Question

Probability question for you guys! -

assuming a fair coin, how many times do I have to flip the coin to have a 99% certainty of getting at least 1 heads?

I have a feeling I do this by multiplying the chances of a tails over and over again until it gets smaller than .01? Is this right?

Answer 1

I think its 1/\[(\frac{ 1 }{ 2 })^7\] which is .007

Answer 2

I think the reasoning is that probability of a heads = 1- probability of a tails... this make sense?

Answer 3

that looks good to me
i think you are solving
\[\left(\frac{1}{2}\right)^n\leq .01\] right?

Answer 4

yes u need to apply log in both sides

Answer 5

basically! My actual problem involves a biased coin, with 5/9 chance of heads. So I figure if chance of tails is 4/9, then 4/9 ^6 = .0077 chance of tails or a 99.3% chance of heads.

Answer 6

u will get 6.6 that is close to 7 like u said

Answer 7

but the method was the same.

Answer 8

Ah, I cheated and just started raising it to powers. Do you think on a question like this I should use a log? It seems like since flips are in whole numbers I would have to round up anyway. But if the correct way to do this is have the answer as a decimal I can do that. Recomendations?

Answer 9

i would guess and check

Answer 10

after all, you can only flip a coin an integer number of times

Answer 11

you could say
\[x\leq\frac{\ln(.01)}{\ln(.5)}\] if you like

Answer 12

I appreciate you trying to help me, I guess Im dense enough to not understand the need to do that. Im guessing x is the value Im looking for, the number of flips needed to get a 99% chance of heads. Why are we doing ln(.01/ln(.5) ? I kind of understand the values used. .01 being the chance I need and .5 being the chance of a heads or tails.

Answer 13

Just plugged that into wolfram alpha, and get 6.64.. So I see that it worked, it matches with my guess and check of 7. Just kind of curious where the ln/ln came from. If you have a moment to explain.

Answer 14

to solve
\[b^x=y\] you can use
\[x=\frac{\ln(y)}{\ln(b)}\]

Answer 15

so to solve
\[\left(\frac{1}{2}\right)^n\leq .01\] you can use
\[x\leq \frac{\ln(.01)}{\ln(.5)}\]

Answer 16

oooooooh ok, duh, thats very clear, thanks for explaining heh, I was baffled :D

Answer 17

this is sometimes called the "change of base" formula

Answer 18

yw

Answer 19

Right, sad to say Im familiar with it, just didnt even see it coming. /sigh hehe