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Saturday, August 8, 2009

Probability: Law of Addition


When you draw cards from the top of your deck, in hopes of drawing into a specific card, say that Warp World, Haunting Echoes, Akroma, Angle of Wrath, or whatever your devious plans call for, you are essentially sampling without replacement. Your deck does not replenish itself as you draw, unless of coarse you are pulling some wicked synergy. The act of drawing actually influences probabilities of future draws. When you are checking two or more separate events, which are independent of one another, and looking for the same outcome, the overall probability is found through the addition of the probabilities. Total Probability, as this is called, looks as follows:

P(A or B) = P(A) + P(B)

However, when the two or more events are dependent, as in the case with a depleting deck size, the equation looks as follows:

P(A or B) = P(A) + P(B) - P(A and B)

We can rearrange this equation in numerous ways; I have two ways which I like to use. P(A and B) is the same as P(A) x P(B); I will cover the Multiplication Law later on.

P(A or B) = P(A) + P(B) - P(A) x P(B)

P(A or B) = P(A) + [P(B) - P(A) x P(B)]

P(A or B) = P(A) + P(B)(1 - P(A))


I like this arrangement, because it is easy for me to remember that, "If I had found the card in the first draw, I wouldn't be looking for it in the second draw." Before I move onto the third arrangement, I should explain P'(A), P prime of A. If P(A) is the probability of successfully finding the card in the first draw, then P'(A) is the probability if failing to find the card in the first draw. Because all possible outcomes add up to 1, when dealing with probabilities we can observe Probability: Law of Subtraction as follows:

P(A) + P'(A) = 1

P'(A) = 1 - P(A)

P(A) = 1 - P'(A)

In second example, the second probability, P(B), must be multiplied by the probability of failure of the first check, P'(A), because we had to fail to get there. Let's see what happens when we substitute 1-P'(A) for P(A):

P(A or B) = P(A) + P(B) -P(A) x P(B)

P(A or B) = P(A) + P(B) - [(1-P'(A)) x P(B)]


P(A or B) = P(A) + P(B) - [P(B) - P'(A) x P(B)]


P(A or B) = P(A) + [P(B) - P(B)
+ P'(A) x P(B)]

P(A or B) = P(A) + [P'(A) x P(B)]


P(A or B) = P(A) + [(1-P'(B)) x P'(A)]


P(A or B) = P(A) + [P'(A) - P'(B) x P'(A)]


P(A or B) = 1 - P'(A) + P'(A) - (P'(A) x P'(B))


P(A or B) = 1 - (P'(A) x P'(B))


P(A or B) = 1 - (1-P(A))(1-P(B))


What this tells us is that if we find the probabilities of failing to draw into the desired card on both the first and second draw, we can multiply them together to get the probability of failing to draw into the card on both draws, P'(A or B). Take one minus that probability, and we have the probability of drawing at least one of the desired card in the first two draws.

So, say I have 4 copies of the desired card, in a 60 card deck, what are my chances of drawing into that card in my opening hand of 7 cards:

P(A or . . . G) = 1 - (1-4/60)(1-4/59)(1-4/58)(1-4/57)(1-4/56)(1-4/55)(1-4/54) = 39.95%



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