I am trying to calculate the chance of specific drops from monsters in game

[pineapplewithmouse]

So in the game, there are monsters that can drop three things, let's call them a b and c. The chance of getting each drop is the same (1/400).
Drop a and b have low value and drop c has high value. I got 11 of drop a, 10 of drop b and 2 of drop c.
I want to calculate the chances that I got 21 of the low value drop and just 2 of the high value drop, and not the other way around or some another option.
So I divided the drops to 2 groups: drop c and not drop c. And I did (2/3*1/400)^21+(1/3*1/400)^2 which is 1 in 1.44m which is... kinda too low.
I do not know where I am wrong, please help me

 

[Cubist]

(2/3*1/400)^21+(1/3*1/400)^2 which is 1 in 1.44m which is... kinda too low.
I do not know where I am wrong, please help me

You're very close . However, the 1/400 doesn't come into the calculation because the 23 drops have already happened. Also, you need to use the binomial distribution. The probability that EXACTLY 2 out of 23 drops will be type C is:-

[math] \binom{21+2}{2} \times \left( \frac{1}{3} \right)^2 \times \left( 1-\frac{1}{3}\right)^{21}[/math][math]= \frac{23!}{21! \times 2!} \times \left( \frac{1}{3} \right)^2 \times \left( \frac{2}{3}\right)^{21}[/math]
This works out to be approximately 0.0056 or 0.56%

It can be very interesting (and educational) to calculate such probabilities, but please see this article should rare events surprise us to gain some real perspective.

 

[pineapplewithmouse]

I do not understand what the 21+2 and the 2 in the brackets means, can you explain please?

 

[BigBeachBanana]

I do not understand what the 21+2 and the 2 in the brackets means, can you explain please?

It's the notation for counting combinations a.k.a binomial coefficient. Basically, it tells you the number of ways of picking 2 (type c) unordered outcomes from 23 possibilities.