# If I put a copper and add 4 more random cards into my hand from my starting deck, does it change the odds of getting the 5:2 split?

I was thinking about a situation where someone accidentally reveals a copper from their starting hand while shuffling their deck or discovers that they only have 9 cards in their starting deck (and they are certain they put the required 3 Estates in).

If they added a copper from the supply to their first hand, and drew 4 more of the remaining 9 cards, would their odds of getting a 5:2 or 2:5 hand versus a 4:3 or 3:4 hand be any different than if they drew a normal 5 card starting hand from a normal 10 card deck? (I realize that a 5:2 hand is slightly different than a 2:5 hand, in that it allows you to change your possible 2:5 coin buy after an opponents first purchase. The thing I am wondering though is if it changes the chances of getting a 5:2 split versus the more likely 4:3 split).

No.

The odds of getting a 5:2 split over the more common 4:3 split is unchanged, even if the first card is intentionally chosen to be a copper.

Mathematically, this can be proven using hypergeometric distribution. This can be calculated using the following function, where `N` is the size of the deck, `n` is the number of cards drawn, and `m` is the number of coppers available:

``````P(X=k) = (m choose k) * ((N-m) choose (n-k)) / (N choose n)
``````

With a normal starting deck, you would use values of `N=10`, `n=5` and `m=7`. Without boring you with the math proper, these would give you a result breakdown as follows:

``````2 copper: 1/12
3 copper: 5/12
4 copper: 5/12
5 copper: 1/12
``````

So you would normally have a `1/12 + 1/12 = 1/6` chance of getting the 5:2 split.

Now, if the first card is preemptively selected to be a copper, this changes our starting numbers. Since the first card is known, your starting draw would effectively be a four card draw from a nine card deck containing only six coppers. Punching in the values `N=9`, `n=4` and `m=6` into our function, we end up with the following table:

``````1 copper: 1/21
2 copper: 5/14
3 copper: 10/21
4 copper: 5/42
``````

Even though the spread of numbers is significantly different, your odds of a 5:2 split work out to `1/21 + 5/42 = 7/42 = 1/6`, which is exactly the same as before.

However, note that although the odds of the split themselves are unchanged, by selecting out the first copper you are now over twice as likely to get the 5:2 draw as you are to get the 2:5 draw.

I believe a probabilistically equivalent question is "If a single card revealed from an opening hand is copper, does it tell you anything about the starting deck?" Order doesn't matter (even in your example situation, as every hand is guaranteed to have copper, that the first card is copper isn't significant).

Bayes' Rule for conditional probability says that

``````P(A|B) = P(B|A) P(A) / P(B)
``````

Let `A` be the event that there is a 5:2 or 2:5 split and `B` be the event that a random card revealed is a copper.

For `P(B|A)`, the number of coppers in the first two hands are 5:2. Which hand is first (the 5 or the 2) is equally likely, so the probability of revealing a copper from the first hand is the average probability of revealing a copper from the two possible first hands, i.e.

``````P(B|A) = (1.00 + 0.40) / 2 = 0.70
``````

Now we must consider the 4:3 case to calculate `P(B)`. If the first two hands are 4:3, same comments about order apply, and the probability of revealing a copper is

``````P(copper revealed | 4:3) = (0.80 + 0.60) / 2 = 0.70
``````

Thus the probability of revealing a copper doesn't change based on a 5:2 or a 4:3 split so `P(B) = P(B|A) = 0.7`. But this is the definition of statistical independence, and going back to the original Bayes' Rule, `P(B)` and `P(B|A)` can cancel, so `P(A|B) = P(A)`. The odds of a 2:5 split are unchanged by reveal of a copper in the hand.

It's initially more counterintuitive, but it follows directly that a first card Estate also provides no information about the breakdown. (In all cases `P(estate revealed) = 1 - P(copper revealed)`, so we'd have .3's instead of .7's.)

Knowing that the first card is a copper (or an estate) doesn't bias the 5/2 distribution at all. Knowing the first two cards, however, does bias the distribution. The best justification I can come up with is to mark a single copper as your special copper. If you divided the ways of getting a 5/2 split into different classes depending on 10 different possible locations of that special copper, each of those 10 sets have equal size and an equal ratio of 5/2 vs 4/3, so you haven't really told me anything about the likelihood of a 5/2 by telling me which of those 10 subsets you are in.

Output of some sims:

``````   0.16722
C  0.16752
CC 0.19536
CE 0.10769
E  0.16728
EE 0.37528
``````

Here is the Python code that produced the numbers.

``````import random
import collections

def split_is_25(start):
cur = start
rest = []
startC = start.count('C')
startE = start.count('E')
rest.extend(['C'] * (7 - startC))
rest.extend(['E'] * (3 - startE))
random.shuffle(rest)
split = start + rest
rand_split = [split[0:5].count('C'), split[5:10].count('C')]
rand_split.sort()
return rand_split == 2

N = 100000
tests = ['', 'C', 'CC', 'CE', 'E', 'EE']
counts = collections.defaultdict(int)

for i in range(N):
for test in tests:
counts[test] += split_is_25(list(test))
for test in tests:
print test, float(counts[test]) / N
``````

I'm a little unclear what you mean by revealing the copper, then keeping on shuffling. If you're still shuffling, then you're not done yet, and any cards revealed should be put back into the 10 cards. If you're already starting to deal, then the cards are fixed, in which case you keep dealing regardless of the revealed card.

Nevertheless, making any kind of decision based on a revealed card will change the odds. Here's what I mean by decision:

I'm trying to game the system, so I shuffle until I accidentally reveal a card, if it's a copper, I let it remain in my starting hand, but if it's an Estate, I shuffle it back in. I've just drastically increased my odds of having more copper in my starting hand.

NOTE: This scenario is equivalent to one where I intentionally reveal copper, as that's simply the same as choosing the first card of your hand, and having the rest random.

In short, getting to choose your first card (instead of it being random) definitely affects the odds of getting any particular hand.

• Actually, as others have shown, choosing deliberately your first card (be it Estate or Copper) and then drawing the remaining 4 cards necessary to make a 5 card starting hand does not change the odds of getting a 5/2 split versus a 4/3 split in your first 2 hands. It does change the odds of getting a 5,4,3,2, or 1 Copper starting hand (but doesn't change the combined first two hands odds) – user1873 Jan 2 '12 at 4:50
• Please read my answer carefully. I said it changes the odds of any particular hand, not the combined odds of one of two hands. A 1st copper will increase the odds of getting any hand with more coppers. The question's been edited anyhow, so this point seems to be moot. – Neal Tibrewala Jan 2 '12 at 6:00