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Treasure Map Probabilty (Can also be applied to other cards)
david707:
In this article: http://dominionstrategy.com/2011/03/09/basic-opening-probabilities/
theory asked "...probability each deck cycle of activating Treasure Maps, assuming n Treasure Maps and p cards in the deck".
In my workings, I've assumed that you have t Treasure maps and n cards in total. It is also assumed n is divisible by 5 for simplicities sake. The probability below is that in at least one hand you have at least 2 treasure maps, it is also one minus the probability of every treasure map being in a different hand. It also assumes you don't have draw cards like lab or Village/Smithy, but you can represent great halls and other +1 Action +1 Card Kingdoms by not counting them when counting the number of cards in your deck.
The answer:
This can be applied to other cards as well, for example, you could make t the number of terminal actions you have and see what the chances are of drawing two together.
Workings:
Involves some pretty advanced mathematics and I expect only a few people will understand the derivation. Basically, I'm too lazy to post them, but if someone requests I will.
If you have any probability questions you would like answered (relating to Dominion), then please say.
theory:
Well, I would like to see the derivation, just to see how over my head it is :)
Say t = 2 and n = 10, 15. What is p?
Could you graph that as well? Graph n vs p, and have different curves for each t.
timchen:
Sorry, but I lol'ed when I saw you said that the math is pretty advanced. :p No offense intended, but I think stuff like Truerank is quite a bit more advanced than that.
theory:
There are different kinds of advanced ... what MIT mathematicians find simple might not be so simple for lowly lawyers :P
timchen:
The problem is to count how many ways there are without more than 1 treasure map in a hand. When t<=n/5, you choose t "bins" out from the n/5 bins to put in the treasure map, and for each bin there are 5 ways to put it. This gives 5^t C(n/5,t). Divide this with the total number of ways to arrange the draw, which is C(n,t), you get the answer.
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