Probability of a Match

Since one team win, another team will lose, so we just have to look at one team (in this solution, team India).

Total number of all different situation: $q$ = 7 $P$ 0 + 7 $P$ 1 + 7 $P$ 2 + 7 $P$ 3 + 7 $P$ 4 + 7 $P$ 5 + 7 $P$ 6 + 7 $P$ 7 (7 $P$ 0 means team India has 0 losses, 7 wins, and 7 $P$ 1 means 1 losses and 6 wins, and so on.)

If team India wins 0, 1, 2 games: $p$ = 0

If team India wins 3 games: $p$ = 5 $P$ 1 = 5 ( Treat three consecutive winnings as a group and perform permutation)

If team India wins 4 games: $p$ = 5 $P$ 2 $\times$ 2 - 4 = 16 ( There are two ways of grouping three consecutive winnings and there can only be once 4 consecutive winnings )

If team India wins 5 games: when team India lost in 2nd and 4th games, or in 3rd and 5th games, or in 3rd and 6th games, the 3 consecutive winnings are broken. So $p$ = 7 $P$ 2 - 3 = 18

If team India wins 6 games: $p$ = 7 since no matter which game team India loses, the consecutive winning will be at least 3.

If team India wins 7 games: $p$ = 1

Adding all $p$ will give you 47

Then $p$ + $q$ = 47 + 128 = 175

I use dynamic programming to solve this kind of problem, the state is memo[pos][udh] , pos is the position where you are now , and udh is the number of match already won by india consecutively

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We know that the number of possibility of the game is 2^7=128. We want to find the number of game where india won at most 2 consecutive game , and then subtract it from 128.

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the recursive structure is

IF(udh==2) memo[pos][udh]=f(pos+1,0)

else memo[pos][udh]=f(pos+1,udh+1)+f(pos+1,0);

base case is when pos==7 return 1;

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We build the DP table with bottom up manner

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memo[7][0]=1;

memo[7][1]=1;

memo[7][2]=1;

memo[6][0]=memo[7][0]+memo[7][1]=2;

memo[6][1]=memo[7][0]+memo[7][2]=2;

memo[6][2]=memo[7][0]=1;

memo[5][0]=memo[6][0]+memo[6][1]=4;

memo[5][1]=memo[6][0]+memo[6][2]=3;

memo[5][2]=memo[6][0]=2;

memo[4][0]=memo[5][0]+memo[5][1]=7;

memo[4][1]=memo[5][0]+memo[5][2]=6;

memo[4][2]=memo[5][0]=4;

memo[3][0]=memo[4][0]+memo[4][1]=13;

memo[3][1]=memo[4][0]+memo[4][2]=11;

memo[3][2]=memo[4][0]=7;

memo[2][0]=memo[3][0]+memo[3][1]=24;

memo[2][1]=memo[3][0]+memo[3][2]=20;

memo[2][2]=memo[3][0]=13;

memo[1][0]=memo[2][0]+memo[2][1]=44;

memo[1][1]=memo[2][0]+memo[2][2]=37;

memo[1][2]=memo[2][0]=24;

result=memo[1][0]+memo[1][1]=81;

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128-81=47 47/128 , p+q=175.

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sorry if the solution is messy ^_^

There are $2^7$ = 128 total arrangements since every match is either a win or a loss for each side

Assigning i for team India & a for team Australia, 3 consecutive winning for team India is form of:

• iiixxxx with $\frac {1}{2} ^ 3 \times 128$ = 16 arrangements
• aiiixxx with $\frac {1}{2} ^ 4 \times 128$ = 8 arrangements
• xaiiixx with $\frac {1}{2} ^ 4 \times 128$ = 8 arrangements
• xxaiiix with $\frac {1}{2} ^ 4 \times 128$ = 8 arrangements
• xxxaiii with $\frac {1}{2} ^ 4 \times 128$ = 8 ${minus}$ 1 case of iiiaiii (already counted) mean 7

Total number of arrangement is 47, with probability $\frac {47}{128}$ $\Rightarrow$ $\boxed{p + q = 47 + 128 = 175}$