What are the odds that I'm such a loser?
May 26, 2006 7:47 PM Subscribe
Probability filter: How do you determine the odds of being extremely unlucky?
I know I've seen this explained, but my google-fu fails me, as does my brute-force mathematical figuring. Here's my problem -- if there's a game in which you're supposed to win 12% of the time, what are the odds that after 150 plays you only win 3 times (i.e. 2% of the time)?
I know I've seen this explained, but my google-fu fails me, as does my brute-force mathematical figuring. Here's my problem -- if there's a game in which you're supposed to win 12% of the time, what are the odds that after 150 plays you only win 3 times (i.e. 2% of the time)?
The exact answer is
129105822506510328235079035870803502997446159322335938209859903517778016592610656673611677829545693385926686940661856618755002678254299558255332801294973419393667922397147762715322224910896287172213604352
over
19636373861190906212383087819945102571900862049997798260446196468748421018886863425840391536149327809108734286260527514052671674441698875611230059296435853809012421857839447714155767243937589228153228759765625
posted by iconjack at 4:29 AM on May 27, 2006
129105822506510328235079035870803502997446159322335938209859903517778016592610656673611677829545693385926686940661856618755002678254299558255332801294973419393667922397147762715322224910896287172213604352
over
19636373861190906212383087819945102571900862049997798260446196468748421018886863425840391536149327809108734286260527514052671674441698875611230059296435853809012421857839447714155767243937589228153228759765625
posted by iconjack at 4:29 AM on May 27, 2006
Ugh. Neither of those appears to be repeating...
Google says the answer is 6.57483013 × 10-6.
posted by klangklangston at 6:47 AM on May 27, 2006
Google says the answer is 6.57483013 × 10-6.
posted by klangklangston at 6:47 AM on May 27, 2006
Response by poster: Thanks guys. I've also found this binomial distribution calculator, which may be helpful to anyone else interested in this question:
Binomial Disribution Utility (for Bernoulli Trials)
posted by TonyRobots at 6:02 PM on May 27, 2006
Binomial Disribution Utility (for Bernoulli Trials)
posted by TonyRobots at 6:02 PM on May 27, 2006
I just thought it would amusing to give the exact answer, which I did in my post above. The approximate answer is 6.57483013 × 10^-6.
posted by iconjack at 6:27 PM on May 27, 2006
posted by iconjack at 6:27 PM on May 27, 2006
This thread is closed to new comments.
p = 0.12
n = 150
x = 3
posted by MzB at 8:02 PM on May 26, 2006