Three chessplayers compete in tournament by the following scheme:
Players A and B play match, than winner plays with C, then new winner plays with the loser of the previous match. Tournament ends at the event of two consecutive wins of some player.
It's given, that players are equal in skills, hence outcome of the match should be considered as coin flip.
We have to find winning probabilities of chessplayers A, B and C, given that first match was between A and B, and A scored it.
Any ideas?;
Edit I misread task terribly. It's two wins required, not three. Already made an edit. Now it's easier significantly.
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$\begingroup$ @CharlesNosbig Yeah. It's pure statistical question, so it don't make sense from the chessplayer point of view) It isn't a $\frac 1 {2^3}$, because we have a priori information that player A is one point ahead. And Player C under circumstances might not even receive a chance to play 3 games, so he's probability obviously lower. Concerning your willingness to play for black twice, I believe I have read somewhere that it's the white who have slight advantage, isn't it? $\endgroup$– Dmitri KovalenkoCommented May 11, 2014 at 9:21
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$\begingroup$ @CharlesNosbig If you are going to work on this, review my edit. I just fixed a huge mistake in the question description. $\endgroup$– Dmitri KovalenkoCommented May 11, 2014 at 9:26
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2 Answers
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There are four states to consider, namely $AC$, $CB$, $BA$, and the ending state $E$. Here $XY$ encodes that the next match is between $X$ and $Y$, whereby $X$ has scored immediately before. At the start we are in state $AC$. Denote the probability that $Z$ will win when the game is in state $XY$ by $p_Z(XY)$.
From a figure one easily reads off that one has $$p_A(AC)={1\over2} +{1\over2}p_A(CB),\quad p_A(CB)={1\over2}p_A(BA),\quad p_A(BA)={1\over2}p_A(AC)\ .$$ Solving this system gives $$p_A(AC)={4\over7},\quad p_A(BA)={2\over7},\quad p_A(CB)={1\over7}\ .$$ Using the inherent circular symmetry of the problem we then obtain $$p_B(AC)=p_A(CB)={1\over7} ,\quad p_C(AC)=p_A(BA)={2\over7}\ .$$
answered May 11, 2014 at 10:36
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$\begingroup$ I really wish I had thought of that before I typed up my solution. That would have saved an incredible amount of time. At least, it appears I got the right answer despite the horribly lengthy process. Good work. $\endgroup$– VincentCommented May 11, 2014 at 10:53
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I'm going to assume you meant 'two wins required' and not who.
Assuming A wins the first game:
The probability of A beating C is $.5$. If A wins, the tournament ends and thus A has a 50% chance to win the tournament.
The probability of B: In order for B to have a chance, C needs to win his/her game against A. This will occur 50% of the time. Subsequently, the chance for B to beat C in their first match is 50%, and the same goes for his/her rematch against A. Thus, $(.5)(.5)(.5)=.125$. B has a 12.5% chance.
The probability of C: C will beat A 50% of the time. Then C will beat B 50% of the time. Thus, the probability of C to win the tournament is 25%.
After typing this up I'm realizing that these may just be the probabilities of a player winning on the first time around. Perhaps it doesn't matter though, if it does, that would explain the missing 12.5%. Let's see!
If A losses his/her first time around and B beats C allowing A a second opportunity to win, his/her chance to win the tournament on second time around is $(.5)(.5)(.5)(.5)= .0625$. Or, 6.25%. This is: A losses his/her first game against C, C losses his/her first game to B, then A beats B and C consecutively. Taking this even further we see that the chance for A to win on their third time around would be $(.5)(.5)(.5)(.5)(.5)(.5)(.5)=.0078125$. Or, $0.78125$%. We can recognize a pattern here. Every subsequent round is hit by $(0.5)^3$.
We can see that A winning on a particular round is represented by: $(.5)^1(.5)^{3n}$ where $n=$ the winning round (starting with round $0$). This means if we sum up:
$$\sum_{n=0}^{\infty}(.5)^1(.5)^{3n}$$ we will find that player A has approximately $57.1428$% chance to win.
On to player B!
We saw before that player B had an initial probability of $12.5$%. Now, taking into account this player winning on subsequent rounds, let's find their total probability.
We can use the same idea from player A. B's initial chance to win is $(.5)(.5)(.5)=.125$. The next opportunity to win yields a probability of $(.5)(.5)(.5)(.5)(.5)(.5)=.015625$. Or, $1.5625$%. We can see that player B follows the same pattern except the initial chance is $(.5)^3$ instead of $(.5)^1$. Thus we get:
$$\sum_{n=0}^{\infty}(.5)^3(.5)^{3n}$$ where n = the winning round (starting with round $0$). Summing this up we see that player B has approximately $14.2857$% chance to win.
Finally player C. Well, we know that player C will have a $28.5714$% chance to win, but let's be sure. Player C's initial chance to win was $(.5)(.5)=.25$. Moving onto the subsequent round we see that for C to win on the second opportunity it will be $(.5)(.5)(.5)(.5)(.5)=.03125$. Or, $3.125$%. Continuing with business as usual we see that player C's total chance of winning is represented by:
$$\sum_{n=0}^{\infty}(.5)^2(.5)^{3n}$$ where n = the winning round (starting with round $0$). Summing this up we see that player C has approximately $28.5714$% chance to win, which is just what he was suppose to have!
In conclusion:
Player A has a $57.1428$% chance to win.
Player B has a $14.2857$% chance to win.
Player C has a $28.5714$% chance to win.
A + B + C $\approx .999999 = 1$
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$\begingroup$ Wow. Christian just schooled me and completed that in a few easy steps! $\endgroup$– VincentCommented May 11, 2014 at 10:51
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1$\begingroup$ Actually, I personally like this working a lot more.. $\endgroup$ Commented May 11, 2014 at 12:04