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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

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closed as off-topic by 900 sit-ups a day, glace, Hakim, RecklessReckoner, Kirill Jul 20 at 2:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 900 sit-ups a day, glace, Hakim, RecklessReckoner, Kirill
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I assume that you have already tried to solve this problem; if not, then I strongly suggest you do so. Once you have done so, please add the answers to some of the following questions: What have you tried? Where did you get stuck? Which relevant concepts are difficult for you? What is the context of this problem? –  Amitesh Datta Jul 8 '11 at 12:57
Textbook writers and mathematics instructors are usually fairly precise in describing a problem. Your description of the problem is not very precise. It will be helpful if you quote the problem more exactly. Then people on this site can show you how to approach the problem. –  André Nicolas Jul 8 '11 at 14:36
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1 Answer 1

up vote 2 down vote accepted

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!

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thanks very much –  dato datuashvili Jul 8 '11 at 19:59
So much for @Amitesh's and @user6312's efforts to lead the OP to actually think about this problem (or, more modestly, to state it precisely). –  Did Jul 8 '11 at 20:41
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