Example: White Token Out Of A Bag

Example: White Token Out Of A Bag#

(MacKay Exercise 3.12)

A bag contains one token, known to be either white or black. A white token is put in, the bag is shaken, and a token is drawn out, which proves to be white. What is now the chance of drawing another white token?

Analytical Solution#

Given that we have picked a white token, the probability that it had already contained another white, thus after adding the white it became WW:

P(WW|W)=P(W|WW).P(WW)P(W)=1×1/2P(W)=1/2P(W)

and the probability that it had a black token beforehand, so after adding the white it became WB is:

P(WB|W)=P(W|WB).P(WB)P(W)=1/2×1/2P(W)=1/4P(W)

We can evaluate the P(W) in the denominator via two ways:

  • By normalization:
    Since there is no other possibility, P(WW|W)+P(WS|W) should be equal to 1. Hence:

1/2P(W)+1/4P(W)=3/4P(W)=1⇒P(W)=34

  • By inference:
    As P(A)=∑iP(A|Bi)P(Bi)
    So, expanding and summing over all the possible cases (WW and WB) yield:

P(W)=P(W|WW).P(WW)+P(W|WB).P(WB)=(1×1/2)+(1/2×1/2)=34

Substituting P(W) in the above equations:

P(WW|W)=1/2P(W)=23
P(WB|W)=1/4P(W)=13

Monte Carlo Solution#

import numpy as np

N = 10000000

A = np.random.randint(0,2,(N,3))
# 0 : Black
# 1 : White

A[:,1] = 1
A[:,2] = 999

A[A[:,0] == A[:,1],2] = 1 # If both are white, outcome will be white
B = A.copy()

# If one is black, one is white, the outcome can be either
B[B[:,2]==999,2] = np.random.randint(0,2,np.sum(B[:,2]==999))

# Filter out the outcomes with white:
W = B[B[:,2] == 1]

num_TOT = W.shape[0]
num_W = np.sum(W[:,0] == 1)
num_B = num_TOT - num_W
print("White: {:d}/{:d} ({:5f})\t|\tBlack: {:d}/{:d} ({:5f})".format(num_W,num_TOT,
                                                                     num_W/num_TOT,
                                                                     num_B,num_TOT,
                                                                     num_B/num_TOT))

White: 5000604/7500747 (0.666681)	|	Black: 2500143/7500747 (0.333319)
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