turbowagon
Member
In case anyone was curious:
An easy way to compute the chances of winning AT LEAST ONE of the remaining contests, is to compute the chances of winning NONE of the remaining contests, and subtracting that probability from 1.
There are 13 remaining drawings.
2 winners each day.
Assume 120 correct entrants.
The probability of LOSING each drawing is 118/120 or 98.3%
The probability of LOSING all the remaining 13 drawings is: 98.3 ^ 13,
or 80.4%
Therefore, the probability of winning AT LEAST ONE of the remaining contests is 1 - 0.804 = 0.196 or about 20%.
Good luck!
An easy way to compute the chances of winning AT LEAST ONE of the remaining contests, is to compute the chances of winning NONE of the remaining contests, and subtracting that probability from 1.
There are 13 remaining drawings.
2 winners each day.
Assume 120 correct entrants.
The probability of LOSING each drawing is 118/120 or 98.3%
The probability of LOSING all the remaining 13 drawings is: 98.3 ^ 13,
or 80.4%
Therefore, the probability of winning AT LEAST ONE of the remaining contests is 1 - 0.804 = 0.196 or about 20%.
Good luck!