Question

Setup: 3 different people take turns rolling dice. They all get doubles WITHIN 3 ROLLS. NOT IN A ROW. Each person waits their turn to roll. On each persons turn, they each roll doubles WITHIN 3 ROLLS.

GIVEN THE FOLLOWING CONDITIONS:

- There is a 25% chance that the dice are loaded so that:  Each die is loaded so that 6 comes up half the time P(6) = .5 for each die. The other numbers (1-5) are equally likely to show up. ( P(1) = P(2) = P(3) = P(4) = P(5) = .1)

- Otherwise (75 % chance): The dice are normal, fair, and un-weighted.

QUESTION : WHAT IS THE PROBABILITY THAT THEY ARE USING THE LOADED DICE? (All dice are the same for all experiments. They either use the loaded OR the fair dice for all rolls).

Again, they each TAKE TURNS to roll a pair of dice up to 3 times. They each get doubles WITHIN 3 TURNS. NOT IN A ROW. **What is the probability that they are using the loaded dice?**

**Please show your work and note the names of any theorems used**

Answer 1

*Probability of getting doubles once on a fair dice = 6/36 = 1/6 (We got 6/36 because out of 3 combinations of numbers only 6 pairs make double. That are 1,1 2,2 3,3 ...6,6)

Probability of getting doubles 2 times(not in a row) within 3 rolls= P = (6/36)*(30/36)*(6/36) = 5/216 = 0.0232

*Probability of getting doubles in a loaded dice = probability of getting (1,1)+probability og getting (2,2)+..........+probability of getting(6,6) = PL = 0.1*0.1+0.1*0.1+0.1*0.1+0.1*0.1+0.1*0.1+0.5*0.5 = 0.3

Probability of getting doubles 2 times(not in a row) within 3 rolls(with loaded dice) = 0.3*(1-0.3))*0.3 = 0.063

Therefore probability that it is loaded dice = = 0.063/(0.063+0.0232) = 0.731