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 019.
From Steven Krupin's email to Dan:
I figured that even with the worst team in the bigs, this 019 record couldn't have been an easy task. So I asked my cousina Ph.D. who works as an economist at the Department of Labor and moonlights as a statistics analyst/columnist on ESPN.com and Baseball Prospectusto crunch some numbers. He and his mathematician friends figured out that the odds of my going 019 this year at Nats Park were 1 in 131,204.
By
Alexa Steele

October 8, 2009; 12:15 PM ET
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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 02.
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 190 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 3348 at home
Posted by: Sec3mysofa  October 8, 2009 2:20 PM  Report abuse
I was 1012 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 3348 = ~~ .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 81 !
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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Is Steinberg hurting for clicks or something?