## Basic Opening Probabilities

This post analyzes opening probabilities for Dominion.  It assumes that you purchase two cards that provide +\$2 on Turns 1 and 2, classifying them into “Silvers” or “terminal Silver equivalents” (e.g., Chancellor, Cutpurse), and calculates the probabilities of various Turn 3 and Turn 4 combinations.  There is also a brief section on Chapel probability.

Unfortunately, I am both lazy and bad at math.  So instead of deriving the actual mathematical probabilities, I just wrote code to simulate the expected distribution 1,000,000 times.  So I can’t give you the theoretical justification for these figures, nor can I promise that they’re perfectly precise.  But I can assure you that they’re pretty close.  And if one of the many math grad students who read this blog would like to do some derivations if they have the free time (either repeating this work, or doing more advanced calculations, like opening probabilities with Warehouse, or the probability each deck cycle of activating Treasure Maps, assuming n Treasure Maps and p cards in the deck), they’re entitled to a guest post explaining their methodology and results 🙂

Let’s start by assuming you purchased two Silvers (or one Silver and one terminal Silver equivalent).  You’ll be able to maximize your Turn 3 and 4 buying power because there is no risk of collision between two terminal Actions.  The following probabilities are for Turns 3 and 4 combined:

 Drawing \$7 8.8% At least one \$6 or \$7 42.4% Not drawing \$5 on either turn 8.8% Drawing \$5 or better twice 14.9% Average total money \$9.17

The 8.8% probability of not drawing \$5 on either turn is basically a death knell unless there’s worthwhile \$3 or \$4 cards around that are better than Silver.

Now let’s assume you purchased two terminal Silver equivalents.  Although you gain some benefit from an extra terminal (for instance, an extra Swindler attack), your buying power suffers whenever you draw them together:

 Collision of the two terminals 30.3% At least one \$6 17.7% Not drawing \$5 on either turn 27.8% Drawing \$5 or better twice 12.6% Average total money \$8.56

These probabilities should give you a good illustration of the tradeoff you make when you open double terminal.  In particular, opening with two terminals greatly decreases the chance you can get to \$5 or \$6 on Turns 3 and 4; accordingly, if you do, you should be prepared for the fact that you may have to settle for \$3’s and 4’s on those critical first two turns (e.g. Caravan is available, or you’re building a Village/Smithy chain).

Finally, let’s assume you opened Chapel/Silver.  The probability of drawing your Chapel with:

 3 Coppers, 1 Estate 26.5% On turn 5 16.6% 1 Silver, 2 Coppers, 1 Estate 16.0% 2 Coppers, 2 Estates 15.9% 4 Coppers 8.9% 1 Silver, 3 Coppers 8.8% 1 Silver, 1 Copper, 2 Estates 5.3% 1 Copper, 3 Estates 1.8% 1 Silver, 3 Estates 0.3%

This also suggests that the probability of drawing your Baron (or Salvager) with at least one Estate on Turns 3 or 4 is about 65.7%.

Let me know if there’s additional questions that you’d be interested in answering.  The full dataset for most likely hands can be found below:

Probability Turn 3 Hand Coin
26.6% ccces \$5
16.0% ccees \$4
13.3% cccee \$3
13.2% cccce \$4
8.9% ccccs \$6
7.9% ccess \$6
4.4% cccss \$7
2.7% cceee \$2
2.6% ceess \$5
2.6% ccccc \$5
1.8% ceees \$3
0.1% eeess \$4

Note that for the following table, mirrored hands are merged. So drawing “ccces” and “ccees” is combined with drawing “ccees” and “ccces”.

Probability Hands Coin Coin
15.2% ccces|ccees \$5 \$4
10.1% ccces|ccces \$5 \$5
10.1% cccee|ccces \$3 \$5
7.6% cccce|ccess \$4 \$6 (\$4)
7.6% cccee|ccess \$3 \$6 (\$4)
7.6% ccccs|ccees \$6 \$4
7.6% cccce|ccees \$4 \$4
5.1% cccce|ccces \$4 \$5
5.0% cccee|cccss \$3 \$7 (\$5)
3.8% cccce|ceess \$4 \$5 (\$3)
2.5% ccccs|cceee \$6 \$2
2.5% ccccs|ceees \$6 \$3
2.5% ccccs|cccee \$6 \$3
2.5% cccss|cceee \$7 (\$5) \$2
2.5% ccccs|ccces \$6 \$5
1.5% ccccc|ccees \$5 \$4
1.5% ccccc|ceess \$5 \$5 (\$3)
1.3% cccce|cccee \$4 \$3
1.2% cccce|cccss \$4 \$7 (\$5)
1.0% ccccc|ceees \$5 \$3
0.7% ccccc|ccess \$5 \$6 (\$4)
0.3% ccccc|cceee \$5 \$2
0.2% ccccc|eeess \$5 \$4 (\$2)
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### 36 Responses to Basic Opening Probabilities

1. timchen1017 says:

On the first two items, I think one important stat is missing: the chance of drawing at least one \$5 or better in turn 3 or four. Specifically, I think you need this number to justify your conclusion of the second part. When I open double terminal, I am not exactly looking for getting \$6 or better or 2 \$5’s on turn 3 or 4. More often than not I can be satisfied with a \$5. It is only disastrous if I have to settle for \$3s or \$4s on both turns.

• timchen1017 says:

Ah! You did list 1-x for this number. My bad. Still, 63% getting a \$5 is not that bad if the return is good. (Like double swindler on BSW in the old days…)

• Steven says:

Philosopher’s Stone doesn’t work very well with large handsizes, which is unruftonate because it’s otherwise always a good thing to improve your average handsize. Why is it always a good thing to improve your average hand size? That makes no sense.PStone with 30 cards = a \$5 card. Get 3 or 4 of them and you’re kicking ass.

• Illy says:

Bigger hand size => More treasures in hand => smaller treasures needed to buy provinces.

Look at Embassy, I rarely need to buy Gold in Embassy decks in order to buy Provinces. Compared to Smithy decks, which really need the Gold.

2. timchen1017 says:

By the way, great article! It is exactly what I want to see for a long time.

3. david707 says:

And if one of the many math grad students who read this blog would like to do some derivations if they have the free time (either repeating this work, or doing more advanced calculations…), they’re entitled to a guest post explaining their methodology and results 🙂

Okay, I’m not a grad yet, but I’m confident I can repeat this work using my probability knowledge rather than simulations.

• david707 says:

Drawing \$5 or better twice is 14.9%, it seems you forgot 1.5% from ccccc|ceess

• theory says:

Fixed!

4. Aaron says:

Great analysis. These are the kind of articles I want to see, because in my imagination at least it is the kind of insight I need to improve.
It also may explain why my wife, who usually buys silver with her 3, experiences much less early game frustration and much more end game winning than I do.

• Captain_Frisk says:

Double terminal openings are very dangerous, especially in a 2p game. Also important to note is that % chance of collision, is also your chance of hitting a combo (throne room / swindler for example). Devastating if you pull it off, but an unpaired throne room is just as bad as an estate otherwise.

Now there are some situations in which you would prefer to “shoot for the moon”. These are:

– You are playing multiple players where there is no reward for 2nd place. In this case playing high variance strategies may make sense.

– You are seriously outmatched. For example, if you are a bad to medium player, and you’re playing against theory who has 1st play, you’re probably going to lose. Might as well go crazy.

The classic example of this is treasure maps. Unassisted against single smithy big money, they lose most of the time (I want to say something like 35% – 65%) However, if they hit, they are almost impossible to beat, as they buy 4 provinces really early (turn 12-13). So if you’re sitting in 4th in a 4p game, and everyone in front of you opens smithy, you might as well go for the map, because you improve your chances of winning from ~20-25% to 30-35%. Granted, if your maps don’t work, you will lose the game in 4th place.

Note that if you play on isotropic, this logic does not apply. The rating system treats each pair of participants as a seperate game, so getting 2nd is still a net win (win against 2 people, loss against 1). If you’re in a tournament where standings matter, then the high variance strategy might not make sense.

5. guided says:

I think you’ve made a miscalculation. The probability of “1 Silver, 3 Coppers” with Chapel is at least 80-90% in my experience 😉

• Axxle says:

And the probability of terminals clashing is ~100%? 😛

Great read, theory. It’s great to get a sense of these numbers.

• Anonymous says:

Another egregious failure to take into account Murphy’s Law. 😉

6. Epoch says:

Hmmm. In the silver/silver case, of the 8.8% chance of not drawing \$5 on either turn 3 or turn 4, how much of that is because of a chance that your turn 6 draw is silver/silver/*/*/*?

Let me answer that question. The odds of getting ss in those positions is 2 in 1/12 * 1/11 * 2, so 1/66, or 1.5%

That suggests that, with a silver/silver opening,

91.2% of the time, you get a \$5 hand in turn 3, turn 4, or both.
7.3% of the time, you do not get a \$5 hand in turn 3 or turn 4, and your turn 5 is not particularly likely to be good.
1.5% of the time, you have a bad turn 3 and turn 4, but a very likely good turn 5. Probably quite good, \$6 or better.

I’m not sure that a good turn 5 saves you from a bad turn 3 and 4, since the bought card in turn 5 takes so much longer to re-enter your hand. Perhaps if you get some deck cycling in turn 3 or turn 4.

• guided says:

A good hand at turn 5 is very cold comfort for no \$5 buy before the 2nd shuffle. I will trade all the \$6 turn 5s in the world for one \$5 turn 4.

• Epoch says:

Yeah, I think you’re right.

Actually, an interesting Dominion strategy concept that I’ve never tackled head on: the potential value of buying early +cards not because you expect to get a ton of buying power out of them, but just as a way to your bought cards into your draw deck quicker. You only need +2 cards in turn 3 or 4 in order to get a chance at your turn 3 and 4 buys one turn earlier. And, indeed, your turns 5/6 buys can also come earlier.

If you assume broadly that the value of your buys steadily increases through the early game, it seems like even a 1-2 turn advantage in drawing what you buy might make a big difference.

• rrenaud says:

Careful, if you continue down this train of thought, you’ll soon be in the position of loving the chancellor ;P.

• Anonymous says:

Keep in mind that if you get a turn 5 draw of silver/copper/*/*/* that you can still get turns 3 & 4 draws of cccce | ccees. This would put you at \$4 | \$4 for the two turns

7. Tejayes says:

Since you’ve admitted to be lazy about the math (which actually isn’t too hard in this case), I’ll do it, if you don’t mind. At least for the double-Silver second deck, which is pretty easy.

The total number of permutations for your second deck — 12 cards consisting of 7 Coppers, 3 Estates, and 2 Silvers – will be 12!/7!3!2! = 7920. To get a \$7 hand, you’ll need both Silvers and 3 Coppers in one of the 2 consecutive sets of five cards that will be your third or fourth hands. Each theoretical hand would have 10 permutations (5!/3!2), with the other 4 Coppers and 3 Estates falling into one of 35 permutations (7!/4!3!). That means in a double-Silver second deck, there are 2*10*35 = 700 different ways to have a \$7 hand, and 700/7920 = 0.088383838… The result of 8.8% given was spot on.

To not draw \$5 on either turn 3 or 4, the possible ways to do so are: A) both Silvers at the very bottom of the deck, basically meaning a second first deck without the 1/6 possibility of a 5/2 or 2/5; and B) one Silver and one Copper at the bottom, and the second Silver in a hand-set with exactly two Estates. Basically, if you get both Silvers in your top half, you’ll have a \$5 hand to work with.

Anyway, since the 2 Silvers at the bottom have just 1 distinct order possible for possibility A, we can use the classic math for opening hand permutations (10!/7!3!) less 5/2 or 2/5 possibilities (5!/3!2!) to get 100 different permutations that work. For possibility B, the Silver-with-two-Estates hand has 5!/2!2! = 30 permutations, the Estate-and-four-Coppers hand has 5!/4! = 5 permutations, and the straggling Silver-and-Copper has 2! = 2 permutations. Including the 2 permutations of said hands, we get 2*30*5*2 = 600 total permutations for this outcome. Added together, we once again get 700 no-\$5-for-you permutations, once again verifying theory’s 8.8% theory.

I have a headache now (and no, it’s NOT from the math), so I’ll do the rest later.

• Tejayes says:

What the heck, I feel okay enough to do a little bit more.

For “at least one \$6 or \$7,” first of all, you’ll never have two \$6 or \$7 hands right away. You’ll only have \$11 total in your second deck, and you’d need at least \$12 to make two \$6 hands possible. But anyway, the total number of permutations for this scenario requires adding the possibilities for both a \$6 hand and a \$7 hand. We already have the number for \$7 from my previous post, so let’s work on \$6.

A second-deck \$6 hand would have one of two combos: A) 1 Silver and 4 Coppers; or B) 2 Silvers, 2 Coppers, and 1 Estate. For A, the \$6 hand has 5!/4! = 5 permutations possible, with the rest of the deck (since we don’t care about the other hand) having 7!/3!3! = 140 permutations possible. Remembering hand permutations (2), we get 2*5*140 = 1400 permutations for a deck containing an Estate-free \$6 hand.

For B, the \$6 hand has 5!/2!2! = 30 permutations possible, with the rest of the deck having 7!/5!2! = 21 permutations possible. With 2 hand permutations possible, we get a total of 2*30*21 = 1260 double-Silver-\$6-hand decks.

In total, that’s 700 + 1400 + 1260 = 3360 deck permutations giving a \$6 or more hand. 3360/7920 = 0.4242… What did the million simulated games give us? 42.4%. Ah, math. I love you.

• guided says:

Forgive me for making a nitpick that doesn’t matter or affect your analysis in any way, but when you say “permutations” you mean “combinations” 😉

• cpflames says:

I’m pretty sure “permutation” is exactly what he means.

A combination, in this starting hand discussion, is the set of 5 cards you can make out of the 12 cards you have. It’s a combination because you don’t care about the order, so we give it the name “scccc” for instance, because it has 1 silver and 4 coppers and we don’t care what order they’re in.

But if you want to know about the likelihood of “scccc”, you want to know how many permutations it has (different orderings the cards could be in) because more possible orderings -> more ways it could occur -> more likely.
So, the combination “scccc” has 5 permutations: scccc, csccc, ccscc, cccsc, and ccccs.

• guided says:

Embarrassing. I was distracted and skimming the math too quickly (there are formulas of the form n!/r!(n-r)!) and rushed to judgment. Ignore me!

• guided says:

(and on account of skimming also said to myself “these permutation numbers are just way too small to be permutations”, failing to realize they were taking multiple instances of each card as indistinguishable)

• Tejayes says:

Time for more double-Silver math.

Getting \$5 or better twice is only possible if at least one Estate is among the bottom two cards. If two Estates go bottom-feeding, then the only way to NOT get \$5/\$6 (or vice versa) is to get a \$7 (\$11 total, and all). So let’s just compute the number of permutations with two Estates in the bottom — 10!/7!2! = 360. Then, we’ll subtract the number of Estate-bottomed decks with \$7 hands. We’ll use the 10 permutations for such a hand from earlier, multiply that by 2 hand permutations, and then multiply by 5 (permutations for the 4-Copper-1-Estate hand remaining) to get 100. 360 – 100 = 260.

Now, if there were one Estate and one Copper on the bottom, both hands would need exactly \$5. With 2 Silvers, 6 Coppers, and 2 Estates remaining, we can accomplish this by: A) 1 Silver, 3 Coppers, and 1 Estate each hand; or B) 5 Coppers in one hand and the rest in the other. For both, we’ll have to multiply by 2 to count for the permutations of the bottom cards.

For A, each hand (having the same combination) would yield a possible 5!/3! = 20 permutations for a total of 20*20*2 = 800 permutations (doubling for hand permutations not done since the hand combos are the exact same). For B, there’s just one distinct perm for the 5-Copper hand, but 30 (5!/2!2!) for the other, giving us 30*1*2*2 = 120 perms.

That’ll give us a total of 260 + 800 + 120 = 1180 permutations for “I can buy two Counting Houses right away!” 1180/7920 = 0.148989… = 14.9%, just as the million run-throughs predicted.

• Tejayes says:

Finally, let’s tackle the average coin count for hands 3 and 4 for a double-Silver deck. We’ll do this by determining how many ways we can get each total of money among those two hands.

The most, of course, is \$11 by having two Estates on the bottom. We’ve already determined this number earlier — 360. We’ll multiply this by 11 to get 3960.

Let’s next go to the decks where we’ll only get \$7 among those two hands. Again, we determined this already by calculating the permutations for a deck with both Silvers on the bottom — 120. Times 7, and we get 840.

For \$10, we’ll need one Copper and one Estate on the bottom. With 2 ways to do that bottom, the top ten can be done 10!/6!2!2! = 1260 ways, which means 2*1260 = 2520 perms. 2520 * 10 = 25200.

Onto \$8, which can be done with a Silver and Copper on the bottom. Again, 2 ways to do that bottom times 10!/6!3! = 840 ways to do the top equals 1680 total \$8 perms. 1680 * 8 = 13440.

Finally, for the \$9. This time, we have two ways to do such: A) two Coppers on the bottom; and B) 1 Silver and 1 Estate. For A, the bottom has just one unique perm while the top has 10!/5!3!2! = 2520. For B, two distinct bottoms times 10!/7!2! = 360 gives us a total of 720. Added, we get 3240. 3240 * 9 = 29160.

To check our permutations numbers, let’s add them quick. 360 + 120 + 2520 + 1680 + 3240 = 7920. Checks out. That just leaves the final step of adding those perms multiplied by their values — 3960 + 840 + 25200 + 13440 + 29160 = 72600 — then dividing that by the total number of permutations — 72600/7920 = 9.1666… What did theory get? \$9.17. Go math!

• david707 says:

I’d worked all this out, but you posted your solutions first. I also worked out the two terminals and Chapel Probabilities, so I’ll post what I did:

The first thing I did is work out the number of possible arrangements for our 12 card deck. We have:
{7 Coppers, 3 Estates, 2 Silvers} (in the future I’ll abbreviate Copper to C, Estate to E and Silver to S).
The number of unique arrangements is 12!/(7!x3!x2!)=7920

Drawing \$7
To draw \$7 in turn 3, we require {SSCCC|3E, 4C}, so the number of unique ways to draw this in turn 3 or 4 is:
(5!/(2!x3!))x(7!/(3!x4!))x2=700
Here I needed to multiply by 2 to get turn 3 or turn 4, since SSCCC can be the first five card block or the second 5 card block.
The Probability is the number of ways to draw \$7 divided by the total arrangements:
700/7920=35/396=8.83% (All percentages given to two decimal places)

At least \$6
One way to draw \$6 is {SSCCE|5C,2E}, number of ways to get this is (5!/(2!x2!x1!))x(7!/(5!x2!))x2=1260, since 1! = 1 and we are multiplying, in the future I will not write 1! In my calculations.
The other way to draw \$6 is {SCCCC|3C,3E,S}, (5!/4!)x(7!/(3!x3!))x2=1400
We also need to include the 700 ways of drawing \$7: 1260+1400+700=3360
Probability is: 3360/7920=14/33=42.42%
Note that since we have a total of \$11 in our deck it’s impossible to draw \$6 or more on turn 3 and turn 4.

Not drawing \$5+ on either turn
There are two cases to consider, we can draw \$4 then \$4 by drawing {CCCCE|CCSEE|SC}, which we can draw in (5!/4!)x(5!/(2!x2!))x(2!)x2=600 ways. Alternatively we could draw \$4 then \$3 by drawing {CCCCE|CCCEE|SS}, which we can draw in (5!/4!)x(5!/3!)x(2!/2!)x2=100 ways, for a total of 700 ways of not getting \$5+ on either turn.
Probability is: 700/7920=35/396=8.83%

Draw \$5+ twice
We could draw \$6 then \$5 in two cases:
{SSCCE|CCCCC|EE}, which can be drawn in (5!/(2!x2!))x2=60 ways.
{SCCCC|SCCCE|EE}, which can be drawn in (5!/4!)x(5!/3!)=200 ways.
We could also draw \$5 then \$5 in two cases:
{SSCEE|CCCCC|CE}, which can be drawn in (5!/(2!x2!))x(2!)x2=120 ways.
{SCCCE|SCCCE|CE}, which can be drawn in (5!/3!)x(5!/3!)x(2!)=800 ways.
Notice that for {SCCCE|SCCCE|CE} I exclude the last x2, since turn 3’s hand is the same as turn 4’s hand, so we can’t double the number of ways by switching them around.
We can draw \$5+ twice in a total of 1180 ways.
Probability is: 1180/7920=59/396=14.90%

Average total money
Here I assumed by average, Theory meant the mean.
I decided to look at each amount separately:
Money Draw Calculation Ways
\$11 {7C,2S,E|EE} 10!/(7!x2!) 360
\$10 {6C,2S,2E|EC} (10!/(6!x2!x2!))x(2!) 2520
\$9 (Case 1) {5C,2S,3E|CC} 10!/(5!x3!x2!) 2520
\$9 (Case 2) {7C,S,2E|SE} (10!/(7!x2!))x(2!) 720
\$8 {6C,S,3E|SC} (10!/(6!x3!))x(2!) 1680
\$7 {7C,3E|SS} 10!/(7!x3!) 120
– – Total: 7920

The mean (average) is ((360×11)+(2520×10)+(3240×9)+(1680×8)+(120×7))=55/6=9.17

Collision of two terminals
For this I decided to temporarily view our deck as {2T,10X}, which can be drawn in 12!/(10!x2!)=66 ways. We require {TTXXX|7X}, which can be drawn in 5!/(3!x2!)=20 ways. The probability is 20/66=10/33=30.30%

At least \$6 (two terminals)
Here, we look at our previous at least one \$6:
{SSCCE|5C,2E} had 1260 ways to draw it, but we no longer include those ways since it isn’t worth \$6 anymore. {SCCCC|3C,3E,S} had 1400 ways to draw it and we still include it.
Probability: 1400/7920=35/198=17.68%

Not drawing \$5+ on either turn (two terminals)
Here, we need to count the 700 ways we had previously, but add on some new ways:
There is another \$4 then \$4, {SSCCE|CCCCE|CE}, which can be drawn in (5!/(2!x2!))x(5!/4!)x(2!)x2=600 ways.
There are two more cases of \$4 then \$3:
Case 1 is {SSCCE|CCCEE|CC}, which can be drawn in (5!/(2!x2!))x(5!/(3!x2!))x2=600 ways.
Case 2 is {SSCEE|CCCCE|CC}, which can be drawn in (5!/(2!x2!))x(5!/4!)x2=300 ways.
We have a total of 2200 ways for a probability of: 2200/7920=5/18=27.78%

Draw \$5+ twice (two terminals)
Again we look at our previous calculations; we had 1180 ways, but now {SSCCE|CCCCC|EE} no longer works, which was 60 ways. Also, {SSCEE|CCCCC|CE} no longer works and that was 120 ways. This leaves us with 1000 ways and gives us a probability of: 1000/7920=25/198=12.62%

Average total money (two terminals)
Sometimes our two terminals will collide, causing us to earn \$2 less. We know that they will collide with probability 10/33. Using this we can work out the new average money (remember 55/6 was the previous average money):
New Average: (10/33)*((55/6)-2)+(23/33)*(55/6)=\$8.56

Chapel Probabilities

Note that we now have a different deck (K represent chapel):
{7C,3E,S,K}, giving us 12!/(7!*3!)=15840 ways total as opposed to the old 7920.
For drawing the Chapel on turn 5, quite simply there are 12 places it could be in the deck (with equally likely probability) and 2 of those 12 places mean you draw it on turn 5, giving us 2/12=1/6.
The percentages add up to 100.01% due to slight rounding error.

Draw Chapel with: Draw Calculation Ways Probability
3 Coppers, 1 Estate {KCCCE|4C,2E,S} (5!/3!)x(7!/(4!x2!))x2 4200 35/132=26.52%
On turn 5 {10X|XK} See above 2640 1/6=16.67%
1 Silver, 2 Coppers, 1 Estate {KSCCE|5C,2E} (5!/2!)x(7!/(5!x2!))x2 2520 7/44=15.91%
2 Coppers, 2 Estates {KCCEE|5C,E,S} (5!/(2!x2!))x(7!/5!)x2 2520 7/44=15.91%
4 Coppers {KCCCC|3C,2E,S} (5!/4!)x(7!/(3!x3!))x2 1400 35/396=8.84%
1 Silver, 3 Coppers {KSCCC|4C,3E} (5!/3!)x(7!/(4!x3!))x2 1400 35/396=8.84%
1 Silver, 1 Copper, 2 Estates {KSCEE|6C,E} (5!/2!)x(7!/6!)x2 840 7/132=5.30%
1 Copper, 3 Estates {KCEEE|6C,S} (5!/3!)x(7!/6!)x2 280 7/396=1.77%
1 Silver, 3 Estates {KSEEE|7C} (5!/3!)x2 40 1/396=0.25%
– – Total 15840 100.01%

Baron or Salvager with at least one Estate:

Simply add up the ways for all the cases with at least one estate: 4200+2520+2520+840+280+40=10400
Probability: 10400/15840=65/99=65.66%

I’ve solved the “Treasure Map” question theory asked for when p is divisible by 5. I think after I’ve solved it in general along with the warehouse question I might ask for a guest article on it.

• Oxeador says:

Actually, there is a much easier way to calculate the average total money.

In the case of two silvers, you have total of 11 coins. Each coin will make it into (round 3 or round 4) with probability 10/12. Hence the average total money should be 11*10/12 = 55/6, which equals 9.17 approximately as theory found.

In the case of two terminal silvers, the expected value should be 55/6-2p, where p is the probability of the two terminal silvers clashing. Luckily we have calculated above p = 10/33, so the expected total money is 55/6-2*10/33=565/66, which is 8.56 as theory found.

The fact that expected values are linear makes life so much simpler.

8. stencil says:

http://stattrek.com/Tables/Hypergeometric.aspx is a useful tool for generating odds should you ever want to. You do have to do some work though to combine the odds for different situations into a useful and comprehensive picture – that’s something a sim does better (like the chances of getting \$5, \$6 or 7\$ for turns 3 and/or 4).

If you wanted to use it to calculate the chances of the baron getting you \$5 on either turn 3 or 4 you could use the tool the following way (using format: population size / sample size / # of successes in population / # of successes in sample / probability used). Note: this is assuming we don’t have a second terminal and that even with the baron and 3 estates in one hand, that we’ll have the coin we need to hit \$5.

baron in turn 3 = 12/5/1/1/P(X=1) = 42%
at least 1 estate in rest of turn 3 = 11/4/3/1/P(X>=1) = 79%
chance of \$5 in turn 3 with baron = 42% * 79% = 33%

baron in turn 4 = chance not drawn in turn 3 * 7/5/1/1/P(X=1) = (100% – 42%) * 71% = 41%
at least 1 estate in turn 4 if 0 were in turn 3 = 6/4/3/1/P(X>=1) * 12/5/3/0/P(X=0) = 100% * 16% = 16%
at least 1 estate in turn 4 if 1 were in turn 3 = 6/4/2/1/P(X>=1) * 12/5/3/1/P(X=1) = 93% * 48% = 45%
at least 1 estate in turn 4 if 2 were in turn 3 = 6/4/1/1/P(X>=1) * 12/5/3/2/P(X=2) = 67% * 32% = 21%
at least 1 estate in turn 4 if 3 were in turn 3 = 6/4/0/1/P(X>=1) * 12/5/3/3/P(X=3) = 0% * 5% = 0%
chance of \$5 in turn 4 with baron = 41% * (16% + 45% + 21%) = 34%

cummulative chance = 67%

Note that those are just the odds for making \$5 using the baron; I’ve omitted hands making \$5 through coins alone. Figuring out the numbers for things in turn 3 is usually pretty easy, its when you have to use those to start figuring out the chances of stuff in turn 4 that it becomes a complicated mess. Which is why I really need to get around to finding time to put some more work into my own simulator 🙂

9. chris says:

I was surprised to see such a high probability of terminal collision — and that’s not even counting the possibility that one of them is a terminal *draw* like Smithy, which can kill the other action by drawing it even if you didn’t draw them together. (Does it include the chance that the two actions are cards 11 and 12? In that case you would know on turn 4 that the collision was coming and would have some chance of saving it by buying a +2 Action, if there are any on the board, which could, with some luck, be in the top 3 cards of your end-of-turn-4 reshuffle. Of course if you open two terminals on turns 1/2, an action booster might be a reasonable turn 3-4 buy anyway, with an appropriate amount of money. Especially if you have a good reason for avoiding Silver, such as Pirate Ship.)

The one opening probability I didn’t see, though: what is the chance of a 5/2 split? It seems that this is often overlooked in “how would you open on this board” type discussions so I assume it must be fairly rare, but it’s not clear to me how to calculate exactly *how* rare.

The chance that your first card is copper is 7/10. Given that, the second card is 6/9, then 5/8, 4/7, and 3/6 likely to be copper, so the total probability of drawing 5 copper (followed by 2 copper, 3 estates) is 7*6*5*4*3/10*9*8*7*6 = 5*4*3/10*9*8 = 60/720 = 1/12. By symmetry, the chance for your *last* five cards to be copper is the same, so the total chance of 5/2 split is 1/6.

…is that correct? If so, in a two player game the chance that exactly one player will have a 5/2 is 10/36 or almost 1/3, and 1/36 of the time both players will open 5/2, and with more players it is even more likely that at least one 5/2 will be present, so ISTM that it should be taken into account in analyzing opening strategies. (On which boards does having a different opening split create a significant swing in win probability? Witch comes to mind…)

• Personman says:

Yes, 1/6 is correct.

I don’t believe the number posted includes the chance that your two terminals are the last two cards, since the post states that these are probabilities for turns 3 and 4. The probability of your two purchases being cards 11 and 12 1/12 * 1/11 =~ .75%, so it doesn’t make a huge difference.

• rrenaud says:

1/6 is correct, but .75% is not. It’s actually double that, since there are two ways to order your terminals.

• Personman says:

Ha, I remembered to account for that when I was originally calculating it in a more complex, incorrect fashion, but then forgot after I realized I was being stupid 😛

Statistics is hard.

• rrenaud says:

Yeah. But I’d consider this just straight up probability. There is no observation and data collecting involved.

• Personman says:

Look, stop being right about things. I didn’t get that much sleep last night!

😀

10. filovirus says:

Just did the math. You’re results are pretty accurate.