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Dominion Articles / Basic Dominion stat. mech.
« on: January 11, 2013, 10:13:25 am »
all pretty basic stuff, just thought i'd refresh everyone's minds of the probability approach to strategy:
Dominion is based off 5-draw hand with an increasing deck population size.
INDIVIDUAL HAND MECHANICS
Probabilities are determined by the Hypergeometric distribution:
http://stattrek.com/online-calculator/hypergeometric.aspx
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Example 1: Third-turn draw probabilities of bought cards (during first two turns)
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Population size = 12 (deck)
Number of successes in population = 2 (bought cards)
Sample size = 5 (hand)
Number of successes in sample = {0,1,2} (this VARIABLE with three possible values represents the possible number of bought cards that can be in your third-turn hand)
We may proceed by using the Hypergeometric Probability calculator linked above. When we put Number of successes (X) equal to 0, we are finding the probability that our third-turn hand will have NEITHER of the two cards purchased in the previous turns. For X=1, you will draw one bought card on your third-turn, and for X=2, you will draw both bought cards on your third-turn.
Here is a summary of the different outcome probabilities using the calculator:
X=0… Probability: 0.318 (probability of drawing neither)
X=1… Probability: 0.530 (probability of drawing either, but not both— exactly one of them)
X=2… Probability: 0.152 (probability of drawing both)
Thus the probability of drawing AT LEAST ONE of the two bought cards is:
P(X=1) + P(X=2) = 0.682, or 68.2%
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Example 2: First-turn or Second-turn, probability of 5-2 buy (Christian hand)
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Population size = 10 (deck)
Number of successes in population = 3 (victory cards)
Sample size = 5 (hand)
Number of successes in sample = {0,1,2,3} (this VARIABLE with four possible values represents the possible number of victory cards that can be in your first-turn hand)
Here we use victory cards as a measure of how many coppers are drawn. We know that in order to obtain a 5-2 buy, our first hand must either have 3 estates or 0 estates.
Here is a summary of the different outcome probabilities using the calculator:
X=0… Probability: 0.083
X=1… Probability: 0.417
X=2… Probability: 0.417
X=3… Probability: 0.083
Hence the probability of having a 5-2 buy for the first two hands is:
P(X=0) + P(X=3) = 0.167, or 16.7%, or 1/6
GROUP HAND MECHANICS
Suppose we now want the probability of any player in a game to start with a 5-2 buy. THIS IS ESSENTIALLY DIFFERENT FROM THE INDIVIDUAL CASE. Here, we will use the individual probability from the hypergeometric distribution (0.167) and evaluate the group probabilities from the binomial distribution: http://stattrek.com/online-calculator/binomial.aspx
Probability that AT LEAST ONE player will start with a 5-2 buy.
2 Players… 30.6%
3 Players… 42.2%
4 Players… 51.9%
5 Players… 59.9%
6 Players… 66.6%
Conclusion: A random player getting a 5-2 buy first two hands is rather common for a game, but the same player getting a 5-2 buy for more than one game in a set of 5 games is more unlikely (~20%, using binomial distribution).
Note: the only way these probabilities wouldn't hold on average is if someone is intentionally controlling their order (aka cheating) AND a complication for late-game dynamics comes up when a mid-turn shuffle of the discard pile becomes the new deck.
-squid
Dominion is based off 5-draw hand with an increasing deck population size.
INDIVIDUAL HAND MECHANICS
Probabilities are determined by the Hypergeometric distribution:
http://stattrek.com/online-calculator/hypergeometric.aspx
==========================================
Example 1: Third-turn draw probabilities of bought cards (during first two turns)
==========================================
Population size = 12 (deck)
Number of successes in population = 2 (bought cards)
Sample size = 5 (hand)
Number of successes in sample = {0,1,2} (this VARIABLE with three possible values represents the possible number of bought cards that can be in your third-turn hand)
We may proceed by using the Hypergeometric Probability calculator linked above. When we put Number of successes (X) equal to 0, we are finding the probability that our third-turn hand will have NEITHER of the two cards purchased in the previous turns. For X=1, you will draw one bought card on your third-turn, and for X=2, you will draw both bought cards on your third-turn.
Here is a summary of the different outcome probabilities using the calculator:
X=0… Probability: 0.318 (probability of drawing neither)
X=1… Probability: 0.530 (probability of drawing either, but not both— exactly one of them)
X=2… Probability: 0.152 (probability of drawing both)
Thus the probability of drawing AT LEAST ONE of the two bought cards is:
P(X=1) + P(X=2) = 0.682, or 68.2%
=============================================
Example 2: First-turn or Second-turn, probability of 5-2 buy (Christian hand)
=============================================
Population size = 10 (deck)
Number of successes in population = 3 (victory cards)
Sample size = 5 (hand)
Number of successes in sample = {0,1,2,3} (this VARIABLE with four possible values represents the possible number of victory cards that can be in your first-turn hand)
Here we use victory cards as a measure of how many coppers are drawn. We know that in order to obtain a 5-2 buy, our first hand must either have 3 estates or 0 estates.
Here is a summary of the different outcome probabilities using the calculator:
X=0… Probability: 0.083
X=1… Probability: 0.417
X=2… Probability: 0.417
X=3… Probability: 0.083
Hence the probability of having a 5-2 buy for the first two hands is:
P(X=0) + P(X=3) = 0.167, or 16.7%, or 1/6
GROUP HAND MECHANICS
Suppose we now want the probability of any player in a game to start with a 5-2 buy. THIS IS ESSENTIALLY DIFFERENT FROM THE INDIVIDUAL CASE. Here, we will use the individual probability from the hypergeometric distribution (0.167) and evaluate the group probabilities from the binomial distribution: http://stattrek.com/online-calculator/binomial.aspx
Probability that AT LEAST ONE player will start with a 5-2 buy.
2 Players… 30.6%
3 Players… 42.2%
4 Players… 51.9%
5 Players… 59.9%
6 Players… 66.6%
Conclusion: A random player getting a 5-2 buy first two hands is rather common for a game, but the same player getting a 5-2 buy for more than one game in a set of 5 games is more unlikely (~20%, using binomial distribution).
Note: the only way these probabilities wouldn't hold on average is if someone is intentionally controlling their order (aka cheating) AND a complication for late-game dynamics comes up when a mid-turn shuffle of the discard pile becomes the new deck.
-squid