[sputum]

(Original post by mashmammad)
guys you all possibly know this. but i have a problem on game theory. the thing is that it assumes none of the people involved in the game will work collaboratively with the other people involved. what if they do. are we sure that they will get the highest profit ( none goes to prison)?

What's the game?

[poyyo]

(Original post by mashmammad)
not all the calculators have that ability. but out of the interest remember to use 3.1416 as pi. it helps accuracy.

thanks, it's just that i need to give some of my answers in terms of pi (eg 3[pi]/2)

so do you think i will have to use another calculator for the exam? (not something i really want to do since i am more used to the way my calculator works)

[mashmammad]

(Original post by poyyo)
thanks, it's just that i need to give some of my answers in terms of pi (eg 3[pi]/2)

so do you think i will have to use another calculator for the exam? (not something i really want to do since i am more used to the way my calculator works)

ok. if you want my opinion dont go for a new calculator. find your answer in radians and divide them by 3.1416( eg if you get 1.5708 radians divide it by 3.1416 and what youll get is 1/2) then times it by pi (e.g. 1(pi)/2) but if you want the method of working with a fancy calculator ill get one of my friends to post it tonight on this thread. hope it helps
by the way tell me what exam you have. itll be better.

[mashmammad]

(Original post by sputum)
What's the game?

ok this is the game theory:
two people have stolen something, police can not arrest any of them if they dont confess.if one of them confesses the other one is arrested and jailed for one year, and the one that has confessed is freed. its the same argument for the other person. but if they both confess each will be prisoned for 6 moths. what do you think they will do?

[sputum]

(Original post by mashmammad)
ok this is the game theory:
two people have stolen something, police can not arrest any of them if they dont confess.if one of them confesses the other one is arrested and jailed for one year, and the one that has confessed is freed. its the same argument for the other person. but if they both confess each will be prisoned for 6 moths. what do you think they will do?

Cooperatively they will shut the something up

[mashmammad]

(Original post by sputum)
Cooperatively they will shut the something up

thats exactly what they dont do. look the first one will think:
if i shut up he'll confess and ill be prisoned, if he doesnt however, if i confess and he'll be prisoned and ill be free, how ever if he has already confessed Ill confess so that my sentence is halved. so for these reasons the first one confesses. the second guy will think the same way, so he'll confess as well, so they'll both end up in prison, for a period of 6 months. its called john nash's game theory.

[sputum]

Your original question was

(Original post by mashmammad)
guys you all possibly know this. but i have a problem on game theory. the thing is that it assumes none of the people involved in the game will work collaboratively with the other people involved. what if they do. are we sure that they will get the highest profit ( none goes to prison)?

if they work collaboratively they both 'win'
unless I'm not understanding something about the question

[mashmammad]

(Original post by sputum)
Your original question was



if they work collaboratively they both 'win'
unless I'm not understanding something about the question

yes if they do work collaboratively. but the whole point is that people are selfish and they go for personal profit.
and end up with something less than what they can get

[sputum]

(Original post by mashmammad)
yes if they do work collaboratively. but the whole point is that people are selfish and they go for personal profit.
and end up with something less than what they can get

well we can all take something good from it then

[poyyo]

(Original post by mashmammad)
ok. if you want my opinion dont go for a new calculator. find your answer in radians and divide them by 3.1416( eg if you get 1.5708 radians divide it by 3.1416 and what youll get is 1/2) then times it by pi (e.g. 1(pi)/2) but if you want the method of working with a fancy calculator ill get one of my friends to post it tonight on this thread. hope it helps
by the way tell me what exam you have. itll be better.

oh wow i just tried it and that's much better, thank you so much!!!!!!!!!!!!!!!!!!!!!!!! i have aqa maths core 3 and core 4 exams by the way

honestly thank you really i am so grateful i was freaking out before about this but really thanks!!!!

[mashmammad]

(Original post by poyyo)
oh wow i just tried it and that's much better, thank you so much!!!!!!!!!!!!!!!!!!!!!!!! i have aqa maths core 3 and core 4 exams by the way

honestly thank you really i am so grateful i was freaking out before about this but really thanks!!!!

No worries

This was posted from The Student Room's Android App on my HTC Wildfire

[BabyMaths]

BMO1 1966

A regular polygon ABCD.... is such that .

How many sides does the polygon have?

I'm hoping to see a better solution than mine. I ended up using De Moivre's theorem which seems inappropriate for BMO1.

[TheMagicMan]

(Original post by BabyMaths)
BMO1 1966

A regular polygon ABCD.... is such that .

How many sides does the polygon have?

I'm hoping to see a better solution than mine. I ended up using De Moivre's theorem which seems inappropriate for BMO1.

There is no need for anything other than basic geometry here...you should be able to express AC and AD in terms of AB using trig. If that seems a bit like a slog, you might make use of the fact that if this is true in a regular polygon, in particular, a certain triangle, consisting or sides with length AB,AC,AD has special properties that you can analyse to get the answer. (this latter method might not work...my brain isn't good enough to run through the whole solution in my head, but I've got it down to something that looks easy to solve)

[BabyMaths]

Hmmm,

With I arrived at the equation .

That much was easy. I then disposed of the possibilities n=3, n=4 and n=6.

It turns out that there are 7 sides.

[BabyMaths]

Anyone interested?

I gambled on looking at n=7 rather than n=5.

Letting De Moivre's theorem gives



Spot the factor c+1 and get



and factor a bit more



etc.

But is there an easier way?

[BabyMaths]

(Original post by TheMagicMan)
There is no need for anything other than basic geometry here...

Any chance you could post this basic geometry solution. I confess I'm not too good at geometry.

[TheMagicMan]

(Original post by BabyMaths)
Any chance you could post this basic geometry solution. I confess I'm not too good at geometry.

neither am I...it was always my great olympiad weakness

I'll have a go later at it. There is definitely a simple solution as I've done the question before.

[swbp]

Hey guys, i've got a very simple percentage I need to work out associated with my A2s - it isn't anywhere near as complex as A2 mathematics, but I still can't do it haha (arts student here)

BASICALLY in one of my A levels:
I achieved 83% on the section of the A level worth 50%
I achieved 87% on the section of the A level worth 20%

How much must I get in the final section, woth 30%, to achieve 80% overall?

Thanks in advance !

[jack.hadamard]

(Original post by swbp)
...

Are you interested how to work it out, or just in the actual number?
If these are rough percentages, anything more than should do.

[swbp]

(Original post by jack.hadamard)
Are you interested how to work it out, or just in the actual number?
If these are rough percentages, anything more than should do.

No, I just needed a figure. Thanks !