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Three players a,b,c take turns in a game according to the following rules:

At the start A and B play (so C does not play). The winner of the first trial plays against C and so on until one of the players wins two trials in a row. Possible outcomes are aa,acc,acbb,acbaa, bb,bcc,bcaa,bcabb etc.

We have to prove that probability of A winning equals p(A) = 5/14, p(B) = 5/14, p(C) = 2/7

I have been stuck on this problem for a long time. The only thing I have been able to find out so far is that C can never win on even turns.

Re-edit

Players continue to play until one of them wins two times consecutively and all players are equally good at playing the game.

Apologies for leaving such crucial information.

I have finally solved this problem with a different approach

Sample space

aa ,acc ,acbb ,acbaa ,acbacc ,acbacbb ,acbabcaa ,...

bb ,bcc ,bcaa ,bcabb ,bcabcc ,bcabcaa ,bcabcabb ,...

Each point in the sample space has a probability (1/2^k) associated with it where k is the number of turns.

For example probability of point (aa) = 1/4

Now let us construct a table enumerating probabilities - Table of probabilities

Let us consider the event that C wins overall. But C can only win if k = 3,6,9,12,...

P(C) = P(C3) + P(C6) + P(C9) ......

where P(Ck) = Probability of C winning overall after k turns.

P(C) = 1/4 + 1/32 .....

P(C) = a / (1 - r) ( Sum of an infinite GP )

a = 1/4 ( First Term )

r = 1/8 ( Ratio )

P(C) = 2/7

P(A) = 5/14 = P(B)

zer0cube
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    How can $acbb$ be a possibile outcome? That would mean there are more than 2 trials. The way I read the problem, there should only be 2 trials in total. – dreamer May 15 '13 at 13:46
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    Presumably the game goes on until one player wins two trials in a row? – Dilip Sarwate May 15 '13 at 13:46
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    Is this all of the question, or are some of these players better than the others? – AnonymouseCat May 15 '13 at 13:46
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    @DilipSarwate Either that, or there are just 2 trials. The question is very much ambiguous. – dreamer May 15 '13 at 13:47
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    You haven't told us when play stops, nor what the probability of, say, A beating B is in any given trial. Question unanswerable without much further information. – Gerry Myerson May 15 '13 at 13:47
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    How many times will this game be played? – AnonymouseCat May 15 '13 at 13:47
  • Presumably the winner stays each round and plays the one who sat out. $C$ can win on any turn after the first, as some of your examples show. – Ross Millikan May 15 '13 at 13:47
  • @CameronBuie, I am not sure whether your edit preserves the content of the original question. We first need a confirmation from zerOcube to be sure. – dreamer May 15 '13 at 13:51
  • I have added the condition for the game ending that I have inferred from your examples. If this is incorrect, please include the correct conditions. – Cameron Buie May 15 '13 at 13:52
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    I may have found the actual question --- see my "answer". – Gerry Myerson May 15 '13 at 13:52
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    @cruise: True, that may not be what was intended, but it seemed a reasonable inference, based on the listed example outcomes. – Cameron Buie May 15 '13 at 13:53
  • @CameronBuie It seems as if you were right after all :). I just wasn't sure, but your guess was indeed correct :). – dreamer May 15 '13 at 13:54
  • The question needs a better title – Hagen von Eitzen May 15 '13 at 15:45
  • One clarification in the above solution: there are two paths for each k where c always wins. Each has probability $1/2^k$. So the probability for c winning at such a level is twice that: $1/2^{k-1}$. For example k=3 has probability of c winning as 1/4. – Anand Mar 11 '22 at 18:38

3 Answers3

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First, let us assume $A$ wins the first game. Let $a$ be the probability that $A$ wins overall, $b$ the probability that $B$ wins overall, and $c$ the probability that $C$ wins overall. Then we can write (in this case) $a=\frac 12 + \frac b2$ because $A$ either wins the second game (and wins overall) or loses the second game and is now in $B$'s position. Similarly, $c=\frac 12a$ because if $C$ wins his first game he is in $A$'s position, while if he loses his first game $A$ wins overall. Finally $b=\frac 12c$ because $B$ needs $C$ to win and he is now coming in, which is $C$'s position. This gives $$a=\frac 12 + \frac b2\\c=\frac a2\\b=\frac c2\\a=\frac 12+\frac c4==\frac 12+\frac a8=\frac 47\\c=\frac 27\\b=\frac 17$$ This is correct for $C$, but we assumed $A$ won the first one. Clearly $A$ and $B$ have the same winning probability at the start so we can split their total chances evenly, giving $$P(A)=\frac 5{14}, P(B)=\frac 5{14},P(C)=\frac 27$$

Ross Millikan
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Perhaps the problem is supposed to be,

3 players A, B and C take turns at a game according to the following rules. At the start A and B play while C is out. The loser is replaced by C and at the second trial the winner plays against C while the loser is out. The game continues in this way until a player wins twice in succession, thus becoming the winner of the game. What is the probability that Player A will be the winner? What is the probability that player C will be the winner? (assume in each match, A has .6 chance to beat B, and .4 chance to beat C, B has .5 chance to beat C.)

which I found at http://www.chegg.com/homework-help/questions-and-answers/3-players-b-c-turns-game-according-following-rules-start-b-play-c--loser-replaced-c-second-q2427231

(and, yes, I know this is not an answer, but I don't think it would fit as a comment).

Gerry Myerson
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  • A variant is at http://www.chegg.com/homework-help/questions-and-answers/players-albert-bonnie-charles-turns-game-chess-according-following-rules-start-trial-alber-q53131 – Gerry Myerson May 15 '13 at 13:54
  • The OP has added info to the question. – Cameron Buie May 15 '13 at 13:55
  • Yes that is the question I want to ask however in my case all players have equal chances of winning a single match against another player. – zer0cube May 15 '13 at 13:56
  • So I think your question is the one in my comment on my answer. – Gerry Myerson May 15 '13 at 14:04
  • @GerryMyerson : Yes the link you have provided in the comment points to the question that I want to ask however I need subscription at chegg.com to view the answer which I do not have. – zer0cube May 15 '13 at 14:12
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Here's my crack at the question:

Since both players $A,B$ satisfy:

$$M_k(p) = \{\{1, 1\},\{1,0,1,1\},\{1,0,1,0,1,1\},...\}$$ where $M_k(p)$ is the $k$th way for player $p$ to win the game. $\{1\}$ indicates winning the trial, and $\{0\}$ indicates loss.

As player C, and only $C$, satisfies; $$N_k(p) = \{\{0,1, 1\},\{0,1,0,1,1\},\{0,1,0,1,0,1\},...\}$$

Trials needed to have elapsed for $C$ to win the game is $\mid N_1(C)\mid = (\mid M_1(A) \mid +\mid M_1(B)\mid)-1 = 3$

Thus, chances of player $p$ winning the game in this case is:

Define players $A,B,C$ as $0,1,2$

$t=$Trials needed to win the game.

$n=$Number of players.

$k=$Number of players with the same chances of winning($A,B$), .

$$ P(p) = \left\{\begin{matrix} {kt-1 \over k((M_1(p)\;t)+1)} & p < 2\\ {t \over (N_1(p)\;t)+1} & p = 2 \end{matrix}\right. ,\quad -1 < p < n $$

Where $P(p)$ is the probability of player $p$ winning.

Edit:

To clarify: The reason that the chance of player $C$ winning is highest, is because there is a greater chance that player ($A,B$) won't win because they share the same set of combinations needed to win.

Disclaimer: I have not tested this thoroughly in any way, and drafting was started prior to question being answered.

JohnWO
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