Like. I know they are different and that one is less likely. But could someone explain this in a way that I can explain it to my partner? I know it is true but cannot remember anything about why and how to explain my point.
In: 6
I’ll focus just on the ten 7% chances and then compare to a single 70% chance.
Having something happening after ten times is the opposite of avoiding something happening ten times.
If you have one go, the chance of something happening is 7%, and the chance of avoiding it is 93%. The chance of avoiding it with two goes, means avoiding it the first AND second time, which is 93% x 93%. So this is 86.49% chance of avoiding it after two goes. This means after two goes, the chance of something happening is 13.51%. Crucially, the chance of something happening with two attempts has gone up because you could have either have the something happen on the first *or* the second go.
With more and more attempts, the odds of avoiding the thing happening goes down. Therefore, the odds of it happening goes up. After ten attempts, your odds of something happening is about 52%. So you have a 52% chance, rather than a 70% chance from the single go.
It turns out that you would need 16-17 attempts for a 7% chance to be equal to the single 70% chance.
You can carry this logic on. If you had infinitely many goes at something happening, with each attempt having a probability of 7%, you would expect the odds of the something happening to be 100%. (Likewise, if you had no attempts, your chance of the something happening would be 0%.)
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