What are the odds that I'm such a loser?

May 26, 2006 7:47 PM

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)?

posted by TonyRobots to technology (7 comments total)

For the exact case (3 wins in 150 trials) you want the binomial distribution (or, M in N).

P(3 in 150)  =  ((150!)/((150-3)!*3!))  * 0.12^3 * (1 — 0.12)150 — 3 

I'll.. leave the math to you.
posted by neustile at 8:01 PM on May 26, 2006

two words: Binomial distribution
p = 0.12
n = 150
x = 3
posted by MzB at 8:02 PM on May 26, 2006

I screwed up the formatting. That last bit is (1-0.12)^(150-3). I tried computing it in bc and got 0.00000657, or 0.000657%
posted by neustile at 8:15 PM on May 26, 2006

The exact answer is

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

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

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