Flipping an unfair coin

veronica1999

Member
An "unfair" coin has a 2/3 probability of turning up heads. If this coin is tossed 50 times, what is the probability that the total number of heads is even?

I set up an equation but I am having trouble with the calculation.

50C0 X (2/3)^0 X (1/3)^50 + 50C2 X (2/3)^2 X (1/3)^48 +............ 50C50 (2/3)^50X(1/3)^0

I tried splitting it up thinking of the sigma notation.

(2/3)^0 + (2/3)^2 .....................+ (2/3)^50 = { 1- (4/9)^25} / (1- 4/9)

(1/3)^0 ......................... + (1/3)^50 = { 1- (1/9)^25}/ (1-1/9)

It doesn't seem to work...

MarkFL

Staff member
Re: coins

As you have realized, we need the binomial probability formula:

$\displaystyle P(x)={n \choose x}p^x(1-p)^{n-x}$

Identifying:

$\displaystyle n=50$

$\displaystyle p=\frac{2}{3}$

we then need to compute:

$\displaystyle P(X)=\sum_{k=0}^{25}P(2k)=\sum_{k=0}^{25}\left[{50 \choose 2k}\left(\frac{2}{3} \right)^{2k}\left(\frac{1}{3} \right)^{50-2k} \right]$

Relying on technology, we find:

$\displaystyle P(X)=\frac{358948993845926294385125}{717897987691852588770249}\approx0.5$

Can you think of a way to obtain the approximation from the binomial theorem and the complementation rule?

veronica1999

Member
Re: coins

As you have realized, we need the binomial probability formula:

$\displaystyle P(x)={n \choose x}p^x(1-p)^{n-x}$

Identifying:

$\displaystyle n=50$

$\displaystyle p=\frac{2}{3}$

we then need to compute:

$\displaystyle P(X)=\sum_{k=0}^{25}P(2k)=\sum_{k=0}^{25}\left[{50 \choose 2k}\left(\frac{2}{3} \right)^{2k}\left(\frac{1}{3} \right)^{50-2k} \right]$

Relying on technology, we find:

$\displaystyle P(X)=\frac{358948993845926294385125}{717897987691852588770249}\approx0.5$

Can you think of a way to obtain the approximation from the binomial theorem and the complementation rule?
There is i/2 and 1/2( 1+ 1/3^50) in the answer choices. Which one would be the answer?
i am trying to work backwards from the answer.

oops, i see the second one is the answer

Last edited:

MarkFL

Staff member
Re: coins

I see we aren't meant to either use technology, or use an approximation.

Intuition tells us that since there are 26 favorable outcomes and 25 unfavorable that the probability will be slightly greater than 1/2. But how much greater?

Perhaps if we rewrite the sum:

$\displaystyle P(X)=\frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]$

Now, let's let Y be the event of getting an odd number of heads, and write:

$\displaystyle P(X)-P(Y)=\delta$

$\displaystyle \frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]-\frac{1}{3^{50}}\sum_{k=0}^{24}\left[{50 \choose 2k+1}2^{2k+1} \right]=\delta$

$\displaystyle 3^{50}\delta=\sum_{k=0}^{50}\left[{50 \choose k}2^{k}(-1)^{50-k}\right]$

By the binomial theorem, we have:

$\displaystyle 3^{50}\delta=(2-1)^{50}=1$

and so:

$\displaystyle \delta=\frac{1}{3^{50}}$

Hence:

$\displaystyle P(X)-P(Y)=\frac{1}{3^{50}}$

$\displaystyle P(X)+P(Y)=1$

$\displaystyle 2P(X)=1+\frac{1}{3^{50}}$

and so finally, we have:

$\displaystyle P(X)=\frac{1}{2}\left(1+\frac{1}{3^{50}} \right)$

veronica1999

Member
Re: coins

I see we aren't meant to either use technology, or use an approximation.

Intuition tells us that since there are 26 favorable outcomes and 25 unfavorable that the probability will be slightly greater than 1/2. But how much greater?

Perhaps if we rewrite the sum:

$\displaystyle P(X)=\frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]$

Now, let's let Y be the event of getting an odd number of heads, and write:

$\displaystyle P(X)-P(Y)=\delta$

$\displaystyle \frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]-\frac{1}{3^{50}}\sum_{k=0}^{24}\left[{50 \choose 2k+1}2^{2k+1} \right]=\delta$

$\displaystyle 3^{50}\delta=\sum_{k=0}^{50}\left[{50 \choose k}2^{k}(-1)^{50-k}\right]$

By the binomial theorem, we have:

$\displaystyle 3^{50}\delta=(2-1)^{50}=1$

and so:

$\displaystyle \delta=\frac{1}{3^{50}}$

Hence:

$\displaystyle P(X)-P(Y)=\frac{1}{3^{50}}$

$\displaystyle P(X)+P(Y)=1$

$\displaystyle 2P(X)=1+\frac{1}{3^{50}}$

and so finally, we have:

$\displaystyle P(X)=\frac{1}{2}\left(1+\frac{1}{3^{50}} \right)$

Thanks!!!
Now it is absolutely clear.

MarkFL

Staff member
I found this problem quite interesting and thought about it while I was out and so wanted to generalize a bit and cut out some of the unnecessary hand-waving...

Let's say the probability of getting a heads is $\displaystyle p$ and we flip the coin $\displaystyle 2n$ times where $\displaystyle n\in\mathbb{N}$. What is the probability that the total number of heads is even?

Let $\displaystyle P(X)$ be the probability that the total number of heads is even and $\displaystyle P(Y)$ be the probability that the total number of heads is odd.

$\displaystyle P(X)+P(Y)=1$

$\displaystyle P(X)-P(Y)=\sum_{k=0}^{2n}\left[{2n \choose k}p^{2n-k}(p-1)^k \right]=(2p-1)^{2n}$

$\displaystyle 2P(X)=1+(2p-1)^{2n}$
$\displaystyle P(X)=\frac{1}{2}\left(1+(2p-1)^{2n} \right)$