Probability / cost

Hello,

i'm pretty horrible at math, but i need to solve a problem. Please excuse any layman terms i might use since english is not my first language and i don't know all the english terminology used in mathematics.

Basically i need someone to calculate an odds/cost type of problem, where the goal is to determine which out of 2 available gambling methods should be statistically cheaper than the other if both achieve the same goal.

So here's the situation:

1. The first method uses 4 items (let's call them items A) to create the desired result (let's call the result item C). The chance of this happening is 0.1% and costs 76 dollars.
2. The second method uses 4 items B to create the same result (item C). The chance of this happening is 0.5%, but costs 496 dollars.
3. When using either method, you will always get a result whether it be the desired one or not:

• If the first method fails in producing the desired item C, you get back 1/4th of the original investment of 76 dollars in 77.5% of cases. In the other 24.99% you produce an item B valued at 124 dollars.
  
• If the second method fails you always get back 1 item B valued at 124 dollars.

Questions:

1. Which of the above two methods should be statistically cheaper in acquiring the desired item C?
2. Is it possible to work out approximately how much it would cost to obtain item C on average and/or how many times you'd have to repeat either method on average?
3. Let's say item C costs 50.000 dollars. Is it statistically better to gamble for it using the above figures or are you more likely to make a loss (spending more than 50k) acquiring item C?

I hope i made everything as clear as possible.
Thanks in advance!

Gil231 said:

So here's the situation:

1. The first method uses 4 items (let's call them items A) to create the desired result (let's call the result item C). The chance of this happening is 0.1% and costs 76 dollars.
2. The second method uses 4 items B to create the same result (item C). The chance of this happening is 0.5%, but costs 496 dollars.
3. When using either method, you will always get a result whether it be the desired one or not:

• If the first method fails in producing the desired item C, you get back 1/4th of the original investment of 76 dollars in 77.5% of cases. In the other 24.99% you produce an item B valued at 124 dollars.
  
• If the second method fails you always get back 1 item B valued at 124 dollars.

Questions:

1. Which of the above two methods should be statistically cheaper in acquiring the desired item C?
2. Is it possible to work out approximately how much it would cost to obtain item C on average and/or how many times you'd have to repeat either method on average?
3. Let's say item C costs 50.000 dollars. Is it statistically better to gamble for it using the above figures or are you more likely to make a loss (spending more than 50k) acquiring item C?

Click to expand...

Call the two games "A" and "B"

For A, you will on average play 1000 times to win C,while for B the average number of tries is 200. The cost of each try is the amount you put in minus the expectation value.

E[A] = 0.775*$76/4 + 0.2499*$124 = ...
..........cost per play = $76 - E[A] = ...
................................*1000 = ...

E = $124
..........cost per play = $496 - E = ...
................................*200 = ...