Game : random variable for net profit

In summary, the conversation discusses a game where a fair coin is tossed up to three times, with a cost of $1 per throw. If a head falls, the player wins $3. The random variable X represents the net profit (profit minus stake) and can take values of -3, 0, 1, and 2 with corresponding probabilities of 1/8, 1/2, 1/4, and 1/8 respectively. The conversation also calculates the probabilities of X being less than 0 and greater than 0, with intermediate steps involving the union and complement of certain sets. The final probabilities are P(X<0) = 1/8 and P(X>0) = 3
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! 😊

You participate in the following game :
You toss a fair coin until heads falls, but no more than three times. You have to pay $1$ euro for each throw. If your head falls, you win $3$ euros. The random variable $X$ describes your net profit (profit
minus stake). Give the values that $X$ can get and the corresponding probabilities. Calculate $P [X <0]$ and $P [X> 0]$. Make a note of the intermediate steps.

I have done the following :
- If we get Head at the first toss, then X = -1 + 3 = 2 EUR.
- If we get Head at the second toss, then X = -1 -1 + 3 = 1 EUR.
- If we get Head at the third toss, then X = -1 -1 -1 + 3 = 0 EUR.
- If we don't get Head at all at the three first tosses, then X = -1 -1 - 1 = -3 EUR.

So are the possible values for $X$ the $\{-3, 0, 1, 2\}$ ? :unsure:

As for the probabilities do we have the following ?
- $P(X=2) = \frac{1}{2}$ (either head or no head at the first toss)
- $P(X=1) = \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$ (either head or no head at the first toss, either head or no head at the second toss)
- $P(X=0) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ (either head or no head at the first toss, either head or no head at the second toss, either head or no head at the third toss)
- $P(X=0) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ (either head or no head at the first toss, either head or no head at the second toss, either head or no head at the third toss)

Is everything correct and complete? :unsure:
 
Physics news on Phys.org
  • #2
Hey mathmari!

All correct. (Nod)

I would rephrase the explanations though.
For instance:
- $P(X=2) = \frac{1}{2}$ (head at the first toss)
- $P(X=1) = \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$ (no head at the first toss, and head at the second toss)
(Thinking)
 
  • #3
mathmari said:
As for the probabilities do we have the following ?
- $P(X=2) = \frac{1}{2}$ (either head or no head at the first toss)
- $P(X=1) = \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$ (either head or no head at the first toss, either head or no head at the second toss)
- $P(X=0) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ (either head or no head at the first toss, either head or no head at the second toss, either head or no head at the third toss)
- $P(X=0) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ (either head or no head at the first toss, either head or no head at the second toss, either head or no head at the third toss)

Ah at the last one Imeant $P(X=-3)$ instead of $P(X=0)$, so we have $P(X=-3) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ ( no head at the first toss, and no head at the second toss, and no head at the third toss ), right? :unsure:

To calculate the probabilities $P(X<0)$ and $P(X>0)$ do we do the following ?
\begin{align*}&\{X<0\}=\cup_{{y\in X(\Omega), y<0}}\{X=y\}=\{X=-3\} \\ & \Rightarrow P[X<0]=P[X=-3]=\frac{1}{8} \\ &\{X>0\}=\cup_{{y\in X(\Omega), y>0}}\{X=y\}=\{X=1\} \cup \{X=2\} \\ & \Rightarrow P[X>0]=P[\{X=1\} \cup \{X=2\}]=P[X=1]+P[X=2]=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\end{align*}
Is that correct? Do we have to mention more intermediate steps or are these the desired ones? :unsure:
 
Last edited by a moderator:
  • #4
Looks correct to me, and I think you have all the desired steps. (Nod)
 
  • #5
Klaas van Aarsen said:
Looks correct to me, and I think you have all the desired steps. (Nod)

I added some more steps. Is that correct now? :unsure:
 
  • #6
mathmari said:
I added some more steps. Is that correct now?
It looks fine to me. (Nod)
 

Related to Game : random variable for net profit

1. What is a random variable?

A random variable is a numerical quantity that takes on different values based on chance or probability. In the context of a game, it represents the potential profit that can be earned.

2. How is a random variable used in games?

A random variable is used in games to determine the potential outcomes and associated probabilities. It allows for uncertainty and risk to be incorporated into the game, making it more exciting for players.

3. What is net profit?

Net profit is the total amount of money earned after all expenses and costs have been deducted. In the context of a game, it represents the amount of money that a player can potentially earn.

4. How is net profit calculated?

Net profit is calculated by subtracting the total expenses and costs from the total revenue or income. In the context of a game, it would be calculated based on the random variable and the associated probabilities.

5. How does understanding random variables and net profit benefit players?

Understanding random variables and net profit can benefit players by allowing them to make more informed decisions and strategize in the game. It also adds an element of excitement and unpredictability to the game, making it more enjoyable.

Similar threads

  • Set Theory, Logic, Probability, Statistics
2
Replies
57
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
1K
  • Set Theory, Logic, Probability, Statistics
2
Replies
45
Views
4K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
939
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
919
  • Precalculus Mathematics Homework Help
Replies
7
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
7
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
970
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
2K
Back
Top