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How does one calculate the possibility of rolling a 20 between multiple people?

My friends claim that between the four of us, if we each roll 1 d20, we have a 20 percent chance of getting a 20, i dont think this is right but i cant prove why.

I cant figure out how to calculate an event happening at least 1 in 4 times when each event has a 5 percent chance of happening.

Derelicta
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  • By your friends’ reasoning, if there were 21 of you, the group would have a 105% chance of rolling at least one 20. – amd Mar 14 '17 at 02:16

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There are two possibilities: Either someone gets a $20$, or no one gets a $20$. Since those two possibilities are exhaustive (there's no way neither could happen) and exclusive (there's no way both could happen at the same time), the sum of their probabilities is $1$.

Now it turns out that "no one gets a $20$" is a much simpler event: The only way this could happen is: You didn't get a $20$, your first friend didn't get a $20$, your second friend didn't get a $20$, and your third friend didn't get a $20$ either.

Now since those events are independent (the result of any of the rolls doesn't depend in any way on the results of the others), the probability of all of them happening is just the product of the individual probabilities. But the individual probabilities here are all the same, namely $19/20$, as there are $19$ other numbers.

Putting it all together, the probability that at least one of you gets a $20$ is $$P(\text{“no $20$”}) = 1 - \left(\frac{19}{20}\right)^4 = 29679/16000 \approx 0.1855$$ Which for most practical purposes is close enough to the claimed $20\,\%$.

celtschk
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  • Yes, the correct value is close to the claimed 20%, but I suspect the leatter number was computed as $4\times5%$. – amd Mar 14 '17 at 02:18
  • @amd: Since no description was given on how they arrived at that conclusion, I'd give them the benefit of the doubt. Especially given that even if they just did that multiplication, their reasoning could have been that the probability is small enough that the linear approximation is sufficient. – celtschk Mar 14 '17 at 02:24