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35 votes
Two people, $P$ and $Q$, decide to independently roll two identical dice, each with $6$ faces, numbered $1$ to $6$. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by $P$ and $Q.$ Assume that all $6$ numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to $3$ decimal places) that one of them wins on the third trial is ____
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Best answer
56 votes
56 votes
One of them win in the third trial i.e. first two trial would be Tie and third should not be Tie.

Probability of Tie $=\frac{6}{36}=\frac{1}{6}$
Probability of NO Tie $=1-\frac{1}{6}=\frac{5}{6}$

Winning in the third Tie $= \text{(First Tie)} \ast \text{(Second Tie)} \ast \text{(No Tie)} = \frac{1}{6}\ast \frac{1}{6}\ast \frac{5}{6}= \frac{5}{216} =  0.023$
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10 votes
10 votes
1st trail
Collision in first will have cases 1,1 2,2 3,3 4,4  5,5 6,6 So total 6 cases
Probability  = 6/36  = 1/6

2nd trail
Collision  will have cases 1,1 2,2 3,3 4,4  5,5 6,6 So total 6 cases
Probability  = 6/36  = 1/6

3rd trIal
Anyone of them, wins means no collision
So Probality = 1-1/6 = 5/6
So
As they all are independent cases so the answer is 1/6*1/6*5/6
0.023
8 votes
8 votes

Win in 3rd trial means tie in first 2.

In an event of rolling a die twice, there are 36 possible outcomes and out of which 6 of them make tie.

So, required probability=P(tie)*P(tie)*P(no tie) 

=(6*6*30)/(36*36*36)

=5/(6*6*6)=5/216

=0.0231 (ANS)

Answer:

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