# Basic coin toss problem to get 1 head, but with a different method

• Master1022
In summary, this formula is used to find the expected value of a discrete probability distribution, which cannot take negative values. The expected value is found by summing up all the probabilities in a triangle, which is rearranged so that the sum of the j-th row equals the j-th term in the sum of all the terms.
Master1022
Homework Statement
What is the expected number of tosses until we get
Relevant Equations
Expected value
Hi,

I was watching a video where the following formula, for expected value, was presented for a discrete probability distribution which cannot take negative values:
$$E[X] = \sum_{j = 1}^{\infty} P(X \geq j)$$
and I never saw this formula before and am trying to develop an intuition for it. I am trying to solve a basic problem with this method just so I understand how it works.

Question:
What is the expected number of times needed to flip a fair coin until we get 1 head?

Attempt:
So the term ## P(X \geq j) ## means that we didn't get a head in the first ## (j - 1) ## throws. Thus, ## P(X \geq j) = 1 - P(X \leq j - 1) = 1 - \begin{pmatrix} j - 1 // 1 \end{pmatrix} \cdot \left( \frac{1}{2} \right) ^{1} \cdot \left( \frac{1}{2} \right) ^{(j - 1) - 1} = \left( \frac{1}{2} \right) ^{j - 1} ##

Thus our summation becomes:
$$E[X] = \sum_{j = 1}^{\infty} \left( \frac{1}{2} \right) ^{j - 1} = 2 \cdot \sum_{j = 1}^{\infty} \left( \frac{1}{2} \right) ^{j} = 2 \cdot \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 2$$

Is that a correct way to use this different formula? Is there any intuition behind this formula?

sysprog
Yes, your application is correct. I can't follow your first line of formulas, which seems to have formatting problems, but your last line is correct, as is your conclusion.

For intuition, consider the following, where ##p(x)## is the probability that ##X=x##.
The expected value of ##X## is
$$\sum_{x=1}^\infty x\cdot p(x) =1\cdot p(1) + 2\cdot p(2) + 3\cdot p(3) + ... = p(1) + p(2) + p(2) + p(3) + p(3) + p(3) + ...$$
We can arrange the terms in that sum in a triangle as follows:

p(1)
p(2) p(2)
p(3) p(3) p(3)
p(4) p(4) p(4) p(4)
.......
etc
The sum of items in the j-th row in the triangle equals the j-th term in the above sum.
And the sum of items in the j-th column is p(j) + p(j+1) + p(j+2) + ...
which equals ##P(X\ge j)##.
So the formula you've been given corresponds to first adding each column and then adding the column sums together.
This contrasts to the usual formula, in which we first add each row, then add the row sums together.

In other words, we've just changed the order of summation in our infinite triangle of terms.
We are allowed to do this because all the terms are positive. We say the sum is "absolutely convergent". If some were negative, changing the order of summation could change the result.

Master1022, pbuk and sysprog
Thanks @andrewkirk !

andrewkirk said:
Yes, your application is correct. I can't follow your first line of formulas, which seems to have formatting problems, but your last line is correct, as is your conclusion.
Apologies, it was supposed to be a combinatorics symbol:
## P(X \geq j) = 1 - P(X \leq j - 1) = 1 - \begin{pmatrix} j - 1 \\ 1 \end{pmatrix} \cdot \left( \frac{1}{2} \right) ^{1} \cdot \left( \frac{1}{2} \right) ^{(j - 1) - 1} = \left( \frac{1}{2} \right) ^{j - 1} ##

I need to double check the algebra that follows after that line, but your response with the intuition was the main thing I was looking for.

andrewkirk said:
For intuition, consider the following, where ##p(x)## is the probability that ##X=x##.
The expected value of ##X## is
$$\sum_{x=1}^\infty x\cdot p(x) =1\cdot p(1) + 2\cdot p(2) + 3\cdot p(3) + ... = p(1) + p(2) + p(2) + p(3) + p(3) + p(3) + ...$$
We can arrange the terms in that sum in a triangle as follows:

p(1)
p(2) p(2)
p(3) p(3) p(3)
p(4) p(4) p(4) p(4)
.......
etc
The sum of items in the j-th row in the triangle equals the j-th term in the above sum.
And the sum of items in the j-th column is p(j) + p(j+1) + p(j+2) + ...
which equals ##P(X\ge j)##.
So the formula you've been given corresponds to first adding each column and then adding the column sums together.
This contrasts to the usual formula, in which we first add each row, then add the row sums together.

In other words, we've just changed the order of summation in our infinite triangle of terms.
We are allowed to do this because all the terms are positive. We say the sum is "absolutely convergent". If some were negative, changing the order of summation could change the result.
That's actually quite interesting. Thank you very much for taking the time to write this out. I will have a think about it for a while, and respond if I have other questions. Otherwise, thanks!

## 1. What is the basic coin toss problem to get 1 head?

The basic coin toss problem to get 1 head is a probability problem where a coin is tossed multiple times and the goal is to get exactly one head. This means that out of all the tosses, only one of them should result in a head while the rest should be tails.

## 2. How is this problem usually solved?

This problem is usually solved using the binomial distribution formula, which calculates the probability of getting a certain number of successes (in this case, one head) out of a certain number of trials (number of coin tosses).

## 3. Can this problem be solved with a different method?

Yes, there are other methods that can be used to solve this problem. One alternative method is using the geometric distribution, which calculates the probability of getting a success (in this case, a head) on the first attempt.

## 4. How does the geometric distribution method work?

The geometric distribution method works by calculating the probability of getting a success on the first attempt, and then multiplying it by the probability of getting a failure (in this case, a tail) on all the subsequent attempts. This gives the overall probability of getting exactly one head in a series of coin tosses.

## 5. Which method is more accurate for solving this problem?

Both the binomial and geometric distribution methods are accurate for solving this problem. However, the method that is more appropriate to use may depend on the specific scenario and the assumptions made about the coin tosses.

• Precalculus Mathematics Homework Help
Replies
1
Views
1K
• Precalculus Mathematics Homework Help
Replies
7
Views
2K
• Precalculus Mathematics Homework Help
Replies
17
Views
1K
• Precalculus Mathematics Homework Help
Replies
4
Views
3K
• Precalculus Mathematics Homework Help
Replies
12
Views
2K
• Precalculus Mathematics Homework Help
Replies
7
Views
1K
• Precalculus Mathematics Homework Help
Replies
4
Views
901
• Precalculus Mathematics Homework Help
Replies
6
Views
1K
• Precalculus Mathematics Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
4
Views
2K