Randomizing with a coin

[iversonian]

You are playing rock paper scissors, and you want to choose which to throw using a coin. You want a probability distribution of 1/3 for each type.

Is it possible to devise an algorithm for choosing what to throw based on a series of coin flips such that you are guaranteed to have a result in a finite number of coin flips? If not, prove. If so, what is the smallest number of coin flips in which you can guarantee a result that has a perfect 1/3 distribution for each type.

e.g. If you flip twice, where head-head = rock, head-tail = paper, tail-head = scissors, and tail-tail = redo, then it may never terminate.

[TomCollins]

[ QUOTE ]
You are playing rock paper scissors, and you want to choose which to throw using a coin. You want a probability distribution of 1/3 for each type.

Is it possible to devise an algorithm for choosing what to throw based on a series of coin flips such that you are guaranteed to have a result in a finite number of coin flips? If not, prove. If so, what is the smallest number of coin flips in which you can guarantee a result that has a perfect 1/3 distribution for each type.

e.g. If you flip twice, where head-head = rock, head-tail = paper, tail-head = scissors, and tail-tail = redo, then it may never terminate.

[/ QUOTE ]

It's impossible to guarantee it in a finite number of coin flips.

[iversonian]

Care to attempt a proof?

[TomCollins]

Not really. But you cannot represent 1/3 exactly in binary, and all coin flips can be represented in binary, so you have a problem.

[alThor]

[ QUOTE ]
Not really. But you cannot represent 1/3 exactly in binary, and all coin flips can be represented in binary, so you have a problem.

[/ QUOTE ]

Well, you tried not to prove it, but then you accidentally proved it (essentially). Better luck not proving things next time!

[TomCollins]

[ QUOTE ]
[ QUOTE ]
Not really. But you cannot represent 1/3 exactly in binary, and all coin flips can be represented in binary, so you have a problem.

[/ QUOTE ]

Well, you tried not to prove it, but then you accidentally proved it (essentially). Better luck not proving things next time!

[/ QUOTE ]

I didn't prove why its impossible to represent it in binary though.

[UATrewqaz]

If this was a question looking for an actual system, not just the math, just roll a fair die (1-2: Rock, 3-4: Paper, 5-6: Scissors)

[BruceZ]

Actually it's easy. Just take a marker and divide each side into three 120 degree sectors. [img]/images/graemlins/laugh.gif[/img]

For the methods that depend on heads and tails, worrying about the game never terminating makes no more sense than worrying about the coin landing on its edge. I suggest flipping the coin until it comes up heads. Then:

Rock: First head is on an even numbered flip.

Paper: First head is on an odd numbered flip, and the next flip is also heads.

Scissors: First head is on an odd numbered flip, and the next flip is tails.

[AlienBoy]

[ QUOTE ]
Actually it's easy. Just take a marker and divide each side into three 120 degree sectors. [img]/images/graemlins/laugh.gif[/img]

For the methods that depend on heads and tails, worrying about the game never terminating makes no more sense than worrying about the coin landing on its edge. I suggest flipping the coin until it comes up heads. Then:

Rock: First head is on an even numbered flip.

Paper: First head is on an odd numbered flip, and the next flip is also heads.

Scissors: First head is on an odd numbered flip, and the next flip is tails.

[/ QUOTE ]





MMMmmmmmmmmm.


Then 50% will be ROCK, and paper or scissors will be 25% each.

AB

[AlienBoy]

It can be done with double flips.

H=Head T=Tail

If the next TWO flips are:

HH = ROCK

HT = PAPER

TH = SCISSORS

TT = NULL (discarded result)


AB