# Backgammon game competition.

Two players $A$ and $B$ compete in a certain backgammon game. The competition is conducted in "Rounds" where each "Round" consists of $5$ single backgammon games. A round is won by the player who has more winnings in $5$ single games and final winner is the one who will first earn $12$ rounds.

Due to previous experience the probability that player $A$ wins in a single game against $B$ is $0,6$ but in order $A$ to challenge his opponent $B$, he offers him $5$ rounds won at the beginning of the competition. Is $A$ still reasonable that he will win the final prize of the competition? More precisely, what is the probability that $A$ will win?

I computed that \eqalign{ P[A\ win\ a\ round] &= \binom{5}{3}0.6^3\cdot 0.4^2 + \binom{5}{4}0.6^4\cdot 0.4^1 \binom{5}{5}0.6^5\cdot 0.4^0 \cr &= 0.68256 \cr }

• The formula is ok, but you now need to compute P(A is final winner) – true blue anil Sep 13 '15 at 8:35
• I know. I had model a solution a long time back. If I remember correctly I used the "Banach's matchbox problem" as a model. – nickchalkida Sep 13 '15 at 8:42
• See my answer. You only need the binomial distribution again. – true blue anil Sep 13 '15 at 8:46
• If you mean the formula $\binom{18}{12}p^{12}\cdot (1-p)^6 + \cdots \binom{18}{18}p^{18} \cdot (1-p)^0$, somehow I disagree – nickchalkida Sep 13 '15 at 8:59

Now use $p = 0.68256$

For A to be the final winner, she must win $\ge12$ of $18$ rounds.

You already know the formula to use.

ps

I had forgotten the Davis Cup analogy, where you play best of 5 matches ti the bitter end even if a team has won the first 3. So as reminded by Joriki, the simplified formula is

$$\binom{18}0p^{18}+\binom{18}1p^{17}q+\cdots+\binom{18}6p^{12}q^6$$

• I think there is a misunderstanding. The competition may end playing just 12 rounds. – nickchalkida Sep 13 '15 at 8:51
• No misunderstanding, I am putting more explanation in answer as a ps. – true blue anil Sep 13 '15 at 8:57
• @trueblueanil: a) The second coefficient should be $\binom{13}2$ (and the third could be equivalently but more clearly written as $\binom{17}6$). b) This is a needless complication; it's equal to $\binom{18}0p^{18}+\binom{18}1p^{17}q+\cdots+\binom{18}6p^{12}q^6$, so your original suggestion to just use the same approach as in the question was correct and better than the P.S. you added. – joriki Sep 13 '15 at 9:26
• @nickchalkida: It doesn't matter whether the remaining games are played or not. In the calculation in your answer, you assumed that all $5$ games would be played, but the winner is the same if the round stops once the winner is determined, i.e. once one player has won $3$ games. Same here the other way around. "Best of five" and "first to win three" are just two different ways of saying the same thing. – joriki Sep 13 '15 at 9:29
• Your edit made it worse -- $\binom{17}{11}$ was correct (though the equivalent $\binom{17}6$ seems clearer). – joriki Sep 13 '15 at 9:39