at a casino last year… I rolled double 6’s at one point. He looked me in the eye, said "I trust you, I know you’ll do it again" and put \$50 on it… I rolled it again. He put another \$50 on it, and I rolled it a third time. He collected his wad of dough and then left, thanking me – with words.

58. Norman Diamond says:

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.]

59. Norman Diamond says:

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. 1*0 == 2*0 therefore 1 == 2. Well, it was true that 1*0 == 2*0.

60. MSDN Archive says:

There is a better way to answer this than by trying to remember your statistics classes:  write a simulation.  Even highly educated people get all bent out of shape about the Monty Hall problem and its ilk, and are only convinced by writing programs to demonstrate the effects of random chance.

Given enough seasons, you should actually be able to come up with a better statistical model than any particular guesses (which are always colored by memory, themselves) would grant.  And given that model, you can find out where the cross-over point with the RISK scenario lies.  That is, if you can demonstrate that the model is reasonable.