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
Probabilities are usually multiplicative. Additionally, when you’re dealing with “at least”, it tends to be better to work out the opposite.
For example, you could figure out the odds of winning game 1, and then add that to the odds of losing game 1 but winning game 2, and then the odds of losing games 1 and 2 but winning game 3, and so on. Or, instead, you could ask “What are the odds of losing 6 games?”
If the odds of losing one game are 80%, then the odds of losing two games are 80% x 80%, or 64%. Do this for 6 games, 80%^(6), you’ll find that the odds of losing 6 games are 26.2%. Therefore, the odds of winning at least one game in 6 are 73.8%.
Typically you multiply probabilities.
So if you had a 20% chance to win, and you wanted to know the probability of winning twice in a row, you would do 20% x 20% = 4%. For more wins in a row you would simply add more multiplication.
On the other hand, you have an 80% chance to lose. If you wanted to know the probability of losing twice in a row you would do 80% x 80% = 64%.
You’ll notice that that only adds to 68%, meaning you have a 32% chance of winning *only *once.
So the short answer is that the easiest way to find winning *at least* once you find the probabilities of winning zero times and subtract that percentage from 100%.
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.
Its easiest if you consider the opposite: what’s the chance he wins 0 times after y rounds? Well, we know he has an 80% chance to lose 1 game, so after 6 games his chance of losing every one is 0.8 ^ 6 = 0.26. And if he has a 26% chance of losing every game, he has a 74% chance of winning at least 1
The alternative to winning at least once is losing every round. The odds of that are (80%)^(6), or 26.2%. Subtract that from 100, and your odds of winning one or more rounds are 73.8%.
In problems like this, it’s easiest to look at the probability he never wins. For each game, he has a 20% of winning, so he has an 80% chance of losing. If he plays 6 games, then that’s (80%) * (80%) * (80%) * (80%) * (80%) * (80%) = 26.2% that he loses all of them. Then there’s a 73.8% chance that he *doesn’t* lose them all, which is just the probability that he wins at least one.
The probability of him losing is 80% or 0.8. With six attempts the probability of losing them all is 0.8^6 = 0.262 or 26.2%. So the probability of winning at least once is 73.8%.
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