ugly psyducky
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I saw someone asking on the Ask the Master's Forum about card Probabilities and having read an older thread I thought a few clarifications might be needed to understand how this can be easiliy calculated. I had also seen some somewhat erroneous discussion of the hypergeometric distribution in an older thread, but didn't want to revive it so instead I started this one:
Obviously the idea is, to ask questions like:
If I have 6 basics in my deck what is the likelihood that I will have a basic in my openning hand
OR
If I have two of a particular trainer what is the likelihood that both cards will be prized
In essence each of these questions is slightly different in terms of the mathematical solution. However both rely on the fact that the deck is completely randomized before each shuffle. Let us take the following
n -- the total number of cards in a deck at this point in the game state
k -- the total number of instances of the same card (e.g. four basics)
t -- the total number of cards to be drawn (e.g. seven cards to start the game)
p -- the probablity of at least one card of type k being in the cards drawn
Then:
p = 1 -[(n-k)!(n-t)!]/[n! (n-k-t)!] {where n! = n x (n-1) x (n-2) x (n-3) .... x (2) x (1)}
So for example if you have 4 basics (k =4)in your deck at the start of the game (i.e. n = 60), and you draw your starting hand (t=7) you have:
p = 1 -[(60-4)!(60-7)!]/[60! (60 - 7 -4)!] = 1 -0.6005 ~ 40%
I have seen reference to the fact that this is the hypergeometric function, but it is not. The hypergeometric function has one more arguement which is the number of cards in the drawn set (we'll call this "q") which are of type "k". So we could ask what is the chance of having exactly one basic in our starting hand. In which case:
P = HYPGEOMDIST* (q,t,k,n) = HYPGEOMDIST (1,7,4,60) ~33.6%
* The factorial breakout of this is pretty ugly, but if you check out Excel help it for this function it will steer you in the right direction
This is the chance that you will start with exactly one of the basics in your hand. If one were to add up the probablities of having exactly 1, exactly 2, exactly 3 and exactly 4. You would get
p = P(q=1) + P (q=2) + P (q = 3) + P (q=4) = 33.6% + 5.9% + 0.4% + 0.007% ~ 40%. The same as given in the formulation above.
The first function is much more useful for asking the questions about muligans or what the likelihood of drawing a particular card in a particular gamestate. Whereas the Hypergeometric distribution is more useful for asking a question like, what is the chance that all my rare candies will be prized etc.
Hopefully this is useful to some. I can do more math if you like, but hopefully this suffices.
Regards
Obviously the idea is, to ask questions like:
If I have 6 basics in my deck what is the likelihood that I will have a basic in my openning hand
OR
If I have two of a particular trainer what is the likelihood that both cards will be prized
In essence each of these questions is slightly different in terms of the mathematical solution. However both rely on the fact that the deck is completely randomized before each shuffle. Let us take the following
n -- the total number of cards in a deck at this point in the game state
k -- the total number of instances of the same card (e.g. four basics)
t -- the total number of cards to be drawn (e.g. seven cards to start the game)
p -- the probablity of at least one card of type k being in the cards drawn
Then:
p = 1 -[(n-k)!(n-t)!]/[n! (n-k-t)!] {where n! = n x (n-1) x (n-2) x (n-3) .... x (2) x (1)}
So for example if you have 4 basics (k =4)in your deck at the start of the game (i.e. n = 60), and you draw your starting hand (t=7) you have:
p = 1 -[(60-4)!(60-7)!]/[60! (60 - 7 -4)!] = 1 -0.6005 ~ 40%
I have seen reference to the fact that this is the hypergeometric function, but it is not. The hypergeometric function has one more arguement which is the number of cards in the drawn set (we'll call this "q") which are of type "k". So we could ask what is the chance of having exactly one basic in our starting hand. In which case:
P = HYPGEOMDIST* (q,t,k,n) = HYPGEOMDIST (1,7,4,60) ~33.6%
* The factorial breakout of this is pretty ugly, but if you check out Excel help it for this function it will steer you in the right direction
This is the chance that you will start with exactly one of the basics in your hand. If one were to add up the probablities of having exactly 1, exactly 2, exactly 3 and exactly 4. You would get
p = P(q=1) + P (q=2) + P (q = 3) + P (q=4) = 33.6% + 5.9% + 0.4% + 0.007% ~ 40%. The same as given in the formulation above.
The first function is much more useful for asking the questions about muligans or what the likelihood of drawing a particular card in a particular gamestate. Whereas the Hypergeometric distribution is more useful for asking a question like, what is the chance that all my rare candies will be prized etc.
Hopefully this is useful to some. I can do more math if you like, but hopefully this suffices.
Regards
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