Romeo and Juliet play a dice game in which the two participants take..

You don't need to use the GP formula if you don't want to. Here is how:

\(P(Romeo) = \frac{1}{6} + (\frac{5}{6}*\frac{5}{6}*\frac{1}{6}) + (\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}) + ...\) ... (I)

\(P(Juliet) = (\frac{5}{6}*\frac{1}{6}) + (\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}) + ...\) ... (II)

Substituting (II) into (I), we get

\(P(Romeo) = \frac{1}{6} + (\frac{5}{6})*P(Juliet)\)

But, \(P(Romeo) + P(Juliet) = 1\)
So,
\(1 - P(Juliet) = \frac{1}{6} + (\frac{5}{6})*P(Juliet)\)

\(P(Juliet) = \frac{5}{11}\)

Answer (D)
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(Lets consider the SIMPLE CASE first) the probability of Juliet winning the game considering Romeo looses in first turn and Juliet wins in second turn was \(\frac{5}{6}*\frac{1}{6}\) = \(\frac{5}{36}\)

But what if Juliet does not win in the her first turn but wins in her second turn. Then we have a scenerio where Romeo looses, Juliet also looses, again Romeo looses & finally Juliet wins which translates into \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\)

The total probability that juliet wins in either of her first two chances will be sum of the individual probabilities of she either winning in her first or second chance.

= \(\frac{5}{6}*\frac{1}{6}\)+ \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\)

Similarly:-

The total probability that juliet wins in either of the first THREE chances is

= \(\frac{5}{6}*\frac{1}{6}\) + \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\) + \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6} *\frac{5}{6}*\frac{1}{6}\)

We can continue like this towards infinity & if we take out \(\frac{5}{6}*\frac{1}{6}\) from this series, we realise that this is a geometric progression with common ratio, r = \(\frac{5}{6}*\frac{5}{6}\)

If we add the probabilities of all individual possible events... we have....
\(\frac{5}{6}*\frac{1}{6}\) + \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\) + \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6} *\frac{5}{6}*\frac{1}{6}\) + ..... so on

Note the formula for a sum of GP is:-


Since common ratio of this GP is less than one then \(r^n\) (where n is infinite) means \(r^n\) will become very minute and we can ignore it.


Therefore substituting the values of a = \(\frac{5}{6}*\frac{1}{6}\) and \(r^n\) = 0 and r = \(\frac{5}{6}*\frac{5}{6}\) we get:-

\(\frac{5}{6}*\frac{1}{6}\)*\(\frac{36}{11}\) = \(\frac{5}{11}\) (Ans)

Hence option D is the correct option.
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Official Answer from Veritas prep:

The key to this difficult probability question is focusing not on Romeo’s win probability, not on Juliet’s win probability, but instead on the relationship between the two. Consider Juliet’s chances as the sum of two cases: Either, with probability \(\frac{1}{6}\) , Romeo gets a 6 on the first roll – Juliet loses; or, with probability \(\frac{5}{6}\) , Romeo does not get a 6 on the first roll – Juliet becomes Romeo. Of course, Juliet does not literally become Romeo in the latter case, but the point is that Romeo’s and Juliet’s situations and probabilities are then reversed: Juliet gets to roll with a chance to win on a 6 or else pass the turn to her opponent, so her probability at that moment is exactly whatever Romeo’s win probability was at the beginning of the game.

This may sound messy, but algebraically it’s quite neat:


\(PJ=\frac{1}{6}∗0+\frac{5}{6}∗PR=\frac{5}{6}PR\)


Of course, someone will win the game eventually, which means that the two probabilities are complementary, and we can also write


\(PJ+PR=1\)


Now we have a system of two equations and two variables, and substitution will clean it up nicely:


\(\frac{5}{6}PR+PR=1\)


\(\frac{11}{6}\)PR=1


PR=\(\frac{6}{11}\)
and
PJ=\(\frac{5}{6}∗\frac{6}{11}=\frac{5}{11}\)


This is the correct answer, D.
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This can't be a two-minute question as per the solutions provided!

Bunuel chetan2u
Hi experts, I was wondering if there was an easier way to solve this problem? The solutions given above will take more than 2 mins to solve.

Hi..

I) Logically if you look at the question, you can eliminate 3 choices..
Both have fair chances, so should have almost equal chances. A and B choice give very less probability to Juliet, so eliminate them
But only difference is that Romeo gets the first try, so his probability should be more and both cannot be equal. hence eliminate E
C and D are close but your probability becomes 50% to choose the correct choice.

II) Geometric Progression
It is a GP question and can be solved in a minute if you know the formula for an infinity....
J can fin in his first chance only if R does not win on first chance.. so R can pick any remaining 5 so \(\frac{5}{6}\), then J has to pick 6 so \(\frac{1}{6}\)..P = \(\frac{5}{6}*\frac{1}{6}\)
J can win in his next throw, if the first three throws have had any of other 5 numbers thus \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\)
So P = \(\frac{5}{6}*\frac{1}{6}\)+\(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\)+.....
so r = \(\frac{5}{6}^2\)
formula is \(\frac{a}{1-r}\)=\((\frac{5}{6}*\frac{1}{6})/(1-\frac{5}{6}^2)=(\frac{5}{36})/\frac{11}{36}=\frac{5}{11}\)
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Romeo and Juliet play a dice game in which the two participants take turns rolling a single fair six-sided die. The first player to roll a 6 wins. If Romeo rolls first, what is the probability that Juliet will win?

A) 1/4
B) 1/3
C) 4/9
D) 5/11
E) 1/2

GMATinsight: Sir is there any other way to solve this probability question apart from using GP expression which other experts have also used?

No, I am unable to think any alternative method to solve this other than coming down to the following calculation

For juliet to win

Case 1: Romeo gets a number other than 6 and juliet in first throw gets 6
Probability \(= (5/6)*(1/6)\)

Case 2: Romeo gets a number other than 6 on two successive throws and juliet in first throw gets other than 6 and in second throw gets a 6
Probability \(= (5/6)*(5/6)*(5/6)*(1/6) = (5/6)^3*(1/6)\)

Case 3: Romeo gets a number other than 6 on three successive throws and juliet in first two throw gets other than 6 and in third throw gets a 6
Probability \(= (5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(1/6) = (5/6)^5*(1/6)\)

and so on...

Total Probability \(= (5/6)*(1/6) + (5/6)^3*(1/6) + (5/6)^5*(1/6) + --- = (1/6)*[(5/6)+(125/216)+---] =\)

which eventually requires sum of an infinite GP which may conveniently be called beyond GMAT scope.

Archit3110

Your thinking is always supercool ...... you always think out of the box!! Needless to say that this post deserves a kudos
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Solution:
P(6) = 1/6
P (winning) = 1/6P (losing) = 5/6
Romeo goes first and we need to find P(Juliet winning)
Juliet wil make 2nd, 4th, 6th, 8th, 10th, etc. number of rolls
P (Juliet winning) = P(Juliet winning in 2nd roll) + P(Juliet winning on 4th roll) + P(Juliet winning on 6th roll ) + .......
P(Juliet winning on 2nd roll) = P(Romeo losing on 1st) * p(Juliet winning on 2nd) = 5/6 * 1/6P(Juliet winning on 4th roll) = P(Romeo losing on 1st)* p(Juliet losing on 2nd) * P(Romeo losing on 3rd) + P(Juliet winning on 4th) = 5/6 * 5/6 * 5/6 * 1/6------
P(Juliet winning) = 5/6 * 1/6 + 5/6*5/6*5/6 * 1/6 + ....This becomes an infinite GP with 1st term = 5/6 * 1/6  and common difference = 5/6*5/6
S(Infinite GP) = a/(1-r) = {5/6* 1/6} / {1 - 5/6 * 5/6}= 5/11
Hope this helps!

Possibilities

1st Case

Ro 5/6 Ju 1/6

5/6 x 1/6 = 5/36


2nd Case

5/6 x 5/6 x 5/6 x 1/6 = 5/36 x 25/36


3rd Case

5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 1/6

5/36 x 25/36 x 25/36


And so on. It is obvious the term forms a GP with common ratio 25/36

Probability that Juliet wins


Sum to infinity

S = a/1-r

5/36/1-25/36

5/36 x 36/11

5/11. Answer choice D
https://www.teacheron.com/tutor-profile/8tG7?r=8tG7

Given: Romeo and Juliet play a dice game in which the two participants take turns rolling a single fair six-sided die. The first player to roll a 6 wins.
Asked: If Romeo rolls first, what is the probability that Juliet will win?

The probability that Juliet will win = \(\frac{5}{6}*\frac{1}{6} + \frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6} + (\frac{5}{6})^5(\frac{1}{6}) + ...\\
= (\frac{5}{36}) * 1/(1-\frac{25}{36}) = \frac{5}{36} * \frac{36}{11} = \frac{5}{11}\)

IMO D

Can someone please help me understand, if "the two participants take turns rolling a single fair six-sided die", why are we considering R and J to take more than 1 turn in each case? What I mean to say is, in the first attempt, R gets one chance and J gets one chance so (5/6)*(1/6), understood. Why in the second attempt R and J bot get 2 consecutive chances?

They don't get two consecutive chances.

For the second attempt to exist both R and J need to lose their first attempts, so the second attempt reflects 5/6 chance R loses first, 5/6 chance J also misses first attempt, then R misses with 5/6 on 2nd attempt, etcetera

Posted from my mobile device

I understand now. Thank you for helping.

The probability of Juliet winning after the first round is 5/6*1/6. However, if both Romeo and Juliet fail to roll a 6 in their first attempt, the game continues. The probability for Juliet to win in her next turn can be calculated as follows: P(Romeo not 6)*P(Juliet not 6)*P(Romeo not 6)*P(Juliet 6) = 5/6*5/6*5/6*1/6. This pattern is repeated for each subsequent round.

Hope it's clear.
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Yes it is clear to me now. Thank you very much.