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
Brent´s solution explores symmetries and it is probably the GMAT-scope way of dealing with this good question.
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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