Trial & Improvement? - Probability.....Again. :P

Gene said:
As I have tried to explain: That is an average per student. Think about rolling a die and getting $1 you can roll a 6. 5/6 of the time you don't and get nothing. 1/6 of the time you do and get a dollar. Your average return per roll is $1/6 = 16 & 2/3¢. You will NEVER get 16 & 2/3¢. It is an average. If you roll 6 times you will (on average) get a 6 once. 6 rolls * (16 & 2/3) average return per roll = the $1 you actually collect. If you roll 12 times you will (on average) get a 6 twice. 12 rolls * (16 & 2/3) average return per roll = the $2 you actually collect.
It is the same idea for the .2p that bothers you so much. You will never collect .2p but if you have 100 customers you will collect 20p. If you have 1,000 customers you will collect 200p. If you have 1,000,000 customers you will collect 200,000p. All of these things are on average.
If you roll that die 6 times you could get five 6s and collect $5 or you could roll it 100 times and get no 6s and collect nothing. But on average you will get 16 & 2/3¢ per roll.
---------------------
Gene
Click to expand...

I see. :shock:

So it's just the average money from one person?

That's why when we times by 50 (the number of people) we get 10, which is how much he would have left when he gives out all the prizes?
 
Gene said:
By George, I think you've got it.
------------------
Gene
Click to expand...

Nearly, my teacher told me to show him what I've done next lesson (monday), so could you just explain the formula so when he asks me I can tell him. :D Step by step. :p

Thanks.

EDIT:

Is this right?

The first bit:

(5p)(38/50)

This is the average money he is getting from each person who looses.

Though he has to take away the cost of the win:

-(q-5p)(12/50)

As he is loosing that money.

2 bits I don't understand are; how can the average probability of a person, win and loose at the same time. Also why does it have to >0.
 
Winning some and losing some is what makes it an average. Sometimes the customer wins (12/50), Sometimes the customer loses (38/50).
(5p)(38/50)-(q-5p)(12/50)>0 That's from PKA's post. Maybe it will make more sense if you expand it.
(5p)(38/50)-(q)(12/50)+(5p)(12/50)>0
5p(50/50)-(q)(12/50)>0
5p-(q)(12/50)>0
He gets 5p every time someone plays, win or lose.
He pays q when they win, 12 times out of 50. He would like that to be > 0 'cause he wants to be ahead when the average person leaves.
--------------------
Gene
 
Gene said:
Winning some and losing some is what makes it an average. Sometimes the customer wins (12/50), Sometimes the customer loses (38/50).
(5p)(38/50)-(q-5p)(12/50)>0 That's from PKA's post. Maybe it will make more sense if you expand it.
(5p)(38/50)-(q)(12/50)+(5p)(12/50)>0
5p(50/50)-(q)(12/50)>0
5p-(q)(12/50)>0
He gets 5p every time someone plays, win or lose.
He pays q when they win, 12 times out of 50. He would like that to be > 0 'cause he wants to be ahead when the average person leaves.
--------------------
Gene
Click to expand...

Ahhh so 0 is the "break even" point. If he puts 0 in ad times by 50 he gets 250...

Expanding it's just going to confuse me even more. :lol: Now I've come all this way learning it like that that one doesn't make sense. How'd you do that? Because now you're not taking away the prob of winning...

Would you say what I wrote is correct:

5p)(38/50)

This is the average money he is getting from each person who looses.

Though he has to take away the cost of the win:

-(q-5p)(12/50)

As he is gaining 5p but loosing the prize money. That gives he the average p.p.

Thank you Gene.
 
I'm sorry, but if you don't see what I did, you have more problems than I can imagine. I multiplied it out and gathered like terms.
(5p)(38/50)-(q-5p)(12/50) =
(5p)(38/50)-(q)(12/50)+(5p)(12/50) =
(5p){(38/50+(12/50)}-(q)(12/50) =
(5p)(50/50) - q(12/50) =
5p-q(12/50) > 0
It is still exactly the same equation.

The q*(12/50) is "taking away the cost of the wins."

I think we have done all that we can for you on this problem.
--------------------
Gene
 
Gene said:
I'm sorry, but if you don't see what I did, you have more problems than I can imagine. I multiplied it out and gathered like terms.
(5p)(38/50)-(q-5p)(12/50) =
(5p)(38/50)-(q)(12/50)+(5p)(12/50) =
(5p){(38/50+(12/50)}-(q)(12/50) =
(5p)(50/50) - q(12/50) =
5p-q(12/50) > 0
It is still exactly the same equation.

The q*(12/50) is "taking away the cost of the wins."

I think we have done all that we can for you on this problem.
--------------------
Gene
Click to expand...

I get it! :D

But could you just tell me if what I posted above is correct and I will leave you alone.:p :lol:

Thanks mate.
 
Sorry to keep bumping this I just want to get it finished and it's sliping off the page.
 
Yes, if you prefer it that way and understand it, it is correct.
---------------------
Gene
 
You must log in or register to reply here.
Top