Posted by AccountingWEB on 11/05/2009 - 13:41
The odds of attending 19 Washington Nationals baseball games last season and watching them lose all 19 times were 1-in-131,204. This fact is according to Sports Illustrated in their story about such a man who witnessed these 19 woeful games. This story brought to mind a baseball math challenge that we had several months back.
Back to our Nationals challenge, what if the guy had seen 30 games where the Nationals lost them all? Still pretty good odds given how bad they were! What’s the answer? (The Nationals were 59-103).
The first ten people who post solid answers (not necessarily correct answers) will receive AccountingWEB baseball caps.
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The correct answer is...
The correct answer to the odds of seeing 30 Nats losses out of 30 home games attended would be 1:1,927,524,580 (or roughly 1 in 2 billion). Nats were 33-48 at home, so probability of the first game being a loss is 48/81. Probability of second game attended also being a loss is 48/81 * 47/80. Follow the pattern and you get the SI answer for 19 losses as well as my answer for 30.
Ivan Flaugh from International Falls, Minnesota
Another math challenge-what are the odds?
My answer is 1 in 794,939.
The percentenge of seeing a loss for each game is 103/162=63.58%.
I took .6358 to the 30th power and my result was rounded to the
nearest integer 794,939 from there the result 1 in 794, 939.
My result is assuming the result in each game is independant. Some
of the other results I see posted assume that there is no replacement
in the set of 162 games but I see this situation as being different than
picking marbles out of the jar since each marble taken out of the jar reduces
the number of marbles in the jar so each result is not independant of the
result of the previous draw.
A further refinement might be spliting the probablities of losing at home from
losing on the road. I thjink that 19 game fan only attended home games.
I saw the Nationals lose on the road when I wnt to their game in Houston right
before All-Star break. 1 game and 1 loss
answer to baseball question
I would say 1 in just under 5 million based on numbers crunched in excel!
Math Challenge
Simple formula from a simple mind!
30/19 = 1.578947
1.578947 X 131,204 = 207,164
Chances of attending 30 losing games is 1 in 207,164.
Math Challenge
Yes, I agree with David's answer of approx 1 in 5 million. However, the original published answer, 1-in-131,204, is wrong. The correct answer is 1-in-10,677.
There are 162 ways to pick the first game, 161 to pick the second, ..., 144 ways to pick the 19th game....133 ways to pick the 30th game There were 103 loses, so to pick the first loss, there are 103 ways, 102 for second loss ... 85 ways to pick the 19th loss...74 ways to pick the 30th loss.
It's worked (for we spreadsheet/detail people) as follows:
103/162 = 63.58% probability of picking the 1st loss
102/161 = 63.35% * 63.58% = 40.28% probability of picking the 2nd loss
101/160 = 63.13% * 40.28% = 25.43% probability of picking the 3rd loss.........
85/144 = 59.03% * .016% = .009366% probability of picking the 19th loss 1/.009366% = 10,677 or 1-in-10,677
74/133 = 55.64% * .000036% = .00002% probability of picking the 30th loss 1/.0000201% = 4,963,809 or 1-in-4,963,809
-- P Martin
Answer
1 in 2 times.
- Tommie Williams
I think that the article is incorrect
If you assume that any set of 19 games out of the 162 has the same chance of being chosen, the number of games out of the 19 that are losses has the hypergeometric distribution. Here's the calculation using R (famous statistical software):
> dhyper(0,59,103,19)
[1] 9.365583e-05
> 1/dhyper(0,59,103,19)
[1] 10677.39
It should be odds of 10676 to 1 against.
For attending 30 games, the answer is 4963808 to 1.
> dhyper(0,59,103,30)
[1] 2.014582e-07
> 1/dhyper(0,59,103,30)
[1] 4963809
- Tom
Math Challange
Assuming their chance of losing is constant throughout the season, the probability of losing 1 game is 103/162 = .6358. The probability of going to 2 games and seeing 2 losses (and no wins) would be .6358^2 = .4042. The probability of going to 19 games and seeing 19 losses (and no wins) would be .6358^19 = .000183269 or 1 in about 5,456 (so I disagree with SI's figure!) The probability of going to 30 games and seeing 30 losses (and no wins) would be .6358^30 = .000001257 or 1 in about 795,031. Because the probability of losing is so high (almost 2/3), it's not too unlikely to see that many losses with no wins. If you flipped a coin 30 times, the probability of see 30 heads would be .5^30 = 9.313x10-10 or 1 in more than a billion! Order doesn't matter since all of the games you saw (or coin flips) must be losses (or heads). The total number of games doesn't matter - just randomly pick 30 and see if they were losses.
-- Lyle Rupert
Math Challenge
Odds of going to exactly one game and seeing them lose it is 103/162 = 65.6%
Odds of going to exactly two game and seeing them lose both is 103/162 * 102/161 = 40.3%
Odds of going to exactly 30 games and seeing them lose all thirty is (103!)(162-30)! / (103-30)!162! = 2.0146 x 10^(-7) or about 1 in 5 million....maybe. Go Phillies! Maybe next year.
David D. from Little Rock, AR