@AndrewAAGamer said in AndrewAAgamer (Allies) vs. Germany (Ragnell804):
Odds of at least 1 area area with Bad dice
65.30%
Odds of at least 2 areas with Bad dice
This is slightly more complicated to do. We have to add up percentages of all different possibilities that include this result. Let’s write some of them out. We have four areas, EG - CAEN - St. Lo - Cherbourg. A 1 means bad dice in the area, a 0 means no bad dice. Let’s look at bad dice in two areas first.
EG - CAEN - St. Lo - Cherbourg
1 - 1 - 0 - 0 = 23% * 23% * 77% * 77% = 3.13%
1 - 0 - 1 - 0 = 23% * 77% * 23% * 77% = same calculation = 3.13%
1 - 0 - 0 - 1 = …
0 - 1 - 1 -0 = …
0 - 1 - 0 - 1 = …
0 - 0 - 1 - 1 = …
As you can see, we just have the same calculation 6 times. The 6 comes from the total number of combinations of two numbers out of a total of four numbers. This can be calculated directly on a calculator using the Combination Formula, which is defined as follows.
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Most calculators have a built-in function to calculate this (i.e. you can type in 4C2 instead of the longer expression on the right side)
And not to forget, we also have to include the possibilities that have bad dice in three or four areas, because those are also included from the wording “at least two areas”.
I am not going to write out the combinations for that, now we can simply use the Combination Formula.
4C3 = 4 and 4C4 = 1 (and 4C2 = 6)
The entire calculation is as follows.
(23% * 23% * 77% * 77%) * 4C2 + (23% * 23% * 23% * 77%)*4C3 + (23%^4)*4C4 = 22.84%
Odds of at least 1 area with Very Bad dice
8.70%
Odds of at least 2 areas of Very Bad dice
Same logic, however the percentages change to 2.25% and 97.75%
(2.25% * 2.25% * 97.75% * 97.75%) * 6 + (2.25% * 2.25% * 2.25% * 97.75%) * 4 + (2.25%^4)*1= 0.3%
You can even create a general formula for this if you want, however for now this will do I think.