Dominion + Maths

Something mathematical about Dominion:

What is the probabality for a 4-3 or a 5-2 start?
A simply question, for a 4-3/3-4 start it is 83,33 % and for a 5-2/2-5 start it is 16,67 %, WHY????
Imagine: You have got 0 cards. You draw your first card. What is the probability for a Copper?: Right: There are 7 Coppers and 3 Estates. So it is 7/10, or 70%. Than you draw your second card. What is the probability for a Copper? There are now 6 Coppers, but 3 Estates, too. So it is: 6/9 or 66%, the probability for a third Copper is 5/8 and for a fourth one it is 4/7. Imagine, you want a 4-3 start. So you need a Estate now. Probablity? There are 3 Estates and 3 Coppers…… Right!!!! The probability for this hand: (C – C – C – C – E) is 7/10 * 6/9 * 5/8 * 4/7 * 1/2 = 8,33 %. But we want the probability for a hand with 4 Coppers. What is about this hand?: (E – C – C – C – C), or this one?: (C – E – C – C – C)
Imagine: The probability to draw the Estate as the first card is equal to the probability to draw the Estate as the second/third/fourth/last card. There are 5 different permutations. So we take the 8,33 % and then: 8,55 % * 5. We get 41,67 %, the probability to draw a 4 Copper hand in the first turn. For a 3 Copper hand it is 41,67 %, too. So we add the first 41,67 % to the second one and get: 83,33 % for a 3-4 / 4-3 start. And what is about the 5-2 start? You could have a 5-2 or a 3-4 start. And of course the probability of a 3-4 start + the probability of a 5-2 = 100 %! And Tata 100% – 81,33 % = 16,67%

How many kingdoms are there?
Was there a moment in your Dominion live where you got a random Setup and you thought: ” I played with this Setup a few Weeks/Months/Years ago. Yes? Ok, that is very funny! Why? : READ:
I have 10 children. How many different groups with 4 children could I form?. Ore in Dominion language: There are 205 Cards. How many Set Ups with 10 card could I form? For this we need the “binomial coefficient”.

bin (n;k) = n! / (  ( n-k)! * n! ) He gives us the answer for this kind of questions.
I write: bin(n;k) for: “binomial coefficient with the variables n and k. n is the number of cards, and k is the large of the Setup. What is “!”? The factorial of a non-negative number x is x!. And what is x!. So: 1! = 1 ; 2 ! = 2*1, 3! = 3*2*1, and so far. Our n is 205 and our k is 10.
bin (205;10) = 205! / (195! * 10!) = 2,8902474 * 10^16. A 2 and then 15 other numbers. It is very large, isn’t it?

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