Thread: Basic Probablilty Theory - Expected Value?

Hi all,

Apologies if this is pretty basic, but I would appreciate some help.

Suppose we have a game that cost $1 to play. The game is a fair coin toss. You get returned $2 if the coin lands on heads (win $1) and returned $0 if it lands on heads
I understand that after 1000 trials the EV of the game is still 0, you should be breaking even.
My question is, what is the probaility that you are $100 up after 1000 and 10000 trials respectively?

Now applying this to a wider context, suppose a die is rolled. If it lands on an odd number, you win an amount equal to that number and vice versa for odd number. The average gain for this game is calculated at -0.5 per roll. What is the probaility you are $20 up after 1000 trials and 10000 trails respectively.

Thanks in advance.

The first part you can work out using binomial distribution. I posted a couple of links in another thread earlier. probability of DNA sequence errors

I'm not sure I understand the second part. This is the confusing part:

If it lands on an odd number, you win an amount equal to that number and vice versa for odd number.

I'd guess you mean you win $1,$3 or $5 on those rolls and lose $2,$4 or $6 on those rolls, not sure though.

Assuming I'm right, the only precise ways of calculating this I can think of you really wouldn't want to do by hand. The easiest way I can think of will get you a decent estimate though.

If you calculate the average winning amount and the average losing amount ($3 and $4 respectively), you can calculate how many wins would be required on average to be that amount up. So for example wanting to be $20 up after 1000 trials would give the formulas:


Where x is the number of wins and y is the number of losses.

This gives the answers:



Now you can round x down to 574 and use it as the successful number of trials in the binomial distribution formula and get an estimate.

Hope this helps.
-Matthew