ELI5: what are the chances if you get something with an 88.1% probability twelve times in a row across 3 days
also, separate question, what are the chances of me rolling something with a 0.4% chance? (my luck is garbage, i tried like 50 times)
ELI5: what are the chances if you get something with an 88.1% probability twelve times in a row across 3 days
also, separate question, what are the chances of me rolling something with a 0.4% chance? (my luck is garbage, i tried like 50 times)
Comments
Impossible to answer the first without knowing covariances (ie: if they’re correlated).
Consider:
What are the odds my cat is a girl?
What are the odds my cat is a girl every time over the next twelve days?
For the second, if it has a 0.4% chance then the odds are 0.004. 1/0.004 is 250, so 1/250.
To the second question: 0.4% IS the chance. Not getting it in 50 tries is not unlucky at all
Since you are using percentages I am assuming you understand those. To calculate the chance of something happening multiple times in a row we just have to multiply by the number of tries.
Take a dice for example: What are the chances to throw number 6 for 3 times in a row.
Each throw is 1 in 6 – 16,667 %. To multiply percentages we divide them by 100, do 16,667 % is a chance of 0,16667 and then we multiply like this:
0,16667 x 0,16667 x 0,16667 = 0,004629907413 (chance)
Multiply that x100 to get a percentage again: 0,4629907413 % chance
If we have to multiply something by itself multiple times we can tell our calculator to do this like so: (0,16667)^(3)
So in your example this becomes: (0,881)^(12) = 0,218630577 which translates into 21,8630577%.
It’s as simple as counting the outcomes.
If you do something that has 100 outcomes two times in a row, the total possible outcomes is 100×100=10,000.
For each of the first outcome options you could have any of the 100 2nd outcome options. (1,1 1,2 1,3… 1,100 then 2,1 2,2 2,3 … 2,100 etc.)
Similarly, If there are 88 good outcomes, then for each if the first good outcomes, there are 88 good 2nd outcomes. Total is 88×88. (1,1… 1,88 then 2,1 …2,88)
If you have an 88.1% chance each time then to get a positive outcome 2 times in a row is just (88.1/100)^2. 3 times is (88.1/100)^3.
It’s just counting the number of outcomes that are pass vs total outcomes.
If you have an outcome that is 88.1% likely to happen each time <something> happens, and you want to determine the chance of only that outcome having occurred after 12 instances of <something> happening, then you can simply multiply the chance with itself 12 times: (0.881 ^ 12) * 100 = 21.86%
Your chance of rolling <something> with an 0.4% chance once on one attempt is 0.4%. Determining your chance of rolling <something> at least once across 50 attempts is then basically just the previous question, but flipped around, as you’re trying to determine the odds of rolling Not-<something> 50 times in a row. First, you flip the 0.4% chance to get the chance of Not-<something>, which is 99.6%. Then you do as before: 0.996 ^ 50 = 81.84%
So the chance of rolling Not-<something> 50 times in a row is 81.84%. That can be flipped again, leaving you with a 18.16% chance of rolling <something> at least once across 50 attempts.