If something has a 5% chance of success, and I’ve failed 19 consecutive times, the next time independently still has a 5% chance of success, fairly low, rather than a guaranteed 100% chance of success in view of the previous attempts. How does this relate to updating percentages in light of new evidence, or is that something separate?
In: 6
To.work out the chances of a 5% chance happening at least once in n tries is equal to 1 – (the probably of it not happening at all in n tries)
If we have a 5% chance to succeed on a given attempt (1/20) we have a 95% chance to fail (19/20). If we try 20 times, the odds it doesn’t happen at all is (19/20)^20 or ~35.8% chance. That means the chance of succeeding at least once is ~64.2%. The odds of succeeding exactly once is 20 * 1/20 * (19/20)^19 or ~37.7%. The odds of succeeding exactly twice are 190 * (1/20)^2 * (19/20)^18 , or ~18.9%
The formula for exactly k successes in n tries is nCk * P(win)^k * P(lose)^n-k . If we add together k=0, 1, 2, 3…n, we get 1. If this formula looks familiar, it’s a binomial expansion. As long as the events are independent, this is how things will go. Like rolling a die, the previous outcome does not affect the next. If it were something like a deck or cards, if I pull every card out of the deck one at a time and haven’t pulled the 8 of hearts, the last card must be the 8 of hearts. That is because these are dependent events where they are impacted by previous events. But if I pull out a card, replace it, and then shuffle the deck, the next card pull is independent of the previous.
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