Question

In the games of chance, each player throws two unbiased dice and scores the difference between the larger smaller under which arises. Two players compete and one the player wins if and only if he scores at least 4 more than his opponent. Find the probability that neither wins.

Answer 1

Hint: In the given question two dice are thrown. Now try to find sample space and then calculate what are the conditions in which the given condition of the question satisfies i.e. the score differs by at least 4 points of the two opponents. From that conditions find probability for each case and bring all things together to find the final probability.

Complete step-by-step answer:
Let’s first see what is question in brief:
Suppose the first player throws two dice A and B. Then his score is the absolute difference of numbers which face top on the dice.
Eg. In one throw two number are 3 and 5 then his score his \[|3 - 5| = 2\]
Now first player’s score is \[X\] and second player’s score is \[Y\] then one of the two player wins only when \[|X - Y| \geqslant 4\]
Now
Sample space of throwing two dices: \[6 \times 6 = 36\]
Possible scores of a player = difference of two numbers of dices = \[P(x) = \] \[\{ 0,1,2,3,4,5\} \]
So we need to calculate \[P( \leqslant 4)\]
\[P(5) = \{ (1,6),(6,1)\} = \dfrac{2}{{36}}\]
\[P(4) = \{ (6,2),(2,6),(5,1),(1,5)\} = \dfrac{4}{{36}}\]
\[P(1) = \{ (1,2),(2,1),(2,3),(3,2),(4,3),(3,4),(5,4),(4,5),(6,5),(5,6)\} = \dfrac{{10}}{{36}}\]
\[P(0) = \{ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\} = \dfrac{6}{{36}}\]
\[P( \geqslant 4)\] is the probability that anyone player wins
\[P( \geqslant 4) = (P(5)P(1)) + (P(4)P(0)) + (P(5)P(0))\]
\[P( \geqslant 4) = \dfrac{2}{{36}} \times \dfrac{{10}}{{36}} + \dfrac{4}{{36}} \times \dfrac{6}{{36}} + \dfrac{2}{{36}} \times \dfrac{6}{{36}}\]
\[P( \geqslant 4) = \dfrac{{56}}{{1296}} = 0.043\]
Now from probability rule \[P(A) = 1 = P(A')\]
\[P(neither\_wins) = P( \leqslant 4) = 1 - P( \geqslant 4) = 1 - \dfrac{{56}}{{1296}}\]
\[P( \leqslant 4) = 1 - 0.043 = 0.957\]

Note: In the given we are asked to calculate the probability that no one wins. If we try to solve this question by counting all favorable conditions and then calculating probability for all these conditions, then it will be much more time consuming as compared to the method from which we have solved. For such questions try to visualize number test cases for all odd and even conditions. Sample space is defined as the set of all possible outcomes for the given conditions.