MHB What is the Probability of Getting an Even Number of Heads with an Unfair Coin?

  • Thread starter Thread starter veronica1999
  • Start date Start date
AI Thread Summary
The discussion focuses on calculating the probability of obtaining an even number of heads when tossing an unfair coin with a 2/3 chance of heads, over 50 tosses. The binomial probability formula is applied, leading to the conclusion that the probability of getting an even number of heads is approximately 0.5. The participants explore various mathematical approaches, including the use of the binomial theorem and complementation rule, to derive the formula for the probability. Ultimately, the probability is expressed as P(X) = 1/2(1 + 1/3^50), indicating it is slightly greater than 1/2 due to the nature of the coin. The problem is recognized as interesting and prompts further generalization for different probabilities.
veronica1999
Messages
61
Reaction score
0
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...:confused:
 
Mathematics news on Phys.org
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?
 
Re: coins

MarkFL said:
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 :o
 
Last edited:
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:

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

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

Adding, we find:

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

and so finally, we have:

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

MarkFL said:
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:

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

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

Adding, we find:

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

and so finally, we have:

$\displaystyle P(X)=\frac{1}{2}\left(1+\frac{1}{3^{50}} \right)$
Thanks!
Now it is absolutely clear.:D
 
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.

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

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

Adding, we find:

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

$\displaystyle P(X)=\frac{1}{2}\left(1+(2p-1)^{2n} \right)$
 
Last edited:
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top