Probability HW...

[HardcoreRobot]

Plz help me guys. Heres the question -
Show how to make a fair two-way choice using a coin that is not known to be fair, that is, where "heads" occurs with unknown (but fixed) probability. (Hint: It is not necessary to determine the probability of "heads".

Im stumped...

 

[axelfox]

eh, two headed coin :?

 

[BrunoPuntzJones]

5

 

[dOrKuS]

i'd help but could you make the question any clearer?
maybe i'm just not reading it right?
what course?

 

[axelfox]

[Lloyd Christmas] So you're tellin' me there's a chance....yeah![/Lloyd Christmas]

 

[Radiohead]

Well it doesn't say you have to flip it so maybe you can just make a choice by guessing what hand the coin is in

I really can't see any other way of making a fair decision by flipping the coin when it is not 50/50.

 

[HardcoreRobot]

Well, that is exactly as the question reads off my assignment, but I'll try to translate.

It would be easy to make a fair two-way choice with a fair coin - one choice is "heads" and the other is "tails". You only need to flip it once, and there are equal chances for either to come up. A coin that was NOT fair, would not give both sides equal chances.

So I guess the question is more like - "You have a coin that is weighted to land on one particular side (say 9 times out of 10 or something). This makes it not a fair coin. You dont know exactly what the probability is of it landing on that weighted side, but that is not necessary to find the solution. Come up with some process of flipping this coin to yield one of two equal probable results.

I assume the solution will require some combination of multiple flips and analysis... Mostly im just

 

[BruinEd03]

errr hide the coin in one hand behind ur back and have the person guess which hand the coin is in?

-Ed

 

[HardcoreRobot]

Heres a similar problem (with solution) that may help.
(Q) Show how to make a fair three-way choice using a fair coin.
(A) Flip it twice. 2 heads is one choice, 2 tails is another, and one of each is the final.

 

[BruinEd03]

<< Heres a similar problem (with solution) that may help.
(Q) Show how to make a fair three-way choice using a fair coin.
(A) Flip it twice. 2 heads is one choice, 2 tails is another, and one of each is the final. >>



That's not really all that fair because ther'es four possibilities:

HH
HT
TH
TT

so u have a 50% chance of getting TH or HT.

-Ed

 

[Radiohead]

You have a coin that is weighted to land on one particular side

If you can control the number of rotations of the coin in each flip, you can do a few sample flips with heads facing up, then facing down and calculate the probablitity of the loaded side landing.

Maybe you can derive something out of this cause I don't really have a clue as to what i'm trying to say. I'm having my own world of fun with Multiple regression at the moment...

 

[BrunoPuntzJones]

<< [Lloyd Christmas] So you're tellin' me there's a chance....yeah![/Lloyd Christmas] >>

i dunno who lloyd christmas is (i think they make gift stuff), but that guy in dumb and dumber named lloyd said that

 

[HardcoreRobot]

<< That's not really all that fair because ther'es four possibilities:

HH
HT
TH
TT

so u have a 50% chance of getting TH or HT. >>



OK well you have me even more confused now. I thought I had that one figured out but now im just too tired. I think it had something to do with sequence not being important (meaning HT and TH became only 1). ARG

 

[BruinEd03]

<<

<< That's not really all that fair because ther'es four possibilities:

HH
HT
TH
TT

so u have a 50% chance of getting TH or HT. >>



OK well you have me even more confused now. I thought I had that one figured out but now im just too tired. I think it had something to do with sequence not being important (meaning HT and TH became only 1). ARG >>



A fair way would be to say ok HH = one decision, HT = another decision, TT = another decision, and if you get a TH = redo it again.

-Ed

 

[BruinEd03]

For ur original question:

how about HT = one choice and TH = another and everything else HH/TT u discard?

There's equal probability of HT as TH since u suppose the probability of H = x and probability of T = (1-x), then the probability of HT = x(1-x) = the probability of TH = (1-x)x . Hope that helps!

-Ed

 

[HardcoreRobot]

Thanks for your suggestions ed. Im mulling over them right now, but they look good.