1. ## Coin Toss

Hi everyone

I have a question which I cannot put my finger on..

A symmetric coin with heads on one side and tails on the other side is tossed 491 times after one another. The total amount of times you get tails is either even or odd. Is the probability that you get an even amount of tails exactly 50%? And the question requires a strong mathematical evidence, e.g. a formula.

I'm thankful for all ideas!

2. List all the ways the Heads and Tails can occur. This is the expansion of
(H+T)^491.

The events with even Heads are those terms with H^(even power), in other words every other term.

You can verify that the sum of the even terms in the binomial is the same as the sum of the odd terms in the binomial by expanding (1-1)^491 = 0. Note that the (-1) keeps track of the even and odd terms.

So, yes, it is 50%.

3. thanks for the reply! but I believe that the question requires some kind of formula?

This is a formula that works though I don't know how to explain it properly.

2^(n-1) / 2^n In this case "n" is 491 so it becomes
2^(491-1) / 2^491

Trying to calculate 2^491 makes your calculator go overflow but if you try something smaller like.. n = 2

2^(2-1) / 2^2 = 2^1 / 4 = 2/4 = 0.5 = 50% for an even number of tails?

The formula is correct, but the problem is that I don't know how to explain the 2^(n-1). n-1 takes the previous toss and makes the odd number (491) an even number (490) but what else does it do??

4. Hello kaitoli
Originally Posted by kaitoli
Hi everyone

I have a question which I cannot put my finger on..

A symmetric coin with heads on one side and tails on the other side is tossed 491 times after one another. The total amount of times you get tails is either even or odd. Is the probability that you get an even amount of tails exactly 50%? And the question requires a strong mathematical evidence, e.g. a formula.

I'm thankful for all ideas!
The answer is Yes, it is 50%.

Proof

The probability that a given coin lands tails is $\displaystyle \frac12$, and the probability that it doesn't is $\displaystyle 1-\frac12=\frac12$. Therefore, using the Binomial Distribution Formula, the probability that we get exactly $\displaystyle r$ tails from $\displaystyle n$ tosses is:
$\displaystyle P(r) =\binom n r \left(\frac12\right)^r\left(\frac12\right)^{n-r}=\binom n r \left(\frac12\right)^n$
We must prove that $\displaystyle P(0)+ P(2) + P(4) + ... = \frac12$. So, we need to prove that, if $\displaystyle n$ is odd (and therefore if $\displaystyle n=491$):
$\displaystyle \binom n 0\left(\frac12\right)^n + \binom n 2\left(\frac12\right)^n+\binom n 4\left(\frac12\right)^n + ... +\binom{n}{n-1} \left(\frac12\right)^n = \frac12$
i.e. that:
$\displaystyle \binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1} = 2^{n-1}=S$, say
Now it is well known that:
$\displaystyle \binom n 0 + \binom n 1+\binom n 2 + ... +\binom{n}{n} = 2^{n}$ ...(1)
(The proof of this is to write down the Binomial Expansion of $\displaystyle (1+1)^n\;( = 2^n)$)

And also that:
$\displaystyle \binom{n}{r} =\binom{n}{n-r}$ ...(2)
Now, adding $\displaystyle S$ to itself, we get:
$\displaystyle 2S = \binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1}+\binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1}$
$\displaystyle = \binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1}+\binom n n + \binom{ n}{n- 2}+\binom{ n}{n- 4} + ... +\binom{n}{1}$, using (2)

$\displaystyle = 2^n$, from (1)
$\displaystyle \Rightarrow S = 2^{n-1}$
which completes the proof.

5. Thank you very much Grandad for your help and explanation of 2^n-1

I found everything great except a few things I didn't understand?

What is "S" supposed to be? An variable? It contains the 2^n-1 equation?

Where does the (1+1)^n (=2^n) come from?

Thanks once again!!

6. Hello kaitoli
Originally Posted by kaitoli
Thank you very much Grandad for your help and explanation of 2^n-1

I found everything great except a few things I didn't understand?

What is "S" supposed to be? An variable? It contains the 2^n-1 equation?
$\displaystyle S$ is just a convenient variable that I introduced to represent the sum $\displaystyle \binom n 0 + \binom n 2 + ... + \binom {n}{n-1}$. I wanted it so that I could write this sum down twice, before re-arranging into $\displaystyle \binom n 0 + \binom n 1 + ... + \binom {n}{n}$. Then I could divide by 2 to establish the result I wanted.

Where does the (1+1)^n (=2^n) come from?
This was just an aside, to show where the proof that $\displaystyle \binom n 0 + \binom n 1 + ... + \binom {n}{n}=2^n$ comes from. I needn't have mentioned it, except that you might have wondered how this well-known result is proved.

7. ok! Thanks for the great solution and explanations! I really appreciate it.

Thanks once again!

Hello kaitoliThe answer is Yes, it is 50%.

Proof

The probability that a given coin lands tails is $\displaystyle \frac12$, and the probability that it doesn't is $\displaystyle 1-\frac12=\frac12$. Therefore, using the Binomial Distribution Formula, the probability that we get exactly $\displaystyle r$ tails from $\displaystyle n$ tosses is:
$\displaystyle P(r) =\binom n r \left(\frac12\right)^r\left(\frac12\right)^{n-r}=\binom n r \left(\frac12\right)^n$
We must prove that $\displaystyle P(0)+ P(2) + P(4) + ... = \frac12$. So, we need to prove that, if $\displaystyle n$ is odd (and therefore if $\displaystyle n=491$):
$\displaystyle \binom n 0\left(\frac12\right)^n + \binom n 2\left(\frac12\right)^n+\binom n 4\left(\frac12\right)^n + ... +\binom{n}{n-1} \left(\frac12\right)^n = \frac12$
i.e. that:
$\displaystyle \binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1} = 2^{n-1}=S$, say
Now it is well known that:
$\displaystyle \binom n 0 + \binom n 1+\binom n 2 + ... +\binom{n}{n} = 2^{n}$ ...(1)
(The proof of this is to write down the Binomial Expansion of $\displaystyle (1+1)^n\;( = 2^n)$)

And also that:
$\displaystyle \binom{n}{r} =\binom{n}{n-r}$ ...(2)
Now, adding $\displaystyle S$ to itself, we get:
$\displaystyle 2S = \binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1}+\binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1}$
$\displaystyle = \binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1}+\binom n n + \binom{ n}{n- 2}+\binom{ n}{n- 4} + ... +\binom{n}{1}$, using (2)

$\displaystyle = 2^n$, from (1)
$\displaystyle \Rightarrow S = 2^{n-1}$
which completes the proof.

If it's okay, I would like to understand some things about the formula even better.

What is the general formula called? Does it have a special name?

$\displaystyle P(0)+ P(2) + P(4) + ... = \frac12$ This means that we have to prove that every even toss up to 490 has the probability of 1/2?

Where does this formula come from and how do you calculate it?:

And I was also wondering if you could help me explain 2^n-1 compared to a probability formula which is:

P(Tails becomes even) = Number of outcomes that leads to an even number of Tails / Number of total outcomes.

In my formula 2^n-1 is the "Number of outcomes that leads to an even number of Tails but how can I prove that 2^n-1 has this position?

Thanks!

9. Hello kaitoli
Originally Posted by kaitoli

If it's okay, I would like to understand some things about the formula even better.

What is the general formula called? Does it have a special name?
The formula I used is the Binomial Distribution Formula. If an experiment has a probability of success of $\displaystyle p$, and the experiment is repeated $\displaystyle n$ times, then the probability that there will be $\displaystyle r$ successes out of $\displaystyle n$ is:
$\displaystyle P(r)=\binom nr p^r(1-p)^{n-r}$
where $\displaystyle \binom nr =\frac{n!}{r!(n-r)!}$
$\displaystyle P(0)+ P(2) + P(4) + ... = \frac12$ This means that we have to prove that every even toss up to 490 has the probability of 1/2?
Not quite; it means that we have to prove that the sum of the probabilities of every 'even' toss up to $\displaystyle 490$ is $\displaystyle \tfrac12$. In fact, I proved it in the general case for any odd $\displaystyle n$, with the 'even' tosses from $\displaystyle 0$ to $\displaystyle (n-1)$.

Where does this formula come from and how do you calculate it?:

I am just re-stating what we need to prove here, using the formula for $\displaystyle P(r)$ that I wrote down earlier.
And I was also wondering if you could help me explain 2^n-1 compared to a probability formula which is:

P(Tails becomes even) = Number of outcomes that leads to an even number of Tails / Number of total outcomes.

In my formula 2^n-1 is the "Number of outcomes that leads to an even number of Tails but how can I prove that 2^n-1 has this position?
I think you need to study how the Binomial Distribution Formula is derived. I Google'd it and found this. I'm sure there are lots more.