Source: Princeton Review
Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?
A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3
The OA is C
Three children, John, Paul, and Ringo, are playing a game.
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The main concept here is that all 3 children are equally likely to win this game (unless one of them possesses supernatural powers that allow him to know what numbers the other two boys will choose )BTGmoderatorLU wrote:Source: Princeton Review
Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?
A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3
The OA is C
Also note that, if everything is random, the probability of winning by choosing the number 2 is the same as the probability of winning by choosing the number 1.
So, regardless of what number Ringo chooses, his probability of winning is exactly the same as each of the other boys winning.
Since all 3 boys have the same probability of winning, P(Ringo wins) = 1/3
Likewise, P(John wins) = 1/3 and P(Paul wins) = 1/3
Answer: C
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Brent
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Brent´s solution explores symmetries and it is probably the GMAT-scope way of dealing with this good question.BTGmoderatorLU wrote:Source: Princeton Review
Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?
A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3
In the same "vibe", I prefer the following argument:
1. Ringo´s fixed preference is irrelevant once it is known. (I mean, if he always chooses 1 is exactly the same problem, hence the answer cannot be different.)
2. John and Paul have no edges between them (sure!) and each one has no edge facing Ringo (and vice-versa) . Let´s see from (say) Paul´s perception:
> If John chooses like Ringo (John chooses 2), Paul may win (choosing 1) or may let them go to the next round (choosing 2) :: there is an edge of Paul over Ringo.
> If John does not choose like Ringo (John chooses 1), Paul may be beaten by Ringo (Paul choosing 1) or he may be beaten by John (Paul choosing 2) :: there is an edge of Ringo over Paul.
In short: Paul has no edge over Ringo (and vice-versa), and the same reasoning applies to conclude that John also does not have an edge over Ringo (and vice-versa).
Conclusion: all of them have the same probability of winning the game, hence 1/3 is the correct answer.
Now let´s see a much clearer solution, but in the end I will use the sum of a convergent geometric sequence with infinite number of terms... that´s out of GMAT´s scope...
$${\rm{Total}}\,\,\left( {R,J,P} \right) = \left( {2,J,P} \right)\,\,:\,\,\,1 \cdot 2 \cdot 2 = 4\,\,{\rm{equiprobable}}\,\,{\rm{possibilities}}\,\,{\rm{per}}\,\,{\rm{round}}\,\,\,\left( * \right)$$
$${\left( {R,J,P} \right)_N}\,\,\,{\rm{indicates}}\,\,{\rm{choices}}\,\,{\rm{in}}\,\,{\rm{round}}\,\,N$$
$${\rm{Favorable}}\,\,:\,\,\,\left\{ \matrix{
\,{\left( {R,J,P} \right)_1} = \left( {2,1,1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{1 \over 4} \hfill \cr
\,{\left( {R,J,P} \right)_1} = \left( {2,2,2} \right)\,\,\,{\rm{and}}\,\,\,\,{\left( {R,J,P} \right)_2} = \left( {2,1,1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{1 \over 4} \cdot {1 \over 4} \hfill \cr
\,{\left( {R,J,P} \right)_1} = \left( {2,2,2} \right)\,\,\,{\rm{and}}\,\,\,\,{\left( {R,J,P} \right)_2} = \left( {2,2,2} \right)\,\,\,{\rm{and}}\,\,\,{\left( {R,J,P} \right)_3} = \left( {2,1,1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{1 \over 4} \cdot {1 \over 4} \cdot {1 \over 4} \hfill \cr
\, \ldots \hfill \cr} \right.$$
$$? = {1 \over 4} + {\left( {{1 \over 4}} \right)^2} + {\left( {{1 \over 4}} \right)^3} + \ldots = {{{1 \over 4}} \over {1 - {1 \over 4}}} = {1 \over 3}$$
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Hi All,
We're told that three children - John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number DIFFERENT from those of the other two children, he is declared the winner. If all of the children choose the SAME number, the process repeats until one child is declared the winner. We're told that Ringo ALWAYS chooses 2 and the other children select numbers randomly. We're asked for the probability that Ringo is declared the winner. This question can be approached in a number of different ways. Since the number of possible outcomes is so small, you can actually just list them all out to determine the answer to this question:
Ringo ALWAYS chooses 2...
IF>..
John = 1 and Paul = 1, then Ringo WINS
John = 1 and Paul = 2, then Ringo loses
John = 2 and Paul = 1, then Ringo loses
John = 2 and Paul = 2, then the game is REPLAYED.
We're told that the game is played (and replayed) until there is a winner, so there really aren't 4 outcomes - there are only 3 outcomes. Since each winning outcome is equally likely, Ringo will win 1/3 of them.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
We're told that three children - John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number DIFFERENT from those of the other two children, he is declared the winner. If all of the children choose the SAME number, the process repeats until one child is declared the winner. We're told that Ringo ALWAYS chooses 2 and the other children select numbers randomly. We're asked for the probability that Ringo is declared the winner. This question can be approached in a number of different ways. Since the number of possible outcomes is so small, you can actually just list them all out to determine the answer to this question:
Ringo ALWAYS chooses 2...
IF>..
John = 1 and Paul = 1, then Ringo WINS
John = 1 and Paul = 2, then Ringo loses
John = 2 and Paul = 1, then Ringo loses
John = 2 and Paul = 2, then the game is REPLAYED.
We're told that the game is played (and replayed) until there is a winner, so there really aren't 4 outcomes - there are only 3 outcomes. Since each winning outcome is equally likely, Ringo will win 1/3 of them.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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Since Ringo always chooses the number 2, there are only 4 possible outcomes for any round of the game. We list Ringo's choice first, followed by John's and then Paul's:BTGmoderatorLU wrote:Source: Princeton Review
Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?
A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3
(2, 1, 1) Ringo wins
(2, 1, 2) Ringo loses (John wins)
(2, 2, 1) Ringo loses (Paul wins)
(2, 2, 2) No one wins, so another round is played. When this happens, the same 4 outcomes occur for the next round and for subsequent rounds.
We see that in any given round, each boy has a 1/3 chance of winning. It doesn't matter if Ringo always chooses 2 or just randomly chooses a number, his probability of winning is 1/3.
Answer: C
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