Stephen Lovell
United States Edgewood Washington

So the main reason I'm posting this here is because Dominion is the game I'm thinking about. I have a few questions for you probability experts out there.
First off, if I've figured it right, the probability of drawing a single Estate in your first hand is 50%, right? You draw 5 cards, and by the time you draw your 5th, if the first 4 were Coppers, then there are 3 Coppers and 3 Estates left in your deck, so 50%.
So, if you buy Baron/Silver your first two hands, then what are your chances of drawing your Baron and a single Estate? Me roughly working it out is shown below.
1 of my cards needs to be Baron, 1 needs to be Estate. Therefore, I have to draw Baron within with first 4 cards, and the Estate on my last (I know it doesn't have to work that way but it makes it easier to think about).
So for the Baron, my odds are 4/12 (drawing 4 cards, only 1 Baron out of 12 cards in the deck).
Then, on the 5th card, my chances of drawing an Estate are 3/8. Multiplying the odds of each happening then gives me about a 12.5% chance.
Please let me know if I did that right, and give me a little more insight. Probability is a little more tricky than I remember in high school (though we didn't do a ton of it).



The first round:
Drawing 4 times copper and one estate: 7/10 * 6/9 * 5/8 * 4/7 * 3/6 = 1/12 This has to be multiplied with 5 as their are 5 possibilities (permutations) to draw your estate, so p(4Copper+1Estate,first round) = 5/12.
The second round: Drawing 3 Copper, Baron and one estate: 7/12 * 6/11 * 5/10 * 1/9 * 3/8 = 7/1056 Here 20 possible permutations exist (1st card baron, 2nd card estate / 1st card baron, 3rd card estate / ...) so p(Baron&Estate) = 35/264



Is this math right? I am no statistics expert, but I was looking at the numbers, and I came up with similar answers to yours.
This being said, the baron/estate comes up MUCH MUCH more than 35/264, so I wonder if this isn't wrong in some way?

Johan Berglind
Sweden Kungälv Unspecified

You can choose the Baron in just one way, the estate in three ways and the three treasures in "8 over 3" ways, since there now are 7 coppers and one silver. And since you have 12 cards in all, there are "12 over 5" ways to draw a hand of 5 cards.
Consequently, the probability of drawing a Baron, exactly one estate, and three treasures is 3 times "8 over 3" divided by "12 over 5", or (3*8*7*6*5*4*3*2*1)/(3*2*1*12*11*10*9*8) which should be equal to 7/33 or about 21%
Johan.



elkabong wrote: This being said, the baron/estate comes up MUCH MUCH more than 35/264, so I wonder if this isn't wrong in some way? Actually this is because, I've treated only the special case 3 Coppers + 1 estate + 1 baron:
The full list of probabilities is: 3 Copper + 1 estate + 1 baron: 13.2% 2 Copper + 2 estate + 1 baron: 8.0% 1 Copper + 3 estate + 1 baron: 0.9%  Total: 22.1%
3 Money + 1 estate + 1 baron: 21.2% 2 Money + 2 estate + 1 baron: 10.6% 1 Money + 3 estate + 1 baron: 1.0%  Total: 32.8% Baron and only Money: 8.8%
The difference of the last to percentages to 100 are those cases, where you didn't draw the baron at all.

Werner Bär
Germany Karlsruhe Baden

elkabong wrote: This being said, the baron/estate comes up MUCH MUCH more than 35/264, so I wonder if this isn't wrong in some way? As others wrote, the question answered was the Baron coming up in round 3 (not 4 or 5), and with exactly 1 estate.
If the Baron shows up in round 3 or 4, the odds to draw it together with at least 1 estate are 26/33, or 78,8%. But it might be that the Baron only shows up in round 5, when the deck size hase increased. Odds are a tiny bit lower here.
I calculated a total chance of 844/1089 = 77.5% that you have an estate in hand the first time the baron shows up.

Andrew Hardin
United States Bentonville Arkansas

I have the following probabilities:
Baron, Estate, Estate, Estate, Copper (5 Coin) = 0.008838 = 0.88% Baron, Estate, Estate, Estate, Silver (6 Coin) = 0.001263 = 0.13% Baron, Estate, Estate, Copper, Copper (6 Coin) = 0.079545 = 7.95% Baron, Estate, Estate, Copper, Silver (7 Coin) = 0.026515 = 2.65% Baron, Estate, Copper, Copper, Copper (7 Coin) = 0.132576 = 13.26% Baron, Estate, Silver, Copper, Copper (8 Coin) = 0.079545 = 7.95%
Probability of Getting Baron and As Least 1 Estate On Turn 3 = 32.8% Probability of Getting Baron and Exactly 1 Estate On Turn 3 = 21.2% Probability of Getting Baron and No Estates on Turn 3 = 8.83%
 Lex
(Edit: As an interesting aside it should be noted that this is one of the highest probabilities of getting a 6+ Coin on Turn 3 of any setup. The bad part is that unlike most of the other techniques there is no chance of getting this Gold before Turn 5)

Stephen Lovell
United States Edgewood Washington

Alright, so I understand what Martin is saying. I don't however understand most of what Johan is saying. I never quite wrapped my head around factorials, so if you don't mind enlightening a little bit that would be rad.

Cameron McKenzie
United States Atlanta Georgia

When he says stuff like "8 over 2" what he is refering to is "8 choose 2" which would represent the number of distinct sets of 2 elements that you can pick out of a set of 8 elements. The order doesn't matter.
For example, in a deck of 52 cards, there are "52 choose 5" ways of drawing a hand of 5 cards.
The value of this expression is computed like so:
"n choose m" is equal to "n!" divided by the product of "m!" and "(nm)!"
So what he's doing here is saying you have 12 cards. How many ways can you draw 1 baron, 1 estate, and 3 treasure? It's important to note that every card is a distinct element, even though some are identical. Well, there are 8 treasure, so there are "8 choose 3" ways to draw the treasure. There are three ways to draw the estate (it's "3 choose 1" if you want to think of it that way) and there is one way to draw the baron "1 choose 1". The product of these gives you the total number of possible ways to draw 1 baron, 1 estate, and 3 treasure.
This value is equal to 168.
If you divide this by the total number of ways to draw ANY 5 cards, you end up with the probability of drawing cards fitting those conditions. The total number of ways to draws 5 cards out of 12 is obviously "12 choose 5".
This value is equal to 792.
So of the 792 possible hands you could draw (all of which are equally likely), 168 of those hands have 1 baron, 1 estate, and 3 treasure. This is about 21.2%. Of course, if you add in the hands which have more than 1 estate along with a baron, there are more, but I'm just giving you an example since you were asking about the computations involved.

Stephen Lovell
United States Edgewood Washington

Wow... I think I get it. (Also I love Laura Roslin ^_^)
So let say we're going for a Chapel Trim deck. First two turns we buy Chapel and Silver. That puts our deck at 8 treasure, 3 Estates and 1 Chapel.
Clearly the best draw on turn three is Chapel, 3 Estates and 1 Copper...
I just realized that I don't know how to figure this out, lol. The Chapel is 1 choose 1, but you need to draw all 3 Estates. Is it... (3 choose 1)^3? And then drawing a single Copper out of 8 treasures... 7 choose 1?
189/792 = 23.8%
That seems too high, so I think I did something wrong. Teach me oh wise ones.

Matt Sargent
United States Portland Oregon

tehgr8supa wrote: So let say we're going for a Chapel Trim deck. First two turns we buy Chapel and Silver. That puts our deck at 8 treasure, 3 Estates and 1 Chapel.
Clearly the best draw on turn three is Chapel, 3 Estates and 1 Copper...
I just realized that I don't know how to figure this out, lol. The Chapel is 1 choose 1, but you need to draw all 3 Estates. Is it... (3 choose 1)^3? And then drawing a single Copper out of 8 treasures... 7 choose 1?
189/792 = 23.8%
That seems too high, so I think I did something wrong. Teach me oh wise ones.
The number of ways in which you can select three estates from a pool of three of them is one (3 choose 3 = 1). The number of ways in which you can select one copper from a pool of seven of them is seven (7 choose 1 = 7). So there are only seven possible hands meeting this criterion, out of 792 possible which is a 0.9% chance.
Of course, anything without silver is a pretty good chapel hand. The number of hands for that is 10 choose 4 = 210. 210/792 = 26.5%

Stephen Lovell
United States Edgewood Washington

Ohh... It's a lot simpler than I was making it, haha. Awesome, thanks a ton everybody.

Stephen Lovell
United States Edgewood Washington

Sorry to keep beating this to death, but I've been thinking about it more.
52 card poker deck, the probability of drawing at least 1 Ace in a 5 card hand.
(4 choose 1) * (51 choose 4) = 999,600
999,600 / 2,598,960 = 38.4% chance of drawing at least one ace? Clearly that's wrong, so what am I missing?

Matt Sargent
United States Portland Oregon

tehgr8supa wrote: Sorry to keep beating this to death, but I've been thinking about it more.
52 card poker deck, the probability of drawing at least 1 Ace in a 5 card hand.
(4 choose 1) * (51 choose 4) = 999,600
999,600 / 2,598,960 = 38.4% chance of drawing at least one ace? Clearly that's wrong, so what am I missing?
Hm, don't know. The probability of NOT drawing an ace in five cards is (48 choose 5) / (52 choose 5) = 65.88%. I know this is correct because it is equal to (48/52) * (47/51) * (46/50) * (45/49) * (44/48). So the probability of drawing an ace is 34.12%.
...
Ooh, I figured it out. (4 choose 1) * (51 choose 4) is doublecounting a lot of hands. For example, suppose we choose AS as our ace, and then one of the hands from (51 choose 4) is AD xx yy zz. We then end up counting that hand again when we choose AD as our ace and one of the other fourcard sets is AS xx yy zz.
The right way to count it is to find the number of hands with exactly one ace, two aces, three aces, and four aces, and to add them all together.
one ace: (4 choose 1)(48 choose 4) = 778,320 two aces: (4 choose 2)(48 choose 3) = 103,776 three aces: (4 choose 3)(48 choose 2) = 4,512 four aces: (4 choose 4)(48 choose 1) = 48 total: 886,656
886,656 / 2,598,960 = 34.12%, the expected answer.

Stephen Lovell
United States Edgewood Washington

Wow, so it's just higher than I expected it to be. Thanks a lot! Figuring all this out is fun.

Andrew Hardin
United States Bentonville Arkansas

The other way to calculate a probability of at least 1 is to use the following rule:
P(Not A) = 1  P(A)
Since you can't have a negative number of Aces, the probability of at least 1 Ace is the same as saying the probability of not getting 0 Aces.
So all you have to do is calculate the probability of 0 Aces:
(48*47*46*45*44)/(52*51*50*49*48) = .6588
So the correct number is 1  .6588 = 34.11%
 Lex
(Edit: Corrected a typo from .6558 to 6588)

Dan Schaeffer
United States Unspecified Illinois

LexH wrote: The other way to calculate a probability of at least 1 is to use the following rule:
P(Not A) = 1  P(A)
Since you can't have a negative number of Aces, the probability of at least 1 Ace is the same as saying the probability of not getting 0 Aces.
So all you have to do is calculate the probability of 0 Aces:
(48*47*46*45*44)/(52*51*50*49*48) = .6558
So the correct number is 1  .6558 = 34.11%
 Lex
Typo. Should be 0.6588. Your apparent subtraction error confused me, until I realized it was a transcription error from the previous equation.

Andrew Hardin
United States Bentonville Arkansas

Sorry about that.
There is a reason I check my equations repeatedly and still make those errors.
 Lex


