Hi All,
I'm trying to work through this question and debating with a few friends..
A soccer team plays 3 games, its 3 points for a win, 1 for a draw and 0 for a loss.
There are then two friends, A wagers B 5£$ that his team will gain 6 points or more during these three games. What is the percentage chance that A's team will get 6 points or more and A win's the wager?
The percentage chance of winning the first game is 80.6% (1.24 decimal), winning the second game 72.5% (1.38 decimal) and the third 30.8% (3.25 decimal), I understand they aren't independent events but we are treating them as such.
There are then 7 scenarios where there are 6 of more points achieved so the calculation is as follows:
{3,3,3}; {3,3,1}, {3,3,0}; {3,1,3}; {1,3,3}; {3,0,3}; {0,3,3}
However, we exclude the draws because we need 2 x 3 points to win the wager and we are left with {3,3,3} {3,3,0} {3,0,3} {0,3,3}
((1/1.24)(1/1.38)(1/3.25) + (1/1.24)(1/1.38)(1-1/3.25) + (1/1.24)(1-1/1.38)(1/3.25) + (1-1/1.24)(1/1.38)(1/3.25))^(-1) based on the below
((1/a)(1/b)(1/c) + (1/a)(1/b)(1-1/c) + (1/a)(1-1/b)(1/c) + (1-1/a)(1/b)(1/c)}^(-1)
So the answer i'm left with is 1.43 in decimal or 80.645% chance of winning the bet. Is that correct?
Also, 3 3 3 and 3 3 0 would be the same as the bet would have been won on the second win, but then the answer is still the same.
Thanks in advance for helping confirm whether the above is correct.
I'm trying to work through this question and debating with a few friends..
A soccer team plays 3 games, its 3 points for a win, 1 for a draw and 0 for a loss.
There are then two friends, A wagers B 5£$ that his team will gain 6 points or more during these three games. What is the percentage chance that A's team will get 6 points or more and A win's the wager?
The percentage chance of winning the first game is 80.6% (1.24 decimal), winning the second game 72.5% (1.38 decimal) and the third 30.8% (3.25 decimal), I understand they aren't independent events but we are treating them as such.
There are then 7 scenarios where there are 6 of more points achieved so the calculation is as follows:
{3,3,3}; {3,3,1}, {3,3,0}; {3,1,3}; {1,3,3}; {3,0,3}; {0,3,3}
However, we exclude the draws because we need 2 x 3 points to win the wager and we are left with {3,3,3} {3,3,0} {3,0,3} {0,3,3}
((1/1.24)(1/1.38)(1/3.25) + (1/1.24)(1/1.38)(1-1/3.25) + (1/1.24)(1-1/1.38)(1/3.25) + (1-1/1.24)(1/1.38)(1/3.25))^(-1) based on the below
((1/a)(1/b)(1/c) + (1/a)(1/b)(1-1/c) + (1/a)(1-1/b)(1/c) + (1-1/a)(1/b)(1/c)}^(-1)
So the answer i'm left with is 1.43 in decimal or 80.645% chance of winning the bet. Is that correct?
Also, 3 3 3 and 3 3 0 would be the same as the bet would have been won on the second win, but then the answer is still the same.
Thanks in advance for helping confirm whether the above is correct.