# Probability of flipping a biased coin k times

1. Feb 12, 2009

### mcnkevin

1. The problem statement, all variables and given/known data
I have an unfair coin that comes ups heads 2/3 of the time and tails 1/3 of the time. If i flip this coin k times, what is probability that tails came up more then heads. Assume k is odd. So basically, what is the probability that the number of tails > k/2?

2. Relevant equations

3. The attempt at a solution

I got a mental block right now. Obviously the probability will depend on k, and the larger the the lower the probability. Is it correct if I say the probability is (1/3)^(k/2+1)?

Thanks

Last edited: Feb 12, 2009
2. Feb 12, 2009

### Mezzlegasm

Sorry if this is a stupid question, but how can tails come up more than the times you flip the coin?

If you flip it k times, it can't come up more than k times.

3. Feb 12, 2009

### mcnkevin

it says k/2 times not k.

4. Feb 12, 2009

### Staff: Mentor

0

Mezzlegasm was right, something is wrong here. Should it be

If i flip this coin k times, what is probability that tails came up more then k/2?

5. Feb 12, 2009

### mcnkevin

Oh right, yes. I think i actually meant to say in that sentance, what is the probability that it comes up more then heads. Later i mention k/2. Sorry.

6. Feb 12, 2009

### D H

Staff Emeritus
Editorial note: This should say "what is probability that tails came up more then heads", not k. This is what led to Mezzlegasm's post.

No. This isn't even true for the trivial case, k=1, in which the probability is obviously 1/3. What is the probability for k=3? What kind of distribution is this?

7. Feb 12, 2009

### Mezzlegasm

Yeah, I missed the latter part because I was confused with the former.

Anyhow, I'm pretty sure that it is (1/3)^(k/2). I always had a hard time with counting principles and probability, so I need someone to confirm that for me as well.

EDIT: alright, I've just seen the above post. My answer didn't seem to make sense because it gave a different probability with more coin tosses, which shouldn't happen (I don't think). I couldn't find a single way to make it work, and the probability that it comes up more than heads seems to be a more appropriate question. Regardless, my answer is flawed.

8. Feb 12, 2009

### mcnkevin

Not sure what you mean by what kind of distribution? Basically, each toss is independant and each toss you have 1/3 of a chance of flipping tails and 2/3 a chance flipping heads. If you make k flips, whats the probability tails was flipped more times then heads.

For k = 3 , well you have to flip 2 tails in 3 tries.

So the possibilities are
All three tails = (1/3)(1/3)(1/3)
First two tails = (1/3)(1/3)(2/3)
First and last tails (1/3)(2/3)(1/3)
Second and last tails = (2/3)(1/3)(1/3)

So basically, (1/3)^3 + (3*((1/3)^2)*(2/3))

Correct? It certainly may be very wrong.. but if it is correct, how could i generalize this to k?

9. Feb 12, 2009

### Mezzlegasm

Ah alright, this is starting to feel familiar now.

Like you said there are different possibilities, so you have to add the different probabilities that have tails at the desired amount. I think this is in the right direction, someone correct me if I'm wrong. I'm not sure where to go now, but I believe it starts with

(1/3)^k +

then you would have to add the different possible probabilities when considering how many times you can flip heads and still have desirable results. I just don't know how to do this.