Call a biased coin a p-coin if it comes up heads with probability p and tails with probability 1-p. We say that a p-coin simulates a q-coin if by flipping a p-coin repeatedly (some fixed finite number of times) one can simulate the behavior of a q-coin.

For example, a fair coin can be used to simulate a 3/4-coin by using two flips and defining a pseudo-head to be any two-flip sequence with at least one real head. The chance of a pseudo-head coming up is 3/4, so we have simulated a 3/4-coin.

1. Find a rational value p such that a p-coin can simulate both a 1/2-coin and a 1/3-coin, or prove that no such value exists.

2. Find an irrational value p such that a p-coin can simulate both a 1/2-coin and a 1/3-coin, or prove that no such value exists.

(In reply to Simple solution by Old Original Oskar!)

I think there are two problems with your solution, Oskar:

(a) using a 1/p coin, the probablity of head-tails is not equal to the probability of tails-heads, unless p = 2

(b) Any solution which includes tossing again does not satisfy the the requirement the "fixed finite number of times" requirement.

For instance, one could simulate a 1/2 coin with a 1/3 coin, by flipping twice and calling two tails a tail (probability 4/9), calling a mixed result a head (probability 4/9) and flipping again if two heads (probability 1/9).  However, this algorithm, does not involve a "fixed finite number of times" .

I suspect a 1/6 or 5/12 coin, if flipped x times, might yield two sets of mutually exclusive combinations which each have probability 1/2.   I don't have time right now to find it, however.

If a 1/6 or 5/12 coin, when flipped y times, also yielded two sets of mutually exclusive combinations which each have probability 1/3 and 2/3 respectively, then that is our solution to part 1.

Edited on February 18, 2005, 7:02 pm

Edited on February 18, 2005, 9:15 pm

Posted by Steve Herman on 2005-02-18 19:01:39