The calculation from abracadabra looks good at the first glance. However he ignored the important factor that the player in the current move ALREADY know his cards (how many card and which color they are).

So I try to visit the problem again.

Assuming World Classic map with 42 territories/ cards. 2 are W (Wild), 14 are R (Red), 13 are B (Blue) and 13 are G (Green)

Assuming it is a fog game so you have no idea which cards already turned in (can not count card like in Black Jack)

Let's called the player in the current move is Player A.

He has n cards ( 1 <= n <= 4).

He wants to take out Player B currently has (6-n) cards

I will solve the case n = 1 here first. Later in other posts will be for n=2, 3 and 4

*****If Player A has a W, we all know his chance is 100%

*****If Player A has a R, let's see

The combination choices for 5 cards of Player B will be

C(41, 5) = 749398

(Getting 5 cards randomly from the left over 41 cards)

In these choices, Player A will have 2 sets

Event X0, Player B has a W (in his 5 cards)

The total choice for event X0

(X0) W = C(2, 1) * C(39,4) = 2 * 82251 = 165702

(Getting 1 W card from the 2 W cards and the rest 4 cards randomly from the left over 39 cards)

Event X1, Player B has these cards RR RRR

(X1) RR RRR = C(13, 5) = 1287

(Getting 5 R cards randomly from the rest 13 R cards)

Event X2, Player B has these cards RR BBB

(X2) RR BBB = C(13, 2) * C(13, 3) = 78 * 286 = 22308

(Getting 2 R cards randomly from the rest 13 R cards AND getting 3 B cards randomly from the 13 B cards)

Event X3 is similar to X2

(X3) RR GGG = C(13, 2) * C(13, 3) = 78 * 286 = 22308

Event X4, Player B has these cards RR RGB

(X4) RR RBG = C(13, 3) * C(13, 1) * C(13, 1) = 286 * 13 * 13 = 11154

(Getting 3 R cards from the rest 13 cards AND 1 B card from the 13 card AND 1 G card from the 13 card)

We do similarly for event X5, X6, X7

(X5) BG RBG = C(13, 1) * C(13, 2) * C(13, 2) = 13 * 78 * 78 = 79092

(X6) BG BBB = C(13, 4) * C(13, 1) = 715 * 13 = 9295

(X7) BG GGG = C(13, 4) * C(13, 1) = 715 * 13 = 9295

WHY these is no event (X8) BG RRR. BECAUSE it is exactly the event (X4). Remember the card sequence is NOT important

So the probability to have 2 sets is

{(X0) + (X1)+ (X2) + (X3) + (X4) + (X5) + (X6)+ (X7)} / 749398

= 44.01%

*****If Player A has a B, let's see

I will not write everything again but only the calculation

REMEMBER since we have 14 R cards but ONLY 13 B cards. So the number will be different

(X0) W = C(2, 1) * C(39,4) = 2 * 82251 = 165702

(X1) BB BBB = C(12, 5) = 792

(X2) BB RRR = C(12, 2) * C(14, 3) = 66 * 364 = 24024

(X3) BB GGG = C(12, 2) * C(13, 3) = 66 * 286 = 18876

(X4) BB RBG = C(12, 3) * C(14, 1) * C(13, 1) = 220 * 14 * 13 = 11154

(X5) RG RBG = C(12, 1) * C(14, 2) * C(13, 2) = 12 * 91 * 78 = 14196

(X6) RG RRR = C(14, 4) * C(13, 1) = 1001 * 13 = 13013

(X7) RG GGG = C(13, 4) * C(14, 1) = 715 * 14 = 10010

So the probability to have 2 sets is

34.40%

*****If Player A has a G, it will be same as having a B

So the probability to have 2 sets is

34.40%

SUMMARY

If you have 1 card, the other has 5 cards. Here are your chance

100% if you have a W

44.01% if you have a R

34.40% if you have a B / G

(applying only for the World Map classic)

IN REALITY

- Keeping a R card will be better than B or G IF there are more R cards in that map.

- If the game is not FOG, the more R cards (and obvious W card) still not used (comparing to B and G cards), the more chance you have.

When I have time , I will do the rest n=2, 3 or 4 to see what the probability is

The calculation from abracadabra looks good at the first glance. However he ignored the important factor that the player in the current move ALREADY know his cards (how many card and which color they are).
So I try to visit the problem again.
Assuming World Classic map with 42 territories/ cards. 2 are W (Wild), 14 are R (Red), 13 are B (Blue) and 13 are G (Green)
Assuming it is a fog game so you have no idea which cards already turned in (can not count card like in Black Jack)
Let's called the player in the current move is Player A.
He has n cards ( 1 <= n <= 4).
He wants to take out Player B currently has (6-n) cards
I will solve the case n = 1 here first. Later in other posts will be for n=2, 3 and 4
*****If Player A has a W, we all know his chance is 100%
*****If Player A has a R, let's see
The combination choices for 5 cards of Player B will be
C(41, 5) = 749398
(Getting 5 cards randomly from the left over 41 cards)
In these choices, Player A will have 2 sets
Event X0, Player B has a W (in his 5 cards)
The total choice for event X0
(X0) W = C(2, 1) * C(39,4) = 2 * 82251 = 165702
(Getting 1 W card from the 2 W cards and the rest 4 cards randomly from the left over 39 cards)
Event X1, Player B has these cards RR RRR
(X1) RR RRR = C(13, 5) = 1287
(Getting 5 R cards randomly from the rest 13 R cards)
Event X2, Player B has these cards RR BBB
(X2) RR BBB = C(13, 2) * C(13, 3) = 78 * 286 = 22308
(Getting 2 R cards randomly from the rest 13 R cards AND getting 3 B cards randomly from the 13 B cards)
Event X3 is similar to X2
(X3) RR GGG = C(13, 2) * C(13, 3) = 78 * 286 = 22308
Event X4, Player B has these cards RR RGB
(X4) RR RBG = C(13, 3) * C(13, 1) * C(13, 1) = 286 * 13 * 13 = 11154
(Getting 3 R cards from the rest 13 cards AND 1 B card from the 13 card AND 1 G card from the 13 card)
We do similarly for event X5, X6, X7
(X5) BG RBG = C(13, 1) * C(13, 2) * C(13, 2) = 13 * 78 * 78 = 79092
(X6) BG BBB = C(13, 4) * C(13, 1) = 715 * 13 = 9295
(X7) BG GGG = C(13, 4) * C(13, 1) = 715 * 13 = 9295
WHY these is no event (X8) BG RRR. BECAUSE it is exactly the event (X4). Remember the card sequence is NOT important
So the probability to have 2 sets is
{(X0) + (X1)+ (X2) + (X3) + (X4) + (X5) + (X6)+ (X7)} / 749398
= 44.01%
*****If Player A has a B, let's see
I will not write everything again but only the calculation
REMEMBER since we have 14 R cards but ONLY 13 B cards. So the number will be different
(X0) W = C(2, 1) * C(39,4) = 2 * 82251 = 165702
(X1) BB BBB = C(12, 5) = 792
(X2) BB RRR = C(12, 2) * C(14, 3) = 66 * 364 = 24024
(X3) BB GGG = C(12, 2) * C(13, 3) = 66 * 286 = 18876
(X4) BB RBG = C(12, 3) * C(14, 1) * C(13, 1) = 220 * 14 * 13 = 11154
(X5) RG RBG = C(12, 1) * C(14, 2) * C(13, 2) = 12 * 91 * 78 = 14196
(X6) RG RRR = C(14, 4) * C(13, 1) = 1001 * 13 = 13013
(X7) RG GGG = C(13, 4) * C(14, 1) = 715 * 14 = 10010
So the probability to have 2 sets is
34.40%
*****If Player A has a G, it will be same as having a B
So the probability to have 2 sets is
34.40%
SUMMARY
If you have 1 card, the other has 5 cards. Here are your chance
100% if you have a W
44.01% if you have a R
34.40% if you have a B / G
(applying only for the World Map classic)
IN REALITY
- Keeping a R card will be better than B or G IF there are more R cards in that map.
- If the game is not FOG, the more R cards (and obvious W card) still not used (comparing to B and G cards), the more chance you have.
When I have time , I will do the rest n=2, 3 or 4 to see what the probability is