Question

Consider a dishonest casino that occasionally uses biased dice. The fair die is used 9 times as often as the biased one, with the biased die made so that 6 is rolled with probability 0.18, and the other five numbers are equally likely as each other. Compute the following quantities.

(i) P(6 is rolled)

(ii) P(biased die used — 6 is rolled)

(iii) P(biased die used — 6 not rolled)

(iv) The number of 6's in a row we would need to see before it is more likely that we had been using the biased die

Answer 1

We are given here that:
P( unbiased) = 9P(biased)

Also, P(biased) + P(unbiased) = 1
Therefore P(biased) = 1/10 = 0.1, and therefore P(unbiased) = 0.9

i) P(6 is rolled ) is computed here as using law of total addition as:
= P(biased)*0.18 + P(unbiased)*(1/6)

= 0.1*0.18 + 0.9*(1/6)

= 0.168

Therefore 0.168 is the required probability here.

(ii) P(biased die used, 6 is rolled)

= P(biased)P(6 | biased)

= 0.1*0.18

= 0.018

Therefore 0.018 is the required probability here.

(iii) P(biased die is rolled, 6 not rolled)

= P(biased)*(1 - P(6 | biased) )

= 0.1*(1 - 0.18)

= 0.1*0.82

= 0.082

Therefore 0.082 is the required probability here.

(iv) Let the number of 6 here in a row required be n. Then the probability of getting a 6 n times in a row is computed here as:
= P(6 n times in a row) = 0.168ⁿ

Probability that biased die is used given that 6 occurs n times in a row is given as:
= 0.1*0.18ⁿ / 0.168ⁿ

= 0.1*1.0714ⁿ

As the above probability needs to be greater than 0.5, we have here:

0.1*1.0714ⁿ >= 0.5

1.0714ⁿ >= 5

Taking natural log both sides, we get the value of n here as:

n = Ln(5) / Ln(1.0714) = 23.32

Therefore 24 continuous times 6 is to be achieved here.