2.12.2013

Bayesian probabilities in the Beads task

Same conclusion with the proper calculation as in Bayes' theorem:
Assume that red is the majority in jar A and green in jar B and the distributions of M and m in the two jars are the inverse of one another and the number of beads are equal in the two jars , e.g. if jar A contains 9 reds and 1 green, jar B has 1 red and 9 greens; if A has 68 reds and 38 greens B has 38 reds and 68 greens.

For the condition of the 9 + 1 versus 1 + 9 distribution:
Probability that the bead is red, given that it is from jar A= 9/10.
Probability that the bead is red, given that it is from jar B= 1/10.

Probability that the bead is red AND it is from jar A= (1/2) . (9/10)
Probability that the bead is red AND it is from jar B= (1/2) . (1/10)

Probability that the bead is from jar A given that it is red 
= Probability that the bead is red given that it is from jar A . Probability that the bead is from jar A / Probability in this condition that any bead is red
Probability that the bead is red given that it is from jar A . Probability that the bead is from jar A / (Probability that a bead is red and it is from jar A + Probability that a bead is red and it is from jar B)
= (9/10) . (1/2)  / [(9/10) . (1/2) + (1/10) . (1/2)]
= (9/10) . (1/2)

Probability that the bead is from jar B given that it is red 
= Probability that the bead is red given that it is from jar B . Probability that the bead is from jar B / Probability in this condition that any bead is red
Probability that the bead is red given that it is from jar B . Probability that the bead is from jar B / (Probability that a bead is red and it is from jar A + Probability that a bead is red and it is from jar B)
= (1/10) . (1/2) / [(9/10) . (1/2) + (1/10) . (1/2)]
= (1/2) . (1/10)
   

Probability that the bead is from jar A given that it is red, briefly P(a)
Probability that the bead is from jar B given that it is red, briefly P(b)
Draw 1
(observation 1)
(9/10) . (1/2)
(1/10) . (1/2)
Draw 2
(observation 2)
[(9/10) . (1/2)]²
[(1/10) . (1/2)]²
Draw n
(observation n)
[(9/10) . (1/2)] to the power of n
[(1/10) . (1/2)] to the power of n

If m is smaller than M and all the quantities m, M and n are positive integers
[M/(m+M) . (1/2)] to the power of n is bigger than [m/(m+M) . (1/2)] to the power of n

Therefore, p(a) is bigger than p(b) for any n.

In our case:
Probability that the bead is from the jar where the beads that are of the same color constitute the majority is greater than the probability that it is from the jar where beads that are of the same color are in the minority, and this is independent from the number of observations.

The probability for either of the two cases will decrease with each draw.
Decrease with every new draw in the smaller probability is bigger than the decrease in the larger probability, i.e. the decrease in the likelihood that the bead is from the jar where it is in the minority is sharper compared to the decrease in the likelihood that the bead is from the jar where it is in the majority. However, this is irrelevant, since the difference between the two probabilities is absolute.

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