Probability of winning at least 1 game out of 5 games.

Question

I have a 1/20 chance of winning a game. I play 4 more separate independent games of each other which has the same probability of winning 1/20 chance. What is the probability I win at least one game out of these 5 games?

Answer 1

Sounds like the binomial distribution.

Formula is (N choose K) * p^k * ( 1-p)^n-k

 

where N is the number of trials, in this case n=5

 

p is the probability of success, which in this case is the probability of winning the game, so p = 1/20

 

and (n choose k) = n!/(k!)(n-k)!

 

The probability of winning at least one game is the same as 1 - probability of winning zero games

To see this, using the fact that Prob(0) + Prob(1) + Prob(2) + Prob(3) + Prob(4) + Prob(5) =1

 

                                                        Prob(1) + Prob(2) + Prob(3) + Prob(4) + Prob(5) =1 - Prob(0)

 

where the left side of this equation is the probability we want, and it is equal to 1 - probability of zero wins.

So instead, we will calculate the probability of losing every game, and then subtract that result from 1. 

                               

k is the number of successes, which in this case is the number of games we want to win, so we want k=0.

 

Now we plug the numbers into the formula.

(N choose k) = (5 choose 0) = 5!/0!5! = 1/0! = 1/1 = 1

 

p^n = (1/20)^0 = 1

 

So it comes down to this:

(1-p)^ N-k = (1 - 1/20)^(5-0) = (19/20)^5 =  2476099/3200000 <-- probability of losing all 5 games

 

So the probability of winning at least one game is 1 - 2476099/3200000 = 

 

   3200000 - 2476099         723901

   ------------------------- =   -------------  which is approximately .2262190625 or roughly 22.6%

           3200000                  3200000

 

 

I simulated this experiment in excel.

The formula =randbetween(1,100) will generate a whole number between 1 and 100 randomly.

The copy and paste this formula down to four more rows.

Using conditional formatting to highlight which if any of the 5 cells are between 41 and 45.

To repeat the experiment, just copy and paste one of the cells and the numbers will change.

out of 100 trials, 23 of them contained at least one number between 41 and 45. Of course, you

can use any range of 5 you want, say 76-80  or  20-24

 

Answer 2

The probability of winning at least 1 game is equal to 1- P(winning zero games).

 

The probability of winning zero games is (19/20)⁵ = .774, so the odds of winning at leat 1 game is 1-.774 = .226