It is found that A and B play a game 12 times.A wins 6 times,B wins 4 times and they draw twice.A and B take part in a series of 3 games.What is the probability that they will win alternatively.

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The wording of the question is a bit ambiguous. Are A and B playing just a series of 3 games, or are they playing a series of 3 matches composed of 12 games each?

If the former, then the possible sequences are ABA and BAB. Now given that A wins a game with probability 1/2, B wins a game with probability 1/3 and they draw with probability 1/6, the two sequences occur with respective probabilities (1/2)(1/3)(1/2) = 1/12 and (1/3)(1/2)(1/3) = 1/18, which add to a probability of 5/36 that they will alternate.

If they play 3 matches of 12 games each, then it gets more complicated. I would assume that winning a match entails winning 7 or more games, so there would be more sequences to analyze. I can try doing this if you think this is how the question is to be read, but at first glance I think the first interpretation is the mostly likely one.

@Abhijeet Verma
–
You make a good point, but I think that the default for the term "alternating" is that we are looking for sequences that only involve outright wins without any draws. Your interpretation is valid, though, so it would also need to be clarified what is meant by "alternating" ,(or "alternatively"), before this question can be resolved.

dont know the answer, according to me it should be zero, as winning alternatively is not possible as in 36 games A wins 18 times,B wins 12 times and 6 draws.6 draws can occur in the beginning or at the end,and remaining 30 games A wins 18 of them.If A starts the win then B follows, this will continue upto 25 games and then only A can win. so my answer is zero...,what was ur approach

For three times the question became a bit easy.
I took into account the following possibilities-
ABA , BAB , DAB , DBA , ADB , BDA , ABD , BAD (where A , B , D stand for A winning ,B winning, and match drawn.)
The probability for ABA is $1/2\times 1/3\times 1/2$
For BAB is $1/3\times 1/2\times 1/3$
For the rest six cases the probability is $1/6\times 1/2\times 1/3$
Adding the eight cases, we get $1/12+1/18+6\times 1/36=11/36$

For 36 games winning alternatively for the whole series may not be possible but suppose we have to consider them winning alternatively for short spells , like (ABAB) then starting from (BAB)...in this manner.

damn this ques firstly its a bit confusing and secondly i dont have its answer, i believe now that only senior members can help, hope u share this note so that i cud get help...

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TopNewest@Kyle Finch sir ,@Brian Charlesworth sir plz help...

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The wording of the question is a bit ambiguous. Are A and B playing just a series of 3 games, or are they playing a series of 3 matches composed of 12 games each?

If the former, then the possible sequences are ABA and BAB. Now given that A wins a game with probability 1/2, B wins a game with probability 1/3 and they draw with probability 1/6, the two sequences occur with respective probabilities (1/2)(1/3)(1/2) = 1/12 and (1/3)(1/2)(1/3) = 1/18, which add to a probability of 5/36 that they will alternate.

If they play 3 matches of 12 games each, then it gets more complicated. I would assume that winning a match entails winning 7 or more games, so there would be more sequences to analyze. I can try doing this if you think this is how the question is to be read, but at first glance I think the first interpretation is the mostly likely one.

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Sir, why can't we consider ADB , BDA , DAB , DBA , ABD , and BAD . The winning of matches is still alternate.

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According to me the answer should be $11\div 36$

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can u explain in brief and post ur method

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is it correct?

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dont know the answer, according to me it should be zero, as winning alternatively is not possible as in 36 games A wins 18 times,B wins 12 times and 6 draws.6 draws can occur in the beginning or at the end,and remaining 30 games A wins 18 of them.If A starts the win then B follows, this will continue upto 25 games and then only A can win. so my answer is zero...,what was ur approach

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Sorry, actually I gave the answer thinking that they play only 3 times rather than 36 times. (Missed the word "series" )

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Yeah, winning alternatively in the whole series doesn't seem to be possible.

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For three times the question became a bit easy. I took into account the following possibilities- ABA , BAB , DAB , DBA , ADB , BDA , ABD , BAD (where A , B , D stand for A winning ,B winning, and match drawn.) The probability for ABA is $1/2\times 1/3\times 1/2$ For BAB is $1/3\times 1/2\times 1/3$ For the rest six cases the probability is $1/6\times 1/2\times 1/3$ Adding the eight cases, we get $1/12+1/18+6\times 1/36=11/36$

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Each game is played 12 times and they play 3 such games so makes it 36, i guess

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For 36 games winning alternatively for the whole series may not be possible but suppose we have to consider them winning alternatively for short spells , like (ABAB) then starting from (BAB)...in this manner.

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damn this ques firstly its a bit confusing and secondly i dont have its answer, i believe now that only senior members can help, hope u share this note so that i cud get help...

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Also see this note and provide the necessary help. https://brilliant.org/discussions/thread/coprimes-confusion/?ref_id=671639

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Answer should be 5/36 under the following assumptions : a) They play 3 games (not a series) b) Draws are not allowed ( in playing alternatively)

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