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I've seen recently this question on another site and gave it a try.

There are $3$ players and a dice with $3$ unbiased faces let's call them $a$, $b$ and $c$. The first player bets on the face $a$, the second player on $b$ and the third player on $c$, they already played and rolled the dice and if the score between the three players is already $2:1:1$ what is the probability that the first player takes $3$ wins (first)?

So I've decided to list down all posibilities, I will denote with $x$ the case that the first player win, $y$ the case that the second player wins and $z$ the case that the third player wins. So the possible outcomes are:

\begin{array}{|c|c|c|} \hline xxx & yyy & zzz\\ \hline xyy & \color{red}{yxx} & \color{red}{zxx}\\ \hline xyx & \color{red}{yxy} & \color{red}{zxz} \\ \hline xxy & yyx & zzx\\ \hline xzz & yzz & zyy\\ \hline xzx & yzy & zyz\\ \hline xxz & yyz & zzy \\ \hline xyz & \color{red}{yxz} & \color{red}{zxy}\\ \hline xzy & yzx & zyx\\ \hline \hline \end{array}

Now from the first column, there are $9$ wins from the first player, from the second one and third I coloured them by red and in total $9+6=15$ favourable cases for the first player to get $3$ victories first. So the probability is simple: $$P(1^{st})=\frac{15}{27}=\frac59$$ Is it correct? Can it be done in a nicer way?

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  • $\begingroup$ (1) How do you assure the "distinct choice than the other 2" condition? Are the bettors colluding? If not, what mechanism assures they bet on different choices? By contemplating that issue, you might be able to (greatly) simplify the analysis. (2) What is a "lead"? Would it be the total number of won bets (as suggested by your analysis) or would it be the difference between the number of won bets and the largest number of won bets among the other two players (as suggested by the usual meaning of "lead")? $\endgroup$ – whuber Feb 11 at 15:33
  • $\begingroup$ I have edited it a little, hope that clears the missunderstanding. $\endgroup$ – user227352 Feb 11 at 15:42
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    $\begingroup$ Your solution is almost correct but you forgot to count the events $yzx$ and $zyx$. $\endgroup$ – Jarle Tufto Feb 11 at 16:28
  • $\begingroup$ Generally speaking, when one says a die or dice in probability and statistics, it is assumed to be a standard cubical object with six faces. There can be adjectives such as loaded or unfair indicating that the six faces are not equally likely to show up or tetrahedral indicating a pyramid-shaped die with 4 triangular faces (dreadful physical model since it does'nt roll easily) or octahedral or dodecahedral etc. But a die with three faces? It's got to be a cylinder and how one can make it fair is something I have no intuition about. $\endgroup$ – Dilip Sarwate Feb 12 at 16:04
  • $\begingroup$ @DilipSarwate Indeed, I made up that description (I had to chose between a coin with $3$ faces and a dice with $3$ faces), but I hope it's clear where I want to arrive at. If it's that is misleading one can think like there's a normal dice with two faces named $a$, two faces named $b$ and two faces named $c$. $\endgroup$ – user227352 Feb 12 at 16:26
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The first player reaches 3 wins first if the next events are $x$, $yx$, $yzx$, $zx$ or $zyx$. These events are disjoint so to the the overall probability is the sum of the probability of each event, that is, $1/3 + 1/9 + 1/27 + 1/9 + 1/27=17/27$.

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You can also look at it as 1 minus the probability that the first player does not win. On the first roll, 2/3 rolls result in the game continuing (player one has not won yet). On the second roll 1/3 rolls result in a non-player 1 player winning. There is also 1/3 rolls where the game continues. On the last roll, there is 2/3 rolls where player one does not win.

So we have 1-(2/3*1/3 + 2/3*1/3*2/3) = 1-(2/9 + 4/27) = 17/27

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