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
The expected number of “wins” in either case is the same: 0.7. But they’re distributed differently, so the chance of **at least one** win is different.
The reason is that your ten 7% chances includes the possibility of *multiple* wins. To get the same average number of wins, those multiple wins have to also be balanced out by a higher chance of zero wins.
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More mathematically, the chance of zero wins here is the chance of not winning at each step, multiplied together. So for your 10 7% chances, your chance of not winning is 93% (that is, 1 – 0.07), and you end up with (1 – 0.07) * (1 – 0.07) * … * (1 – 0.07) ten times. That comes out to (1 – 0.07)^10 = (0.93)^10 = about 0.48, or 48%.
It may help to think of things in smaller, easier to intuit numbers. How about one 100% chance and two 50% chances? Obviously two coin flips can come up both tails, so the two must be different, and it’s worth actually listing out every option to see why. The single 100% chance has only one option: “win”. The two 50% chances have four equally likely options: “win win”, “win lose”, “lose win”, and “lose lose”. In both cases, you average exactly one win, but in the second, the “win win” case gets balanced out by a “lose lose” one.
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