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Bog: The Unluckiest Fan

Dan Steinberg's DC Sports Bog tells the tale of a fan who attended 19 Nats games, for which the team managed a record of 0-19.

From Steven Krupin's e-mail to Dan:

I figured that even with the worst team in the bigs, this 0-19 record couldn't have been an easy task. So I asked my cousin--a Ph.D. who works as an economist at the Department of Labor and moonlights as a statistics analyst/columnist on and Baseball Prospectus--to crunch some numbers. He and his mathematician friends figured out that the odds of my going 0-19 this year at Nats Park were 1 in 131,204.

By Alexa Steele  |  October 8, 2009; 12:15 PM ET
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Is Steinberg hurting for clicks or something?

Posted by: NatsNut | October 8, 2009 12:23 PM | Report abuse

On the other hand, I bought one of those "Buy Four, Get One Free" game packs. And I saw the Nats go 3 and 2 in those five. Can't say I'm dissatisfied.

Posted by: MikeH0714 | October 8, 2009 12:38 PM | Report abuse

Guess I'm lucky, I traveled over a thousand miles round trip for two games and was 0-2.

Posted by: cokedispatch | October 8, 2009 1:01 PM | Report abuse

1 in 131,204? Not too shabby.

For comparison, hitting the Jackpot on the D.C. "Hot Lotto" game: 1:10,939,383

Posted by: ihatewalks | October 8, 2009 1:14 PM | Report abuse

OK, but what was the probability of seeing the Nats go 19-0 at Nats Park this season?

Did anyone do that?

If so, let's chip in and get them a full plan for 2010.

Posted by: KenNat | October 8, 2009 1:43 PM | Report abuse

When I read this, I just assumed Steinberg was continuing my comment from the Nats Journal...

"Did anyone else see young Lerner in the stands at the Yankees game last night?

Get a good look at playoff baseball young Lerner."

Posted by: dclifer97 | October 8, 2009 2:08 PM | Report abuse

they were 33-48 at home

Posted by: Sec3mysofa | October 8, 2009 2:20 PM | Report abuse

I was 10-12 this year, and was there for every delicious victory in the sweep over that stinking team from Miami.

Posted by: dand187 | October 8, 2009 3:03 PM | Report abuse

OK, so if I'm doing this right:
they were 33-48 = ~~ .407 winning %
That's the odds of going to one game they won.
Odds of going to 2 games, and they win both
= .407 x .407 = .1656
So the odds of going to exactly 19 games, all of which they win, should be
.407 x .407 ... nineteen times, or about
.00000003894, or about 1 in over 25.6 million.

The odds of seeing a .500 team (or a coin toss) would be about .000001907, or a little over 524,000 to 1.

Stats geeks -- did I do that right?

Posted by: Sec3mysofa | October 8, 2009 3:10 PM | Report abuse

You are correct. My stats class this semester taught me that! Guess I have learned something in college after all...

Posted by: tengoalyrunr30 | October 8, 2009 7:19 PM | Report abuse

Ta, tengo. Now can you explain how the statistician in the bog post got the larger figure, instead of the ~20,000 to 1 that this method gives? They never explained that.

Posted by: Sec3mysofa | October 8, 2009 7:37 PM | Report abuse

OK, somebody did explain it. Makes sense. I should have figured this out.
I guess I'll take a crack at explaining the stats as I understand them.

The 1 in 130,000 number comes from the idea that you are picking the 19 losses out of the 81 total games the Nats played at home this year. Imagine that there are 81 cards face down on a table in front of you, 33 of them are wins, and 48 of them are losses. After you pick a card you can't pick it again, so as you pick each loss, picking a loss becomes harder. After 1 game there are 33 wins and 47 losses available. After 2 games there are 33 wins and 46 losses. And this continues until the odds are actually against you picking a loss, since after 16 games the available choices are 33 wins and 33 losses. So the chance of picking 19 losers isn't .593 to the 19th, instead you have to calculate the odds of getting each loss individually.

I wonder if that explanation helped anyone...

Posted by: Hobes | October 8, 2009 4:25 PM

Posted by: Sec3mysofa | October 8, 2009 7:46 PM | Report abuse

Of course, the "problem" with this is that it assumes the games are already lost, which isn't the case with games (more like coin flips than cards), so I dismissed the idea. But that is where the number comes from.

Posted by: Sec3mysofa | October 8, 2009 7:50 PM | Report abuse

Wait, that's backwards.
In other words, using the .407 win percentage assumes the games are already played -- as in the example, where the cards are already wins and losses when they are on the table.
So they are calculating the W/L percentage as they go, then, and basing the odds of the next game on the outcomes of the previous ones? That would seem to make sense.

Posted by: Sec3mysofa | October 8, 2009 7:56 PM | Report abuse

I was 8-1 !

Posted by: Sal143 | October 9, 2009 9:05 AM | Report abuse

I was 0 - 10 until the last game of the season. But I follow Steve Czaben's ALE (always leave early) Theory, and left in the 8th inning. They rallied and won in the 9th. (While I was listening on the radio). I guess I am now 0 -11?

Posted by: 1of9000 | October 9, 2009 10:35 AM | Report abuse

Ban this guy from the ballpark next year! LOL

Posted by: dmacman88 | October 9, 2009 2:09 PM | Report abuse

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