Winning 1 coinflip out of 16?

[samr]

Took 16 coinflips today (pair v overs or overs v pair) and won one of them. What is the odds of that?

Is this an application of the Bernoulli theorem?

win pct: w = 0.5
lose pct l = 0.5
(w + l)^16, find 16 c 15 w^1 * l^15

= 16 * .5^1 * .5^15 = 2^4 * 2^-16 = 2^-12 = 1 in 4096.

[kyro]

Winning exactly 1 coinflip is 1/4096. Winning 1 or less is 1/4096 + 1/65536 = 17/65536. So yeah, tough luck.

[jba]

I would say it's the same as losing 16

or winning 16

or losing the first eight and losing the last eight

all of these are the same; of all the possible ways flipping a coin 16 times could result, there is only one that fits our description. in this case the answer is:

P=.5^16


EXCEPT: that in your problem there are 16 ways to do it (win the first, lose the next 15; lose the first, win the second, lose the rest, etc). You don't care which one you won.

P=.5^16*16=0.000244140625

1/4096

hey that's what you said


I guess I should go google this bernoulli punk..