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

[Monkeyseat]

Hi, I have this problemo (QUESTION B):

http://img223.imageshack.us/img223/2692/hpim08420en.jpg

Well the fact is I've done it but was wondering is there any other way/a better way of doing it than trial and improvement? Also is it a valid method?

I did it by (12x5)+(38+5)=250

Then gradually put figures in to see what the highest number I could put in that would come out below 250 and came up with 20p.

Thanks.

 

[pka]

I think that you once wrote that you were working from an old book. Well, the problems seem to be poorly put to say the least. It seems to me that we are expected to use the experimental data as the true probabilities: 12/50 to win.

So now we need the expected value from Jason’s point-of-view.
From each player, he can expect (5p)(38/50)-(q-5p)(12/50) where q is the amount he pays out if the player wins.
[Note: we assume that a player pays 5p to play regardless of winning or loosing. He looses the 5p failing to land on red, but wins q-5p if he wins.]
Thus, Jason must set q to an amount that makes (5p)(38/50)-(q-5p)(12/50)>0.

 

[Gene]

You are right about how much the suckers paid. 250 p. You don't need trials. If you look at the board they should win 11/64 and he would make a little more, but I guess you are supposed to use his 12/50. You think they will win 12 times out of 50 so to break even he would pay out 12x = 250
x = 20.83 as you found. Now you know x < 20 p. If he paid 20 p for a win he would make 10 P for every 50 trys. 12*20 + 10 = 250.
He makes (250-12x)/50 per try paying x.
(250-12*20)/50 = .2 p per try
(250-12*15)/50 = 1.4 p per try
(250-12*10)/50 = 2.6 p per try
(250-12*5)/50 = 3.8 p per try but VERY few customers.
(250-12*0)/50 = 5 p per try but NO customers.
It looks like 6 < x < 20

 

[Monkeyseat]

You are right about how much the suckers paid. 250 p. You don't need trials. If you look at the board they should win 11/64 and he would make a little more, but I guess you are supposed to use his 12/50. You think they will win 12 times out of 50 so to break even he would pay out 12x = 250
x = 20.83 as you found. Now you know x < 20 p. If he paid 20 p for a win he would make 10 P for every 50 trys. 12*20 + 10 = 250.
He makes (250-12x)/50 per try paying x.
(250-12*20)/50 = .2 p per try
(250-12*15)/50 = 1.4 p per try
(250-12*10)/50 = 2.6 p per try
(250-12*5)/50 = 3.8 p per try but VERY few customers.
(250-12*0)/50 = 5 p per try but NO customers.
It looks like 6 < x < 20

:?: I didn't really get that sorry; what was this about "(250-12*10)/50 = 2.6 p per try"?

I thought it would be 0<x<20 as any number below 20 and he will make a profit?

 

[Guest]

I thought it would be 0<x<20 as any number below 20 and he will make a profit?

It's hard to tell what answer the book expects. If you were a customer and you saw that you had to pay 5 pence to play the game, and if you win, you get your 5 pence back, then would you play?

 

[Monkeyseat]

I thought it would be 0<x<20 as any number below 20 and he will make a profit?

It's hard to tell what answer the book expects. If you were a customer and you saw that you had to pay 5 pence to play the game, and if you win, you get your 5 pence back, then would you play?

I guess not.

So yeah it must be 6>x>20.

What were you doing here though (I'm a fool XD):

(250-12*20)/50 = .2 p per try
(250-12*15)/50 = 1.4 p per try
(250-12*10)/50 = 2.6 p per try
(250-12*5)/50 = 3.8 p per try but VERY few customers.
(250-12*0)/50 = 5 p per try but NO customers.

:?:

 

[Gene]

He makes (250-12x)/50 per try paying x.

The list is how much he makes per customer if he pays x for a win. The last one looks best. (250-12*0)/50 = 5 p. He pays 0 and gets 5 p for each, but how many people will play if they pay 5 p and if they win get nothing. Same with (250-12*5)/50 = 3.8 p. If they win they get their 5 p back and break even. A very few might play but I doubt it. The rest of the list are possible. The bigger the payment the more customers. He is going to have to guess what the maximum profit is with (250-12*x)/50 times the number of players. It may be that making .2 p per is his best 'cause they may line up and just keep playing.
I would guess that paying 10 or 15 p would generate the most but who knows.
-------------------
Gene

 

[Monkeyseat]

He makes (250-12x)/50 per try paying x.

The list is how much he makes per customer if he pays x for a win. The last one looks best. (250-12*0)/50 = 5 p. He pays 0 and gets 5 p for each, but how many people will play if they pay 5 p and if they win get nothing. Same with (250-12*5)/50 = 3.8 p. If they win they get their 5 p back and break even. A very few might play but I doubt it. The rest of the list are possible. The bigger the payment the more customers. He is going to have to guess what the maximum profit is with (250-12*x)/50 times the number of players. It may be that making .2 p per is his best 'cause they may line up and just keep playing.
I would guess that paying 10 or 15 p would generate the most but who knows.
-------------------
Gene

Ahhh I get it now, but how can you make 3.8p or .2p?

Also would you say trial and improvement is a valid way of working it out, is there a better way of showing what I've done? What could I show for working out?

EDIT: Just saw this sorry:

I think that you once wrote that you were working from an old book. Well, the problems seem to be poorly put to say the least. It seems to me that we are expected to use the experimental data as the true probabilities: 12/50 to win.

So now we need the expected value from Jason’s point-of-view.
From each player, he can expect (5p)(38/50)-(q-5p)(12/50) where q is the amount he pays out if the player wins.
[Note: we assume that a player pays 5p to play regardless of winning or loosing. He looses the 5p failing to land on red, but wins q-5p if he wins.]
Thus, Jason must set q to an amount that makes (5p)(38/50)-(q-5p)(12/50)>0.

Pka, what does e.g. (38/50) represent is it the probability of a loss? Also I thought it would be q + 5p as he is getting the prize and his money.

 

[Monkeyseat]

Help please Gene/Pka. *Nudge Nudge*

 

[pka]

I did say that the problem is very poorly written.
It says “Jason decides to charge a pupil 5p to take part in his game”.
It does not say that he will return the 5p if the pupil wins.
So do we assume that he does? I see no reason to do so.

Now, I took Jason’s own experimental data as the ‘true probability’.
So (38/50) the pupil does not win and (12/50) the pupil wins.
From Jason’s point of view he gets 5p if the pupil loses or otherwise he pays (q-5p).

Actually, this is an absurd problem. In the first place to actually calculate the probabilities we would need to know the size of a penny as well as the size of each square. Don’t forget the requirement that the penny be completely within the red square. So I assume we are meant to use Jason’s experimental data.

 

[Gene]

Ahhh I get it now, but how can you make 3.8p or .2p?

That is an average. Think about 50 customers when he is paying 5p.
12 times he pays 5p
38 times he pays 0p
He takes in 250p
250 - 12*5 = 190p
190/50 = 3.8p per customer.

Also would you say trial and improvement is a valid way of working it out, is there a better way of showing what I've done? What could I show for working out?

It might be valid if there were no other way (Look up Monte Carlo method on the web.) but I worked it out in my first answer.

what does e.g. (38/50) represent is it the probability of a loss? Also I thought it would be q + 5p as he is getting the prize and his money.

Yes, that is the probability of a loss by the customer. Pka and I assume that he pays 5p no matter what happens. It doesn't say he gets his 5p back if he wins. What Jason pays if he does win is a seperate transaction. On average Jason pays (12/50) times whatever he decides the prize should be. If you are giving him his 5p back then

I did it by (12x5)+(38*5)=250

is wrong. It would be
I did it by (12x0)+(38*5)=190

 

[Monkeyseat]

I did it by (12x5)+(38*5)=250

is wrong. It would be
I did it by (12x0)+(38*5)=190

Where did you get the 0 from?

Monte Carlo method.:shock: Looks a bit complex for me. Bah, I'm useless at math, what would you say would be good working out? If you had to show working out; what would you do?

(5p)(38/50)-(q-5p)(12/50)>0.

So any number that brought it out bigger than 0 would be okay?

Thanks.

 

[Gene]

You were figuring out how much he took in. If he is giving back the 5p when they win, then he takes nothing in when they win. That is where the zero comes from.
I would say the way I worked it out in my first reply is what I would say about working it out :twisted:
----------------
Gene

 

[Monkeyseat]

You were figuring out how much he took in. If he is giving back the 5p when they win, then he takes nothing in when they win. That is where the zero comes from.
I would say the way I worked it out in my first reply is what I would say about working it out :twisted:
----------------
Gene

Oh sorry about that I missed the 12x=250 bit just saw the bit underneath.

Can I just ask why you rounded 20.83 to 20.

Right, here's what I'm going to do for working out, this okay?:

(12x5)+(38+5)=250

12x=250p

x=20.83p

x <or equal to 20p

 

[Gene]

Why truncate? 'Cause you would have a hard time slicing a pence into 6 parts just to get the exact break even amount. That is why I computed only the 5, 10, 15 & 20 payouts. If you were running the joint you wouldn't want to mess around with the pence change.
If you want to split it into
(12*5)+(38*5) that's okay but not necessary. 50*5 works just as well but some comments about what you are doing wouldn't hurt.
--------------------
Gene

 

[Monkeyseat]

Why truncate? 'Cause you would have a hard time slicing a pence into 6 parts just to get the exact break even amount. That is why I computed only the 5, 10, 15 & 20 payouts. If you were running the joint you wouldn't want to mess around with the pence change.
If you want to split it into
(12*5)+(38*5) that's okay but not necessary. 50*5 works just as well but some comments about what you are doing wouldn't hurt.
--------------------
Gene

Okay but if you round don't you round up if it's over .5? :?

 

[Gene]

Gene wrote:Why truncate? 'Cause you would have...

I wasn't rounding. If I had been, yes it would have been 21 but that would not have been less than 250 p so I didn't.
------------------
Gene

 

[Monkeyseat]

Gene wrote:Why truncate? 'Cause you would have...

I wasn't rounding. If I had been, yes it would have been 21 but that would not have been less than 250 p so I didn't.
------------------
Gene

I see. Finally.

You rule. :lol:

Well I've just been reading through the thread, trying to get my head around this:

(5p)(38/50)-(q-5p)(12/50)>0.


My teacher was showing it me somthing like this and I don't get any of it.

It's:

(cost to play)(probability of loss) - (prize-cost to play)(probability of win)>0

Say we put 20p in we get 0.2 out????? :shock:

 

[Monkeyseat]

Bumped.

 

[Gene]

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