i have following question from probability theory,problem is stated as follow: suppose there are two player,each player rolls dice one by one,winner will be who's number on dice will be more then 3 otherwise game is continued by the same rule.our task is to find probability of this event by which time game will be finished after each player rolled dice 3 times,please help me how to solve it
The way I am understanding this problem is that player 1 rolls the die, if the number on the die is 4, 5, or 6, the game is over, otherwise player 2 rolls his die and checks to see if the number is 4, 5, or 6. If player 2 did not get a high roll, player 1 goes again. The question is what is the probability that each player rolls the die at least 3 times. The probability of success for each individual roll is 1/2 (Probability of getting a 4, 5, or 6 given that each option of 1, 2, 3, 4, 5, & 6 are equally likely). In order for the game to continue for all 6 rolls, there must have been 6 unsuccessful roles. Thus the probability that it will take more than 6 rolls is the probability that there were 6 failed rolls in a row. This would be (1 - P(Success))^Num Of Rolls which is (1-0.5)^6 which is 0.5^6 which is 1/64 = 0.015625. Now if you are looking at each having 3 failed rolls then on player 1's 4th roll it is a success it would be (P(Success))(1-P(Success))^6 which would be (0.5)(1-0.5)^6 which is 1/128 = 0.0078125. I hope this helps, if I misunderstood the question, let me know!