0
$\begingroup$

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.

enter image description here

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

$\endgroup$
4
  • $\begingroup$ The formula is ok, but you now need to compute P(A is final winner) $\endgroup$ – true blue anil Sep 13 '15 at 8:35
  • $\begingroup$ 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. $\endgroup$ – nickchalkida Sep 13 '15 at 8:42
  • $\begingroup$ See my answer. You only need the binomial distribution again. $\endgroup$ – true blue anil Sep 13 '15 at 8:46
  • $\begingroup$ 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 $\endgroup$ – nickchalkida Sep 13 '15 at 8:59
1
$\begingroup$

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

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

$\endgroup$
9
  • $\begingroup$ I think there is a misunderstanding. The competition may end playing just 12 rounds. $\endgroup$ – nickchalkida Sep 13 '15 at 8:51
  • $\begingroup$ No misunderstanding, I am putting more explanation in answer as a ps. $\endgroup$ – true blue anil Sep 13 '15 at 8:57
  • $\begingroup$ @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. $\endgroup$ – joriki Sep 13 '15 at 9:26
  • 1
    $\begingroup$ @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. $\endgroup$ – joriki Sep 13 '15 at 9:29
  • $\begingroup$ Your edit made it worse -- $\binom{17}{11}$ was correct (though the equivalent $\binom{17}6$ seems clearer). $\endgroup$ – joriki Sep 13 '15 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.