Expected Armies Due Next Turn

[tahitiwahini]

How many armies will a player receive in his next turn?

I'm interested here in a flat rate game, but this idea could be easily be extended to escalating games as well.

I want to quantify the contribution that cashing in a set of cards will have on the number of armies a player will receive in his next turn. This contribution is based on the number of cards currently held by a player.

Let:

T(n) be total expected armies player will receive in his next turn when holding n cards.
B be bonus armies due for owning countries (country owning bonus).
b(d) be bonus armies for owning d countries (basic allotment)
d be countries held by a player.
c be armies due for owning a group of countries (continent bonus).
m be consecutive missed turns by player.
E(n) be expected armies for cashing in a set of cards (card cashing bonus).
e(n) be expected armies for cashing in a particular color set when holding n cards (card set bonus).
H(n) be expected armies for owning countries specified on cards cashed in (country card bonus).
h(n) be probability player has set when holding n cards.
D be total number of countries on game map.

Then, total expected armies when holding n cards is sum of country owning bonus and card cashing bonus:

T(n) = B + E(n)

Where, country owning bonus is sum of basic allotment and continent bonus adjusted by missed turn multiplier:

B = (b + c) * (m + 1)

Basic allotment is 3 armies plus 1 for every additional multiple of 3 armies held above 12:

b(d < 12) = 3
b(d >= 12) = floor (d / 3)

Card cashing bonus is sum of card set bonus and country card bonus:

E(n) = e(n) + H(n)

Expected card set bonus when holding n cards:

Code: Select all

e(0 < n < 3) = 0   // card set not possible
e(3) = 2.889   // 10*(18/81)+(4+6+8)*(3/81); 3^4=81 possibilities,  18+(3*3)=27 sets
e(4) = 5.333   // 10*(18/54)+(4+6+8)*(6/54); 81-27=54 possibilities, 18+(3*6)=36 sets
e(5) = 7.333   // 10*(6/18)+(4+6+8)*(4/18); 81-27-36=18 possibilities, 6+(3*4)=18 sets


Expected country card bonus when holding n cards (3 cards each worth 2 armies if country named on card cashed is owned by player):

H(n) = 3 * 2 * h(n) * (d / D)

Expected probability of having a set when holding n cards:

[Edit]

Code: Select all

h(0 < n < 3) = 0  // card set not possible
h(3) = 0.333                 // 27/81
h(4) = 0.667                 // 36/54
h(5) = 1                      // 18/18


[/Edit]


Examples assuming the Classic map with 42 total countries:

1) Player owns 15 countries, owns North America and South America, and holds 3 cards:

T(3) = (5 + 7) * (1 + 0) + 2.889 + 3 * 2 * 0.333 * (15 / 42) = 15.6 new armies next turn.

He will receive at least 12, at most 28, on average 15.6 new armies next turn.

The least you can receive in flat rate when cashing a set is 4 (for red), the most is 16 (10 for mixed, and 6 for owning all three countries specified on the cards).

2) Player owns 13 countries, owns Australia, and holds 4 cards:

T(4) = (4 + 2) * (1 + 0) + 5.333 + 3 * 2 * 0.778 * (13 / 42) = 12.8 new armies next turn.

He will receive at least 6, at most 22, on average 12.8 new armies next turn.

3) Player owns 8 countries, owns no continents, missed two turns, and holds 5 cards:

[Edit]
T(5) = (3 + 0) * (2 + 1) + 7.333 + 3 * 2 * 1 * (8 / 42) = 14.5 new armies next turn.
[/Edit]

He will receive at least 10, at most 22, on average 14.5 new armies next turn.

[dominationnation]

2...many...numbers...ahhhhhh

[Hologram]

It might help if the variable definitions weren't confusing.

[s.xkitten]

hey hey hey...i come to CC to escape from school...stop being smart

[Lev306]

interesting idea, though the avg armies wont really work because since the amount u get varies greatly depending on continent bonuses and countries held.

I suppose it might be some what useful to calculate the maximum and minimum number of armies, and also a likely number of armies, but it will be very difficult because of the amount of parameters involved. Things to consider would be, if the player just deployed and didn't attack, all out attacks, dice luck, etc. Will be very difficult to program and calculate but if u get it working it would be very cool, especially if its accurate and precise.

[ClessAlvein]

I'm not going to bother doing the math, but the current model assumes two things:

1) Players will not save up cards in hopes of a mixed set
2) Players will not cash in existing low-value sets in order to get the country bonuses

If you like trudging through the probabilities, you can add a psychology factor into it by calculating the lowered probability of a rrr, ggg, or bbb set being cashed in, and also factoring in the heightened probability of a rrr, ggg, or bbb set being cashed in if two or more of the three countries are currently held, despite their lowered probability described by the base psychology of holding out for a better set.

[podge]

How many armies will a player receive in his next turn?

I'm interested here in a flat rate game, but this idea could be easily be extended to escalating games as well.

I want to quantify the contribution that cashing in a set of cards will have on the number of armies a player will receive in his next turn. This contribution is based on the number of cards currently held by a player.

Let:

T(n) be total expected armies player will receive in his next turn when holding n cards.
B be bonus armies due for owning countries (country owning bonus).
b(d) be bonus armies for owning d countries (basic allotment)
d be countries held by a player.
c be armies due for owning a group of countries (continent bonus).
m be consecutive missed turns by player.
E(n) be expected armies for cashing in a set of cards (card cashing bonus).
e(n) be expected armies for cashing in a particular color set when holding n cards (card set bonus).
H(n) be expected armies for owning countries specified on cards cashed in (country card bonus).
h(n) be probability player has set when holding n cards.
D be total number of countries on game map.

Then, total expected armies when holding n cards is sum of country owning bonus and card cashing bonus:

T(n) = B + E(n)

Where, country owning bonus is sum of basic allotment and continent bonus adjusted by missed turn multiplier:

B = (b + c) * (m + 1)

Basic allotment is 3 armies plus 1 for every additional multiple of 3 armies held above 12:

b(d < 12) = 3
b(d >= 12) = floor (d / 3)

Card cashing bonus is sum of card set bonus and country card bonus:

E(n) = e(n) + H(n)

Expected card set bonus when holding n cards:

Code: Select all

e(0 < n < 3) = 0   // card set not possible
e(3) = 2.889   // 10*(18/81)+(4+6+8)*(3/81); 3^4=81 possibilities,  18+(3*3)=27 sets
e(4) = 5.333   // 10*(18/54)+(4+6+8)*(6/54); 81-27=54 possibilities, 18+(3*6)=36 sets
e(5) = 7.333   // 10*(6/18)+(4+6+8)*(4/18); 81-27-36=18 possibilities, 6+(3*4)=18 sets


Expected country card bonus when holding n cards (3 cards each worth 2 armies if country named on card cashed is owned by player):

H(n) = 3 * 2 * h(n) * (d / D)

Expected probability of having a set when holding n cards:

Code: Select all

h(0 < n < 3) = 0  // card set not possible
h(3) = 0.333                 // 27/81
h(4) = 0.778                 // (27+36)/81
h(5) = 1                      // (27+36+18)/81

Examples assuming the Classic map with 42 total countries:

1) Player owns 15 countries, owns North America and South America, and holds 3 cards:

T(3) = (5 + 7) * (1 + 0) + 2.889 + 3 * 2 * 0.333 * (15 / 42) = 15.6 new armies next turn.

He will receive at least 12, at most 28, on average 15.6 new armies next turn.

The least you can receive in flat rate when cashing a set is 4 (for red), the most is 16 (10 for mixed, and 6 for owning all three countries specified on the cards).

2) Player owns 13 countries, owns Australia, and holds 4 cards:

T(4) = (4 + 2) * (1 + 0) + 5.333 + 3 * 2 * 0.778 * (13 / 42) = 12.8 new armies next turn.

He will receive at least 6, at most 22, on average 12.8 new armies next turn.

3) Player owns 8 countries, owns no continents, missed two turns, and holds 5 cards:

T(2) = (3 + 0) * (2 + 1) + 7.333 + 3 * 2 * 1 * (8 / 42) = 14.5 new armies next turn.

He will receive at least 10, at most 22, on average 14.5 new armies next turn.

You have far too much free time on your hands

[tahitiwahini]

I'm not going to bother doing the math, but the current model assumes two things:

1) Players will not save up cards in hopes of a mixed set
2) Players will not cash in existing low-value sets in order to get the country bonuses

If you like trudging through the probabilities, you can add a psychology factor into it by calculating the lowered probability of a rrr, ggg, or bbb set being cashed in, and also factoring in the heightened probability of a rrr, ggg, or bbb set being cashed in if two or more of the three countries are currently held, despite their lowered probability described by the base psychology of holding out for a better set.

The model's assumption is that a player will cash in a card set as soon as possible. This assumption was made for three reasons: it's arguably the optimal strategy, it's probably the thing that happens most frequently, and it simplifies the calculations. So, your first point is correct. But for the same reason your second point is not. The second point is actually an inducement to get a player to behave in the way that the model assumes he will, namely cashing in a set as soon as possible.

As for adding a psychological factor, I would have no idea how to quantify that. There are all sorts of factors that go into why someone would or would not cash in a set, I'm not sure how any of them could be quantified, but I'm open to suggestions.

[tahitiwahini]

interesting idea, though the avg armies wont really work because since the amount u get varies greatly depending on continent bonuses and countries held.

I suppose it might be some what useful to calculate the maximum and minimum number of armies, and also a likely number of armies, but it will be very difficult because of the amount of parameters involved. Things to consider would be, if the player just deployed and didn't attack, all out attacks, dice luck, etc. Will be very difficult to program and calculate but if u get it working it would be very cool, especially if its accurate and precise.

I'll agree that the average armies received in the next turn does embody a good deal of uncertainty, but not for the reasons you gave. The number of armies received due to bonuses for countries and continents held is known exactly. In fact, it's currently calculated in the Conquer Club - Roberts Mods greasemonkey script. What is uncertain are the armies to be received for cashing in a set of cards. That amount does vary greatly: from as little as zero (not cashing a set), to 4 (for a red set), to as much as 16 (10 for a mixed set, plus getting the country card bonus for all three cards). That is quite a range and the average suffers from that much uncertainty. I've modified the Conquer Club - Roberts Mods script to make the new calculations and I'm currently using it in my games.

Your second idea is a good one I think. In addition to knowing the average expected armies due each turn, it would be a good thing to know the minimum and maximum values as well. I don't think the factors you mentioned will affect this calculation at all which is limited to predicting the number of armies a player will be able to deploy on his next turn. Attacks and dice luck are outside of the model I'm making and don't really influence the number of armies a player will be able to deploy on his next turn. I'm looking into calculating the maximum and minimum values.

[RobinJ]

How many armies will a player receive in his next turn?

I'm interested here in a flat rate game, but this idea could be easily be extended to escalating games as well.

I want to quantify the contribution that cashing in a set of cards will have on the number of armies a player will receive in his next turn. This contribution is based on the number of cards currently held by a player.

Let:

T(n) be total expected armies player will receive in his next turn when holding n cards.
B be bonus armies due for owning countries (country owning bonus).
b(d) be bonus armies for owning d countries (basic allotment)
d be countries held by a player.
c be armies due for owning a group of countries (continent bonus).
m be consecutive missed turns by player.
E(n) be expected armies for cashing in a set of cards (card cashing bonus).
e(n) be expected armies for cashing in a particular color set when holding n cards (card set bonus).
H(n) be expected armies for owning countries specified on cards cashed in (country card bonus).
h(n) be probability player has set when holding n cards.
D be total number of countries on game map.

Then, total expected armies when holding n cards is sum of country owning bonus and card cashing bonus:

T(n) = B + E(n)

Where, country owning bonus is sum of basic allotment and continent bonus adjusted by missed turn multiplier:

B = (b + c) * (m + 1)

Basic allotment is 3 armies plus 1 for every additional multiple of 3 armies held above 12:

b(d < 12) = 3
b(d >= 12) = floor (d / 3)

Card cashing bonus is sum of card set bonus and country card bonus:

E(n) = e(n) + H(n)

Expected card set bonus when holding n cards:

Code: Select all

e(0 < n < 3) = 0   // card set not possible
e(3) = 2.889   // 10*(18/81)+(4+6+8)*(3/81); 3^4=81 possibilities,  18+(3*3)=27 sets
e(4) = 5.333   // 10*(18/54)+(4+6+8)*(6/54); 81-27=54 possibilities, 18+(3*6)=36 sets
e(5) = 7.333   // 10*(6/18)+(4+6+8)*(4/18); 81-27-36=18 possibilities, 6+(3*4)=18 sets


Expected country card bonus when holding n cards (3 cards each worth 2 armies if country named on card cashed is owned by player):

H(n) = 3 * 2 * h(n) * (d / D)

Expected probability of having a set when holding n cards:

Code: Select all

h(0 < n < 3) = 0  // card set not possible
h(3) = 0.333                 // 27/81
h(4) = 0.778                 // (27+36)/81
h(5) = 1                      // (27+36+18)/81

Examples assuming the Classic map with 42 total countries:

1) Player owns 15 countries, owns North America and South America, and holds 3 cards:

T(3) = (5 + 7) * (1 + 0) + 2.889 + 3 * 2 * 0.333 * (15 / 42) = 15.6 new armies next turn.

He will receive at least 12, at most 28, on average 15.6 new armies next turn.

The least you can receive in flat rate when cashing a set is 4 (for red), the most is 16 (10 for mixed, and 6 for owning all three countries specified on the cards).

2) Player owns 13 countries, owns Australia, and holds 4 cards:

T(4) = (4 + 2) * (1 + 0) + 5.333 + 3 * 2 * 0.778 * (13 / 42) = 12.8 new armies next turn.

He will receive at least 6, at most 22, on average 12.8 new armies next turn.

3) Player owns 8 countries, owns no continents, missed two turns, and holds 5 cards:

T(2) = (3 + 0) * (2 + 1) + 7.333 + 3 * 2 * 1 * (8 / 42) = 14.5 new armies next turn.

He will receive at least 10, at most 22, on average 14.5 new armies next turn.

You have far too much free time on your hands

Sorry to quote such a large amount of info but I'm lazy unlike you , tahitiwahiti - definitely too much free time - go out and have fun!

Also, I might understand the figures if I tried but there is simply too much, especially for the general discussion forum!

[ClessAlvein]

Well, the "psychology factor" is really just splaying out all 3^4 permutations of posessing 4 cards plus all 3^5 permutations of posessing 5 cards, and adjusting your final probabilities accordingly. I'll do the 4-card one:

rrr + [r, g, b] ** (60% withhold)
rrg + [r, g, b]
rrb + [r, g, b]
rgr + [r, g, b]
rgg + [r, g, b]
rgb + [r, g, b] *
rbr + [r, g, b]
rbg + [r, g, b] *
rbb + [r, g, b]

grr + [r, g, b]
grg + [r, g, b]
grb + [r, g, b] *
ggr + [r, g, b]
ggg + [r, g, b] ** (40% withhold)
ggb + [r, g, b]
gbr + [r, g, b] *
gbg + [r, g, b]
gbb + [r, g, b]

brr + [r, g, b]
brg + [r, g, b] *
brb + [r, g, b]
bgr + [r, g, b] *
bgg + [r, g, b]
bgb + [r, g, b]
bbr + [r, g, b]
bbg + [r, g, b]
bbb + [r, g, b] ** (20% withhold)

Where * indicates an impossible combination, because the player has already turned it in, and ** indicates that the player might be saving up for a better set. I used some rough percentages based on how many more armies you get if you hold out for a mixed set as the likelihood of withholding one's cards.

Final probability of turning in (not having) a set: [3*0.4 + 3*0.6 + 3*0.8 + 2*6*3] / [3*3 + 3*6*3] = 41.4/63 = 0.6571

Compare this with 0.778, which is artificially inflated because it assumes that players will not turn in any set when they first get it.

This becomes somewhat more complicated for the 5-card process, but I can probably find time to do it later tonight.

Also, as my second point is a conditional sub-set of the first point, it's hard to accept only my first point and not the second, as the second point merely refines the first point. Let's say a player has rrr. On his turn, the likelihood of him turning it in is lower than if he had, say, bbb. If, however, he holds all three of those countries on those rrr cards, then he will almost always turn it in (he might lose those countries next round!). If you base it on armies gained vs. a mixed set, the percentages will be 40%, 20%, 0% chance of withholding, for 1, 2, and 3 of those countries owned, respectively. I suspect that this calculation is going to be a very tedious calculation for very little refinement of the algorithm, though.

[Aimless]

Well, the "psychology factor" is really just splaying out all 3^4 permutations of posessing 4 cards plus all 3^5 permutations of posessing 5 cards, and adjusting your final probabilities accordingly.

While this is true, I think it's relatively unimportant, since as tahitiwahini said the optimal strategy is very likely turning in a set of cards of any type as soon as you get them.

I don't want to go through the math again (I did so in an old thread here), but the opportunity cost of turning in a set of four red cards rather than waiting on the hope that you'll make a mixed set is only about 1.333 armies. For the other mixed sets, it's obviously lower.

Since this opportunity cost is spread out over two turns, it is probably not a good idea to wait. Further, if you do wait, and get lucky and cash a mixed set, you will be seeding your next hand with two red (or other matched) cards, lowering the likelihood that the set after will also be mixed.

[Blueoctober]

How many armies will a player receive in his next turn?

I'm interested here in a flat rate game, but this idea could be easily be extended to escalating games as well.

I want to quantify the contribution that cashing in a set of cards will have on the number of armies a player will receive in his next turn. This contribution is based on the number of cards currently held by a player.

Let:

T(n) be total expected armies player will receive in his next turn when holding n cards.
B be bonus armies due for owning countries (country owning bonus).
b(d) be bonus armies for owning d countries (basic allotment)
d be countries held by a player.
c be armies due for owning a group of countries (continent bonus).
m be consecutive missed turns by player.
E(n) be expected armies for cashing in a set of cards (card cashing bonus).
e(n) be expected armies for cashing in a particular color set when holding n cards (card set bonus).
H(n) be expected armies for owning countries specified on cards cashed in (country card bonus).
h(n) be probability player has set when holding n cards.
D be total number of countries on game map.

Then, total expected armies when holding n cards is sum of country owning bonus and card cashing bonus:

T(n) = B + E(n)

Where, country owning bonus is sum of basic allotment and continent bonus adjusted by missed turn multiplier:

B = (b + c) * (m + 1)

Basic allotment is 3 armies plus 1 for every additional multiple of 3 armies held above 12:

b(d < 12) = 3
b(d >= 12) = floor (d / 3)

Card cashing bonus is sum of card set bonus and country card bonus:

E(n) = e(n) + H(n)

Expected card set bonus when holding n cards:

Code: Select all

e(0 < n < 3) = 0   // card set not possible
e(3) = 2.889   // 10*(18/81)+(4+6+8)*(3/81); 3^4=81 possibilities,  18+(3*3)=27 sets
e(4) = 5.333   // 10*(18/54)+(4+6+8)*(6/54); 81-27=54 possibilities, 18+(3*6)=36 sets
e(5) = 7.333   // 10*(6/18)+(4+6+8)*(4/18); 81-27-36=18 possibilities, 6+(3*4)=18 sets


Expected country card bonus when holding n cards (3 cards each worth 2 armies if country named on card cashed is owned by player):

H(n) = 3 * 2 * h(n) * (d / D)

Expected probability of having a set when holding n cards:

Code: Select all

h(0 < n < 3) = 0  // card set not possible
h(3) = 0.333                 // 27/81
h(4) = 0.778                 // (27+36)/81
h(5) = 1                      // (27+36+18)/81

Examples assuming the Classic map with 42 total countries:

1) Player owns 15 countries, owns North America and South America, and holds 3 cards:

T(3) = (5 + 7) * (1 + 0) + 2.889 + 3 * 2 * 0.333 * (15 / 42) = 15.6 new armies next turn.

He will receive at least 12, at most 28, on average 15.6 new armies next turn.

The least you can receive in flat rate when cashing a set is 4 (for red), the most is 16 (10 for mixed, and 6 for owning all three countries specified on the cards).

2) Player owns 13 countries, owns Australia, and holds 4 cards:

T(4) = (4 + 2) * (1 + 0) + 5.333 + 3 * 2 * 0.778 * (13 / 42) = 12.8 new armies next turn.

He will receive at least 6, at most 22, on average 12.8 new armies next turn.

3) Player owns 8 countries, owns no continents, missed two turns, and holds 5 cards:

[Edit]
T(5) = (3 + 0) * (2 + 1) + 7.333 + 3 * 2 * 1 * (8 / 42) = 14.5 new armies next turn.
[/Edit]

He will receive at least 10, at most 22, on average 14.5 new armies next turn.

you could have just counted everything up on the map and figured it out by how many the last guy who used cards got but your way is much simpler

[ClessAlvein]

Well, the "psychology factor" is really just splaying out all 3^4 permutations of posessing 4 cards plus all 3^5 permutations of posessing 5 cards, and adjusting your final probabilities accordingly.

While this is true, I think it's relatively unimportant, since as tahitiwahini said the optimal strategy is very likely turning in a set of cards of any type as soon as you get them.

I don't want to go through the math again (I did so in an old thread here), but the opportunity cost of turning in a set of four red cards rather than waiting on the hope that you'll make a mixed set is only about 1.333 armies. For the other mixed sets, it's obviously lower.

Since this opportunity cost is spread out over two turns, it is probably not a good idea to wait. Further, if you do wait, and get lucky and cash a mixed set, you will be seeding your next hand with two red (or other matched) cards, lowering the likelihood that the set after will also be mixed.

The bit about "seeding" and spreading over two turns are pretty good points, but I'd like to point out that the original calculations made by tahitiwahini do not assume that a player cashes in as soon as he gets a set. Otherwise the denominator would not be 3^4.

[Aimless]

The bit about "seeding" and spreading over two turns are pretty good points, but I'd like to point out that the original calculations made by tahitiwahini do not assume that a player cashes in as soon as he gets a set. Otherwise the denominator would not be 3^4.

Right. I'm actually pretty sure he's using the expectation values I worked out in the other thread; at least, the numbers he's using are the same, and I got those values by considering cases where the player cashed in immediately after making any one set.

I was just try to justify that I'm not sure there is much of an advantage to trying to modify the expectation values based upon a psychological effect of a player wanting to hold onto his cards in hopes of a better set, since that doesn't seem to be an optimum strategy.

[tahitiwahini]

How many armies will a player receive in his next turn?

I'm interested here in a flat rate game, but this idea could be easily be extended to escalating games as well.

you could have just counted everything up on the map and figured it out by how many the last guy who used cards got but your way is much simpler

In flat rate, the redemption value of a set is independent of the value of the last set cashed.

Maybe I'm missing your point, but I don't see how the counting method would help with determining how many armies a player will receive in his next turn, even in an escalating game.

[tahitiwahini]

The bit about "seeding" and spreading over two turns are pretty good points, but I'd like to point out that the original calculations made by tahitiwahini do not assume that a player cashes in as soon as he gets a set. Otherwise the denominator would not be 3^4.

Actually, the original calculations made by me do assume that a player cashes in as soon as he gets a set.

Aimless made the original calculations he referenced in the other thread. I wanted to check those calculations by arriving at the solution through enumeration of all the possibilities. The reason for doing this was to make sure that all the assumptions being made were in fact valid.

To do this I would have had to enumerate all possibilities for five cards, namely 3^5. I started doing this and did the first 81 (3^4) possibilities and then realized I could stop at that point. What I wanted to determine was the number of mixed sets and pure sets that would occur when holding 3, 4, and 5 cards. The table below shows all 81 possibilities for the first 4 cards. The column representing the fifth card is always an A.

I then went through each hand starting with the first three cards. If there was a set present in those three cards, then I made an indication of the set in the column for sets made with the first three cards. If there wasn't a set made in the first three cards, then I looked at the fourth card. If there was a set in present in those four cards, then I made an indication of the set in the column for sets made with the first four cards. And so on for the first five cards.

Code: Select all

5   4   3   2   1   5   4   3
A   A   A   A   A         AAA
A   A   A   A   B      AAA   
A   A   A   A   C      AAA   
A   A   A   B   A      AAA   
A   A   A   B   B   AAA      
A   A   A   B   C         ABC
A   A   A   C   A      AAA   
A   A   A   C   B         ABC
A   A   A   C   C   AAA      
A   A   B   A   A      AAA   
A   A   B   A   B   AAA      
A   A   B   A   C         ABC
A   A   B   B   A   AAA      
A   A   B   B   B         BBB
A   A   B   B   C      ABC   
A   A   B   C   A         ABC
A   A   B   C   B      ABC   
A   A   B   C   C      ABC   
A   A   C   A   A      AAA   
A   A   C   A   B         ABC
A   A   C   A   C   AAA      
A   A   C   B   A         ABC
A   A   C   B   B      ABC   
A   A   C   B   C      ABC   
A   A   C   C   A   AAA      
A   A   C   C   B      ABC   
A   A   C   C   C         CCC
A   B   A   A   A         AAA
A   B   A   A   B   AAA      
A   B   A   A   C      ABC   
A   B   A   B   A   AAA      
A   B   A   B   B      BBB   
A   B   A   B   C         ABC
A   B   A   C   A      ABC   
A   B   A   C   B         ABC
A   B   A   C   C      ABC   
A   B   B   A   A   AAA      
A   B   B   A   B      BBB   
A   B   B   A   C         ABC
A   B   B   B   A      BBB   
A   B   B   B   B         BBB
A   B   B   B   C      BBB   
A   B   B   C   A         ABC
A   B   B   C   B      BBB   
A   B   B   C   C   ABC      
A   B   C   A   A      ABC   
A   B   C   A   B         ABC
A   B   C   A   C      ABC   
A   B   C   B   A         ABC
A   B   C   B   B      BBB   
A   B   C   B   C   ABC      
A   B   C   C   A      ABC   
A   B   C   C   B   ABC      
A   B   C   C   C         CCC
A   C   A   A   A         AAA
A   C   A   A   B      ABC   
A   C   A   A   C   AAA      
A   C   A   B   A      ABC   
A   C   A   B   B      ABC   
A   C   A   B   C         ABC
A   C   A   C   A   AAA      
A   C   A   C   B         ABC
A   C   A   C   C      CCC   
A   C   B   A   A      ABC   
A   C   B   A   B      ABC   
A   C   B   A   C         ABC
A   C   B   B   A      ABC   
A   C   B   B   B         BBB
A   C   B   B   C   ABC      
A   C   B   C   A         ABC
A   C   B   C   B   ABC      
A   C   B   C   C      CCC   
A   C   C   A   A   AAA      
A   C   C   A   B         ABC
A   C   C   A   C      CCC   
A   C   C   B   A         ABC
A   C   C   B   B   ABC      
A   C   C   B   C      CCC   
A   C   C   C   A      CCC   
A   C   C   C   B      CCC   
A   C   C   C   C         CCC


Then I added up the number of mixed and pure sets made with the first 3, 4, and 5 cards as follows:

Code: Select all

          5   4   3
Mixed sets   6   18   18
Pure sets   12   18   9
         
Total sets   18   36   27


I could repeat the above procedure allowing the fifth card to be B and then to be C, instead of A, and I should get the exact same numbers for mixed sets and pure sets each time. Because the number of mixed sets and pure sets will be the same regardless of whether the fifth card is an A, B, or C, I am justified in doing the enumeration for just one of those cases.

I leave the rest of the math to you if you're interested.

[ClessAlvein]

Yes, this is fine, but you also claimed that the denominator for the probability of getting a set with 4 cards in your hand, h(4), was 81. I assume you arrived at this by doing 3^4? Well, the 81 that you get takes into account some impossible combinations for 4-card possession, because it includes all combinations, even the ones that would already be turned in (for example, a set that goes r-g-b-r)

[tahitiwahini]

Yes, this is fine, but you also claimed that the denominator for the probability of getting a set with 4 cards in your hand, h(4), was 81. I assume you arrived at this by doing 3^4? Well, the 81 that you get takes into account some impossible combinations for 4-card possession, because it includes all combinations, even the ones that would already be turned in (for example, a set that goes r-g-b-r)

There are 243 (3^5) possibilities. If you accept the argument in my last post then you will see why I only need to consider 81 (3^4) possibilities.

You ask about h(4).

Let's look at what happens when we start out with 81 possibilities and we consider those cases where the first three cards form a set. There are 27 of those, leaving 54 remaining possibilities (which do not have a set in the first three cards). Of those 54, 36 have a set in the first four cards, leaving 18 possibilities (which do not have a set in the first 4 cards).

So, out of my original 81 possibilities, I find that 27 have formed a set in 3 cards and 36 (not including any of the aforementioned 27) have formed a set in 4 cards, therefore if I want to know the probability that I will have a set in 4 cards:

(27 + 36) / 81 = 63 / 81 = 0.7777778

If the 81 seems suspicious you can multiply the ratio by 1 (i.e., 3/3) to get:

63 / 81 = (3 * 63) / (3 * 81) = 189 / 243 = 0.7777778

I enumerated the first 81 possibilities, I'll leave the other 162 possibilities for someone else to enumerate, or else explain why they would show any different results than the first 81. I contend that if you extend the enumeration I posted above you will indeed get the same number of mixed and pure sets in the next 81 possibilities as you will in the final 81 possibilities. The number of mixed sets will be the same and the number of pure sets will be the same. Exactly what those pure sets will look like will change. In the 81 possibilities in which the fifth card is always A, I expect there will be more AAA's than BBB's or CCC's. In the next 81 possibilities (if you were to continue the enumeration with the fifth card being a B rather than an A), I would expect there to be more BBB's than AAA's or CCC's. But I maintain the number of such pure sets would be the same in each of the 81 possibilities.

[ClessAlvein]

Oh, the calculations are fine. There is, however, a logical discrepancy between these two statements of yours:

1) A player will turn in a set of cards as soon as he/she gets it
2) The possibility that the first three cards form a set in a 4-card set exists, and are tallied as part of the total

If condition 1 was true, then h(4) would be 36/54, not (27 + 36) / 81.

[tahitiwahini]

Oh, the calculations are fine. There is, however, a logical discrepancy between these two statements of yours:

1) A player will turn in a set of cards as soon as he/she gets it
2) The possibility that the first three cards form a set in a 4-card set exists, and are tallied as part of the total

If condition 1 was true, then h(4) would be 36/54, not (27 + 36) / 81.

You're right. condition 1 is true, and condition 2 is incorrect. I've corrected the h(n) calculations in the original post so now they read:

Code: Select all

h(0 < n < 3) = 0  // card set not possible
h(3) = 0.333                 // 27/81
h(4) = 0.667                 // 36/54
h(5) = 1                      // 18/18


Good catch, thanks.

[ClessAlvein]

No prob Hope this algorithm gets into some future version of someone's GM script.

[tahitiwahini]

No prob Hope this algorithm gets into some future version of someone's GM script.

It's in mine now and seems to work well, but then I only play flat rate games. I can't make it publicly available until I extend it to work for escalating and no-card games as well.

Of course, anyone (stocksr?) is welcome to incorporate the algorithm into a new release of the Conquer Club script if they wish.