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Two teams A and B play an one-day cricket match.The match is in the last-over and the situation of the match is as follows:

Four balls are to be bowled by the team B;no wides or no-balls are bowled;the team A has to score 16 runs to win.Runs that can be scored off a ball are 0,1,2,3,4,5 and 6.Find the probability that the team A will win by scoring in the last ball.


Since the 4 balls are left,so the probability of winning in the last ball has 6 cases.
(1)15 runs in first three balls and 1 run in the last ball.
(2)14 runs in first three balls and 2 run in the last ball.
(3)13 runs in first three balls and 3 run in the last ball.
(4)12 runs in first three balls and 4 run in the last ball.
(5)11 runs in first three balls and 5 run in the last ball.
(6)10 runs in first three balls and 6 run in the last ball.

I let $x_1,x_2,x_3$ as the run in the fourth last,third last,second last balls respectively such that $0\leq x_1,x_2,x_3\leq 6$.
$(1)x_1+x_2+x_3=15\Rightarrow y_1+y_2+y_3=3$,here $y_1,y_2,y_3\geq 0$ Probability in this case is $\binom{5}{2}\times \frac{1}{6}$

In the same way,taking all six cases,$\binom{5}{2}\times \frac{1}{6}+\binom{6}{5}\times \frac{1}{6}+\binom{7}{6}\times \frac{1}{6}+\binom{8}{7}\times \frac{1}{6}+\binom{9}{8}\times \frac{1}{6}+\binom{10}{9}\times \frac{1}{6}$

But my answer is coming wrong.Correct answer given is $\frac{405}{7^4}$.Is my logic totally wrong.Where have i gone wrong.Please help me.

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Assuming the second to sixth terms in your final equation are just typos, there are two mistakes in your solution.

(1) Your understanding of the question is wrong. This is not really a mathematical error though. $A$ wins for any score greater than or equal to $16$ instead of only $16$ due to the nature of a scoring based game.

For example, your first case "15 runs in first three balls and $1$ run in the last ball." should actually be "15 runs in first three balls and $1$ ~ $6$ run in the last ball." And so on.

(2) The way you let $y = 6-x$ is a great way of simplifying calculations, however you forgot to restrict $y \leq6$ as well for the last two cases.

When $y_1+y_2+y_3=7$ or larger $y$ is actually possible to be 7 and generate negative $x$ so you need to substract these cases.

For sum of $y$ = $7$, there are {$7,0,0$} {$0,7,0$} {$0,0,7$}, three cases.
For sum of $y$ = $8$, there are permutations of {$8,0,0$} and permutations of {$7,1,0$}, totally nine cases.

Overall there are

${5\choose2}\times6 +{6\choose2}\times5 + {7\choose2}\times4 + {8\choose2}\times3 + ({9\choose2}-3)\times2 + ({10\choose2}-9)\times1$ = 405 cases.

Hence the possibility is $\frac{405}{7^4}$.

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