Ten Coins

We lately discussed a simple looking question, which seems to be more difficult that expected.

You have 1010 coins numbered 11 through 1010 on one side and with a 00 on the other side. You toss all coins and sum up the values that face up. What is the probability that the sum is greater than or equal to 4545?

We found two ways to answer the question. But we are not satisfied with our solutions because you cannot(?) use them for a more general version of the problem. Maybe you have better solutions?

1. solution:

Because there are 10 (perfect, not biased) coins, there are 210{ 2 }^{ 10 } possible outcomes. In order to get the probability, we need to know for how many outcomes the sum is greater than or equal to 4545. If the sum of the sides facing up is greater than or equal to 4545 then the sum of the sides facing down is less than or equal to 1010. Therefore we just need to know for how many outcomes the sum of the sides facing down is less than or equal to 1010. Or in other words: How many ways are there to write a natural number less than 1111 as the sum of distinct natural numbers.

You can find all possibilitys manually or just ask OEIS: There are 1+1+1+2+2+3+4+5+6+8+10=431 + 1 + 1 + 2 + 2 + 3 + 4 + 5+ 6+8+10=43 ways to write a number less than 1111 as sum of distinct natural numbers. (There is 11 way for 00, when all zeros facing down; there is 11 way to write 11 as sum of distinct natural numbers; 11 way for 22; etc.)

Hence the probability is 43210≈0.042\frac { 43 }{ { 2 }^{ 10 } } \approx 0.042

2. solution:

We made a few python simulations and figured out that it is a normal distribution with μ≈27.5\mu \approx 27.5 and σ≈9.8 \sigma \approx 9.8. By taking the area under the curve from 44.544.5 to ∞\infty with the given mean and standard deviation we got a probability of ≈0.0414\approx 0.0414.

But none of these two methods work for a more general question:

You have nn coins numbered 11 through nn on one side and with a 00 on the other side. You toss all coins and sum up the values that face up. What is the probability that the sum is greater than or equal to xx?

How would you solve this general version of the question?

#Combinatorics

Note by CodeCrafter 1
5 months, 1 week ago

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