Average profit of coin tossing game?

Question

The problem from this video is stated as follows:

You play a game with a coin, not necessarily fair:

• If you throw heads, you get 1 unit of money and throw again.
• If you throw tails, you get nothing and the game ends.

What is the average amount of money you will get from this game?

(Let $h$ and $t$ be te probability of throwing heads and tails respectively, and obviously we habe $h+t=1$)

In the video the average profit $P$ is calculated like this: $$ \begin{align} P & = \sum_{n=1}^\infty (n * \text{Probability of winning exactly $n$ units of money} ) \\ & = ht + 2h^2t + 3h^3t+4h^4t + \cdots \end{align} $$ Which is then simplified to a short answer. I do understand and appreciate this approach, however I would do it a little different: $$ \begin{align} P & = \sum_{n=1}^\infty \text{Probability of winning one money unit at the $n$th toss} \\ & = h + h^2 + h^3 + h^4 + \cdots \end{align} $$

I was able to show that both expressions can be simplified to the solution $P=\frac{h}{t}$:

$$ \begin{align} P & = ht + 2h^2t + 3h^3t+4h^4t + \cdots \\ & = ht * (1+2h+3h^2+4h^3+\cdots) \\ & = h(1-h) * (1+2h+3h^2+4h^3+\cdots) \\ & = (1-h) * (h+2h^2+3h^3+4h^4+\cdots) \\ & = (h+2h^2+3h^3+4h^4+\cdots) - (h^2+2h^3+3h^4+4h^5+\cdots) \\ & = h+h^2+h^3+h^4+\cdots \\ & = \frac{h}{1-h} = \frac{h}{t} \end{align} $$

But even though my approach also yields the correct answer, I wonder if it does so for the correct reason, i.e. is my reasoning/approach correct?

Improving Edit: Actually I did not really reason my approach. So the actual question is: What is the correct reasoning for the correctness of my approach that works without any use of intuition?

Answer 1

Your reasonment is correct, and tacitly uses an important theorem about the expected value (the average amount won in the game) denoted $\mathbb{E}(X)$:

If $X$ is a nonnegative integer random variable,

$\mathbb{E}(X)=\sum_{k\in\mathbb{N}}k \mathbb{P}(X=k)=\sum_{k\geq1} \mathbb{P}(X\geq k)$

The first part of the equality is the classic way to compute expected value in the case of discrete random variable, but the second, which is the one you use actually works for every positive integer random variable. It sometimes enables to achieve results using easier reasonning or computation, as was the case in your example.

You can prove the equality above easily :

$\sum_{k\geq1} \mathbb{P}(X\geq k)=\sum_{k\geq1} \sum_{j\geq k}\mathbb{P}(X= j)$

$=\sum_{j\geq 1} \sum_{k=1}^{j} \mathbb{P}(X= j)=\sum_{j\geq 1}j\mathbb{P}(X= j)=\sum_{j\geq 0}j\mathbb{P}(X= j)=\mathbb{E}(X)$

Notice that the first equality of the demonstration works only because $X$ can only take integer values, and the last one only because $X$ is nonnegative.

Answer 2

In fact, both answers are the same and follows the same tree and somehow the same theory.

We may use the probability as a transition factor in case of multiple choices and average them if we consider many events. It's more intuitive to imagine that there is a lot of tests and that they are spread at each crossroads using the unitary probabilities h and t.

Enter at the top and exit by left red gates ( each with its probability ) or by a green gate after an infinite sequence. The probability we are looking for is

• the average sum of each exit with the earned amount : then the sum of the right column product the amount earned at each step ( video solution ) : $0 h + (0 h + 1 h t ) + (0 h + 1 h t + 2 h² t ) + ... $
• the average earn conserved without exiting : $0 h + 1 h + 1 h² + ... $ , which is exactly the same.

Since we consider a limit, exiting or not is just a detail. In the first case, we must add all the exit options and in the latter we must just consider what had been earned without exiting. Many exits or one at the infinity, 2 views with calculus that coincide after simplifications as you showed in the question.