My Math Forum Two players A and B play a game of dice .

 Algebra Pre-Algebra and Basic Algebra Math Forum

 October 30th, 2013, 01:23 PM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Two players A and B play a game of dice . Two players A and B play a game of dice . They roll a pair of dice alternately . The player who rolls 7 first wins . If A starts then find the probability of B winning the game ? please help me how I can solve this question
 October 30th, 2013, 01:47 PM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Two players A and B play a game of dice . Well, the probability of rolling a 7 is 1/6, so the probability of not rolling a 7 is 5/6. If A goes first, he must not roll a 7 in order for B to win. So this has a 5/6 chance of occurrence. Then, if B rolls a 7, B wins. This has a chance of 1/6. So the probability of B winning after one round is $\frac{5}{6} \cdot \frac{1}{6}= \frac{5}{36}$. But this is the probability of B winning in any round, given that neither player has won yet. The probability that neither player wins in a given round is $\frac{5}{6} \cdot \frac{5}{6}= \frac{25}{36}$. Hence, the probability that B wins in the nth round is $(\frac{25}{36})^{n-1} \cdot \frac{5}{36}$. The probability that B wins is then $\frac{5}{36}(1 + \frac{25}{36} + (\frac{25}{36})^2 + (\frac{25}{36})^3 + ... )= \frac{5}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{5}{36} \cdot \frac{36}{11} = \frac{5}{11}$.
 October 30th, 2013, 01:59 PM #3 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: Two players A and B play a game of dice . How you got A ) that the probability of rolling a 7 is 1/6 ? B ) the probability that neither player wins in a given round is 25/36 thanks
 October 31st, 2013, 11:37 AM #4 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Two players A and B play a game of dice . A) There are 36 possible rolls for two dice, since there are 6 possibilities for the first die and 6 possibilities for the second die. Of these, there are 6 combinations that yield a sum of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). This gives a $\frac{6}{36}= \frac{1}{6}$ chance of rolling a 7. B) Since there is a 1/6 chance of either player rolling a 7, there is a 5/6 chance that A will not roll a 7, and a 5/6 chance that B will not roll a 7. Since these two events are independent, the probability of both of them occurring in one round is the product of both chances, which is $\frac{5}{6} \cdot \frac{5}{6}= \frac{25}{36}$.
October 31st, 2013, 12:44 PM   #5
Math Team

Joined: Oct 2011

Posts: 9,457
Thanks: 640

Re: Two players A and B play a game of dice .

Quote:
 Originally Posted by r-soy How you got A ) that the probability of rolling a 7 is 1/6 ?
Example:
1st roll : 3
2nd roll: only a 4 will make 7, so 1/6

Works similarly for any 1st roll result: 1-6, 2-5, 3-4, 4-3, 5-2, 6-1

 October 31st, 2013, 03:49 PM #6 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: Two players A and B play a game of dice . thanks all now is clear

 Tags dice, game, play, players

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# two dices thrown if a gets 5 he win probability of b as winner if a starts firstg

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post BenFRayfield Math Events 0 February 20th, 2014 12:39 PM PrayForStan Algebra 3 November 30th, 2012 07:10 AM DamballahWeddo New Users 3 August 4th, 2010 03:31 PM rv74 Algebra 0 October 30th, 2008 05:16 AM PrayForStan Number Theory 0 January 1st, 1970 12:00 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top