A Couple of Math Questions

[Depression Moon]

Sorry to bring this to you so late. I have a math exam and there are a couple of questions that I'm having trouble understanding how to solve. questions like these will show up. Googling my question doesn't help much.

Here is question 1.
At the casino, a game costs $1 to play. If you win, you get $4. If the chance of winning is 10%, how much money should you lose each time you play (what is the expected value)? How much will I lose?

Question 2
Sum the numbers 10 + 20 + 30 + .... + 500
This is familiar to me and I know there is a simple solution, but I can't remember it.

[Jessweeee♪]

Question 1:

If you win 10% of the time, then you win once out of every ten games.
In ten games, you pay ten dollars and win four, so you lose six dollars every ten games.
If you know how many games you are playing, you can divide that number by ten and multiply the result by six.
But uh, for each game individually, the most you lose is one dollar, as that is the price of the game. I think the biggest question here is what exactly is being asked of you.

[Depression Moon]

I actually did receive some help with someone and this sounds familiar, but I'm trying to understand if what he is saying is the expected value or how much I will lose. I don't think it's how much I would lose. I think that number would be larger.


as for q1, naïvely : 90%*(-1)+10%*(+3)=-0.6$ on average

[Aerith's Knight]

Question 2
Sum the numbers 10 + 20 + 30 + .... + 500
This is familiar to me and I know there is a simple solution, but I can't remember it.

Sum(10n) with n=1,2,...,50

Or alternatively:

25*[(10+500)/2]

or in n-form:

n/2 * (N+(N-(N-1))/2

[Depression Moon]

I appreciate the help.

[Peegee]

Question 2
Sum the numbers 10 + 20 + 30 + .... + 500
This is familiar to me and I know there is a simple solution, but I can't remember it.

Sum(10n) with n=1,2,...,50

Or alternatively:

25*[(10+500)/2]

or in n-form:

n/2 * (N+(N-(N-1))/2

Stole the answer out of my liberal arts major ass. :aimmad: