*There are 24 letters in the Greek alphabet, from alpha, beta, and gamma, through chi, psi, and omega. The names of the letters are of different lengths, from short two-letter names like mu or pi through the seven letters in omicron. We want to generate a random Greek letter, where the probability of a letter is determined by the length of its name, and the following:*

Psi is a special case with a probability of zero (we are actually generating one of 23 letters) Letters whose names are composed of an even number of letters have a probability of 2^length times the probability of the letter Tau

All the other letters (except Psi and Tau) have a probability of 2^(length-2) times the probability of Tau So, for example, the letter mu (with a length of two letters) should appear 2^2 = 4 times more than tau. Omicron (with a length of seven letters) should appear 2^(7-2), or 32 times more often than tau.

The task is to generate the random Greek letter using six coin tosses, in which the coins are chosen from two types: one with a probability of p_1 of getting a tails and 1-p_1 of getting heads; and the other with a probability of p_2 for tails and 1-p_2 for heads.

Psi is a special case with a probability of zero (we are actually generating one of 23 letters) Letters whose names are composed of an even number of letters have a probability of 2^length times the probability of the letter Tau

All the other letters (except Psi and Tau) have a probability of 2^(length-2) times the probability of Tau So, for example, the letter mu (with a length of two letters) should appear 2^2 = 4 times more than tau. Omicron (with a length of seven letters) should appear 2^(7-2), or 32 times more often than tau.

The task is to generate the random Greek letter using six coin tosses, in which the coins are chosen from two types: one with a probability of p_1 of getting a tails and 1-p_1 of getting heads; and the other with a probability of p_2 for tails and 1-p_2 for heads.

One solution has p_1 = 8 / 17

and p_2 = 1 / 3

We can represent the solution as a circle. We divide up the circumfrence of the circle such that the probability is proportional to the length of the section of the cicumfrence. Then we can present a solution as an image:

## No comments:

Post a Comment