#1
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Winning 1 coinflip out of 16?
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. |
#2
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Re: Winning 1 coinflip out of 16?
Winning exactly 1 coinflip is 1/4096. Winning 1 or less is 1/4096 + 1/65536 = 17/65536. So yeah, tough luck.
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#3
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Re: Winning 1 coinflip out of 16?
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.. |
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