Is this probability calculation right? (99% chance of winning a 1 in 20,000 random game)

probability

The odds of winning in a game are 1 out of 20,000. How many times would one have to play the game in order to be sure they would win?

Let n be the number of times you need to play

The probability of not winning the game 1 – 1/20000= 19999/20000.

Using this probability calculation below:

1 – (19,999/20,000)ⁿ

As I understand I should solve for n and equal something like ~0.009 to represent a 99% of winning.

So what I found is n = 94,000

Is it fair to say if there are 94,000 players of the game with a 1 in 20,000 chance of winning (no skill involved) that there is a 99% chance that someone wins?

Best Answer

"Is it fair to say if there are 94,000 players of the game with a 1 in 20,000 chance of winning (no skill involved) that there is a 99% chance that someone wins?"

It would be better to say there is 99% that at least one person wins.

Related Solutions

[Math] Probability of a team winning the series

i will solve a general case , say probability of $A$ winning is $p$ and for a team to win it needs to win $n$ matches. now say $k$ matches have been played , and $A$ won the series , it means $A$ won the last match , thus out of the rest $k-1$ matches $A$ won $n-1$ games. thus for $A$ to win the series in $k$ matches probability is $$\binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$

now $k>=n$ and $k<=2n-1$ ($A$ needs to win before $B$) thus our total probability is $$\sum_{k=n}^{2n-1} \binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$

Probability of winning 3 out of 5 games (with different winning chance on each game)

You do not have to run one million simulations to estimate the probability of winning exactly three games. What you would have to do, is calculate the probability of each of the possible scenarios. Since you need to win three out of five games, the number of valid combinations equals ${5 \choose 3} = 10$. The following Python program takes care of this:

from itertools import permutations


def unique_permutations(iterable, r=None):
    previous = tuple()
    for p in permutations(sorted(iterable), r):
        if p > previous:
            previous = p
            yield p

        
p_win = [0.5, 0.6, 0.05, 0, 0.4]
p_tot = 0
for t in unique_permutations([1, 1, 1, 0, 0]):
    p = 1
    for i in range(5):
        p *= p_win[i] if p[i] else 1 - p_win[i]
    p_tot += p
print(p_tot)

Overall, we find that the probability of winning exactly three games equals $0.133$.

Best Answer

Related Solutions

[Math] Probability of a team winning the series

Probability of winning 3 out of 5 games (with different winning chance on each game)

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