GATE CSE 2018 | Question: 15

Question

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 ____

Comments

I just want to know whether P and Q are rolling two identical dice, meaning total 4 dice or each are given one identical dice, meaning total 2 dice?

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why probability of tie 1/6 why not 5/6

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because there are 36 elements in the sample space and 6 of them are the same. $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and  (6, 6)$ out of $36$ elements. $6/36 = 1/6$.

Answer 1

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 $= \text{(First Tie)} \ast \text{(Second Tie)} \ast \text{(No Tie)} = \frac{1}{6}\ast \frac{1}{6}\ast \frac{5}{6}= \frac{5}{216} =  0.023$

Comments

I marked it 0.231 :(

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and in hurry I marked 0.032 :(

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I marked it 0.032 :(

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Why must the first and second trials be tie? The question didn't say anything about the outcomes of first and second trials.

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It says independently two identical dice.
so doesn't it means both P & Q are rolling two dice ?.
so we have to deal with sums here from 2 to 12

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@ habedo007 u may read the ques one more time..it will be clear.

@ yogi_p .. independently means p has one dice and q has another dice , their throwing of die will not depend on each other.

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yeah that i got it.
but

Two people, P and Q, decide to independently roll two identical dice

Doesn't here independently means that P & Q both would be throwing 2 dice each.
That's my confusion.
I did same in exam, made cases from sum 2 to 12, wasted a lot of time in it. 

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@gari I get it now. The point lies in the clause "In the case of tie, they throw the die repeatedly until there is no tie" meaning it is just one match, consisting of $n$ trials until one of them wins. So proceeding into 3rd trial will mean the first two trials were tie, and when one wins the third trial, the match ends.

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yes @ habedo007 you got it :)

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Rishab bhai aapki to rank 1 aati fir

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@Digvijay Pandey @Mk Utkarsh why the cases of winning in frst and second trials is not added

i.e. why (5/6) + (1/6)*(5/6) + (1/6)*(1/6)*(5/6) is not correct here??

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because in question it is asking only about third trial. if asked that probability that one of them wins at most third trial.

Answer 2

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

Comments

i think u mistyped trial for first two case

Answer 3

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 4

I Think it's 30/ (36)^2 = .0231