In a game where a guy has a 20% win probabiliy in each round. What’s the probabiliy for him to win at least one time after six attempts? It can’t be 120%. I know there is a way to calculate this but idk.
In: Mathematics
This uses what is called a binomial distribution. When you’re learning it at the secondary school level, you’d normally use Pascal’s triangle.
However, in the specific case you’re talking about – win at least one time – there’s a simpler solution based on rephrasing the quesiton to “what is the chance for him to win zero times?”.
For the guy to win zero times, he has to not win in every round. He has an 80% chance of not winning in the first round. If he doesn’t win in the first round, he has an 80% chance of not winning in the second… and so forth.
Mathematically, this translates into .8^6 = 0.2621 chance of not winning after 6 rounds. Since we want to know his chance of winning at least once, it would be the remaining probability – all the other options besides losing every time. So 1 – 0.2621 = 73.79% chance of winning at least once.
Philosophically, probability is just the number of times that your selected option occurs divided by the total number of possible arrangements of the system.
So let’s say instead of 20%, we say he’s rolling 5-sided dice. He has a 20% chance of each option on the first die. Then with the second dice, we look at all the possible two die combination – (1,1),(1,2)…(5,5). There are 25 of them – 5 times 5. Adding another die would yield 25 * 5 = 125 possible ways the system could be arranged. For 6 dice, it would be 5^6 = 15625 different arrangements.
Then you just need to count all the different ways you could possibly have at least one ‘win’ and perform the division.
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