How do I proof, that this algorithm gives out the right probability for finding the euro in the desk drawers?

Assignment
There is a desk in your office that has a 50% chance of holding one euro. This desk has three drawers. If there is a euro in this desk, it is equally likely to be in one of the drawers. What is the probability that the euro is in the third drawer if you have already searched in vain for it in the in the first and second?

Solution:

int iterations = 10000;
int desk;// 0 or 1 - 50%        
int[] foundEuro = new int[5];

for (int i=1; i <= iterations; i++){
    desk = (int) (Math.random() * 2);

    if ( desk == 0){ // found Euro              
        int drawer = (int) (Math.random() * 3);

        if ( drawer == 0){
            foundEuro[drawer]++; // euro in drawer1             
        } else {
            foundEuro[drawer]++; // euro in drawer2                 
            foundEuro[drawer+1]++; // euro in drawer3
        }
    } else {
        foundEuro[desk]++;
    }
}

showResult(foundEuro);
float probability = ( ((float) foundEuro[0]) / iterations) * 100;
System.out.printf("%.2f%%", probability);

Output
Euro in drawer 1: 1638
Euro in drawer 2: 6622
Euro in drawer 3: 3343
16,38%
Note
I think, my code has no errors and is supposed to show the right result, but idk if it is really the right probability for finding the euro in the third drawer, while in the other first two drawers it wasn't there.

Your algorithm and result is simply wrong. What you calculated is basically the probability of the coin being in a given drawer in the desk and that is obviously 1/2 * 1/3, probability of correct desk * probability of correct drawer = 1/6 roughly 16.6

The correct answer however is 25%. You can work that out on paper or you can adjust your program to properly reflect the 'if you have already searched in vain for it in the in the first and second' constraint. You basically have to discard those random coin choices that violate this constraint. The code then becomes:

int iterations = 100000;
int found = 0;
int violateThePrecondition = 0;
for (int i = 1; i <= iterations; i++) {
    int desk = (int) (Math.random() * 2);
    if (desk == 0) { // found Euro
        int drawer = (int) (Math.random() * 3);
        if (drawer == 2) { // coin in drawer 2
            found++;
        } else { // drawers 0 and 1 can by definition not have the coin
            violateThePrecondition++;
        }
    }
}

float probability = (((float) found) / (iterations - violateThePrecondition)) * 100;
System.out.printf("%.2f%%", probability);

>25.05%

The minimal change to your code would have been to change the probability calculation to

float probability = ( ((float) foundEuro[0]) / (iterations - foundEuro[2])) * 100;

The math involved is (cNdM is the coin being in drawer N in desk M = 1/3 * 1/2 = 1/6, d0 is the coin being in desk0 = 1/2):

P(c2d0 | !c1d0 and !c0d0) = 
P(c2d0 and (!c1d0 and !c0d0)) / P(!c1d0 and !c0d0) = 
    with (!c1d0 and !c0d0) = (!d0 or c2d0)
P(c2d0 and (!d0 or c2d0)) / P(!d0 or c2d0) =
P(c2d0) / (P(!d0) + P(c2d0)) =
1/6 / (1/2 + 1/6) =
1/6 / 4/6 =
1 / 4

First, you decide, whether the Euro is in the desk at all or not. If I interpret this correctly, once it is NOT, you increase the counter in foundEuro[1], as you are in the else clause, where desk holds value 1. This exact same array element gets also increased, when there is a Euro in drawer 2, so the 66,22% chance is actually for the Euro being somewhere else OR in drawer 2 and not only for being in drawer 2. Additionally, I don't get why you always increase both counters for drawer 2 and drawer 3, when drawer != 0, you should probably make use of a switch case statement.

Second, I think you might check out the Monty Hall problem, which may be related and give you a hint towards how to approach this kind of statistics problem, as I think your assumptions are already flawed. Your question is less about the code and more about how to solve a probability problem, so maybe take a look at https://stats.stackexchange.com

The chance for finding a Euro in the desk is 50%, 1/2.
If you find one, it is equally likely in any of the three drawers.
I.e. the chance to find it in any given drawer is (1/3)*(1/2)=1/6.

The special situation that you do not find the Euro in two of the drawers has a chance of 4/6. It could be in the third drawer in 1/6 cases and is not at all in this desk in 1/2=3/6. These chances do not overlap and can hence be added, 1/6+3/6=4/6.

The chance that you find the Euro in the third drawer, specially in that situation is (1/6)/(4/6)=1/4, because the chance of the situation, multiplied by the chance for that situation must yield the total chance, i.e. (1/4)*(4/6)=1/6.