What are the exact chance of having 2 sets from 6 cards

Assuming typical World map, no (2 of) wild card are used yet. Any mathematician / statistician want to do the calculation ?

Number of players would have to be included, what would you choose?

Read this topic: http://www.dominating12.com//forum/?cmd=topic&id=2045&board=19&page=2#post-31196 That post sais it's 33% chance to have a double turn in with 6 cards (assuming no wildcards). So something you should never guess upon.

[quote=abracadabra - Jan 30, 02:46 PM]The odds of getting 2 pairs with 6 cards are indeed very small. For your possible enjoyment and future reference, here are the probabilities worked out. The possible ways to get [b]two pairs with six (indistinguishable) cards[/b] are "AAA AAA", "BBB BBB", "CCC CCC" (let's call these events X1, X2 and X3), or "AAA BBB", "AAA CCC" or "BBB CCC" (say X4, X5, X6), and "ABC ABC" (say X7). In total there are Ω = 3^6 = 729 posible combinations. The odds for X1 are simple because there is only one way, so P[X1] = P[X2] = P[X3] = 1/729 ~ 0.14 % The odds for X4 are better, there are 6 over 3 ways to order the cards, so that ω(X1) = 6! / (3! * (6-3)!) = 20, and P[X4] = 20/729 ~ 2.74 % X7 is a bit more complicated. There are 3! = 6 ways to order any ABC pair, so we have (3!)^2 = 36 ways to order two pairs of ABC (like "ABC ABC", "BAC ACB", etc.). Then we can also have pairs like "ABB ACC", in this case there 3 ways to order each pair of ABB or ACC, and there are 3!=6 ways to make an ABB type pair (ABB, ACC, BAA, BCC, CAA, CBB). So this second part gives another 3x3x3! = 54 orderings, which gives a total of ω(X7)=36+54=90, so that finally P[X7] = 90/729 ~ 12.34% Actually 12.345678901234.... nice repeating decimals :) So, in order to get a pair without wildcards the odds are P[2 pairs | 6 cards] = ( 1+1+1 + 20+20+20 + 90 ) /729 = 153 / 729 ~ [b]21 %[/b] In other words, not so much. If you have a [b]wildcard[/b] you are certain to get a pair. If not, the probability for the other player to have one is p (I am not sure how many of the cards are wildcards). If the other player does have one you make a pair. The a priori probability for this to happen (lets say event Xw) is 1-(1-p)^n. Let's take n=3, assuming we take 3 cards from the other player. So the total probability is P[2 pairs | 6 cards with possible wildcards in the last 3] = P[X1...X7] + P[not X1...X7]*P[Xw] = 153/729 + 576/729 * (1-(1-p)^3) Let's say 2 out of 44 cards is a wildcard so p=1/22, then this adds about 10%, so P[2 pairs | ...] ~ 31 % Yaay math, it works! :D[/quote] It should be 21% without wild card. I think it is certain to have 2 turn ins with 6 cards if you have a wild card. Somehow I think the number is much less than 21% because this will be a dependent event, so the odd will be lesser.

Read the other post near the end supi. It should be more than 21% I will do one once I have time

But assuming the dependency of the event, it will be less. Since the cards will be redeal every 7 rounds, for 6 players and if each player gets a card. The probability of it should be less but depends on what are the cards left in the pile tho. I was trying to calculate the probability using odds of 42 cards and 2 black, since in world classic, you have 14 red, 13 blue and 13 green. But then the calculation is too long and there are too many combination for 6 cards and 2 turn ins, also depending on attacker and victim cards and hand. 1+5, 2+4, 3+3 it all contributes to different odd I suppose.

He forgot the AAA ABC possibilities with the 21% calculation, but yes, the calculation has some simplified assumptions.

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

Well done working out the Maths for Classic 2p games, that must have taken some thinking! Regardless that you got the percentage higher than previously calculated, it is still less than 50% in all cases other than having a wild card in your hand. Even if it was 50%, it would be a coin flip, a hit or miss or whatever other analogy we could use, which is too risky to ruin the game, however the odds are not even that good, in 66% of cases the odds are a 1 in 3 chance and always has the best odds on failure! Odds: We could also ask, what are the odds of having 2 wild cards within your first 4 cards? Pretty low I would guess, but it has happened to me once in game 1171 out of 1173 games, which is probably way lower than expected. The odds might say it should have happened to me 2 or 3 times in that many games, but when the coins are flipped, they don't always land on Heads or even land on Heads once in 4 flips. There are Black / Red roulette gambling types and Risk & Strategy types of play.

Very interesting calculation, but aeronatic you don't need to count cards for 2p games right? However taken into consideration where Player B is rational player and if Player B holds 5 cards chances are Player B held on to a 4 cards and no trade situation. Therefore, event X1 and X4 is not relevant because said combination may have an earlier trade at 3 or 4 cards. Given that Player A has 1 card and Player B has 5 cards. Continuing form 123playcard If Player A has a card, 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) Considering combination where Player B has 5 cards combination given Player B cannot trade with 4 cards or earlier. The combination is listed below. **Assuming a randomly assigned card using the existing starting pool to simplified the calculation. However the chances can be lower than the calculated number due to the dependency of the event.** Event | Player B combination | Card need for double trade to happen | Combination X1 | RRR BB | B or G | C(14,3) * C(13,2) = 364*78 = 28392 X2 | RRR GG | B or G | C(14,3) * C(13,2) = 364*78 = 28392 X3 | BBB GG | R or G | C(13,3) * C(12,2) = 286*78 = 22308 X4 | BBB RR | R or G | C(13,3) * C(12,2) = 286*91 = 26026 X5 | GGG BB | B or R | C(13,3) * C(13,2) = 286*78 = 22308 X6 | GGG RR | B or R | C(13,3) * C(14,2) = 286*91 =2 6026 X7 | RR BB G | G | C(14,2) * C(13,2) * C(13,1) = 91*78*13 = 92274 X8 | BB GG R | R | C(13,2) * C(13,2) * C(14,1) = 78*78*14 = 85176 X9 | GG RR B | B | C(13,2) * C(14,2) * C(13,1) = 78*91*13 = 92274 If Player A has R card Odds of double trade {(X3)+(X4)+(X5)+(X6)+(X8)}/749398=24.26% If Player A has B card Odds of double trade {(X1)+(X2)+(X5)+(X6)+(X9)}/749398=26.34% If Player A has G card Odds of double trade {(X1)+(X2)+(X3)+(X4)+(X7)}/749398=26.34% Disclaimer: I assume player B gets card from the exist pool of 14 or 13, Chances are if you hold a Red card, your odds are smaller because player B will need to get from Blue or Green. If you have Blue or Green it is easier for player B to get Red card. However from my calculation the % is not a big difference by what card you are holding, around 25% Unless Player B is a total noobie and don't turn in at 3 or 4 when he/she could, the odds of double trade will be much higher like how 123playcard calculated. So, I don't think holding on to a Red card will gives you a bigger advantage, it is just about the same.

[quote=aeronautic - May 9, 09:25 PM] Odds: We could also ask, what are the odds of having 2 wild cards within your first 4 cards? Pretty low I would guess, but it has happened to me once in game 1171 out of 1173 games, which is probably way lower than expected. The odds might say it should have happened to me 2 or 3 times in that many games, but when the coins are flipped, they don't always land on Heads or even land on Heads once in 4 flips. There are Black / Red roulette gambling types and Risk & Strategy types of play.[/quote] The probability of having 2 wild cards in the first 2 rounds is {(2/42)*(1/41)} = 0.11% The probability of having 2 wild cards in the first 4 rounds is {(2/42)*(40/41)*(39/40)*(1/39)} = 0.12% So, it happens around 1 in every 1000 games, your odds seem to be on the right track. I, however got very lucky, played about 630 games, and if I remember correctly, I hold on to 2 black games, the first 2 rounds, [b]TWICE[/b]! Totally shocker for me. But it was back then when I got the luck on my sleeves. Above calculation didn't consider the fact that other players also randomly get a card which it not wild, so the wild will be in your hand, so the odds should be much much smaller if there are more players in the game.

Yes supi. Your assumption is good " However taken into consideration where Player B is rational player and if Player B holds 5 cards chances are Player B held on to a 4 cards and no trade situation. Therefore, event X1 and X4 is not relevant because said combination may have an earlier trade at 3 or 4 cards. " However I have played long enough to see this scenario couple times Here is one example, Player D kills Player C, E etc (after someone makes a bad move) and now has a huge bonus. However double turn in can still take Player C capital. Player B has 5 cards because he does not think Player C will make a bad move Player A has 1 card and now it is his turn. *** If Player A take 1 more card, Player B will attack Player A to get double turn in and win the game *** If Player A does not take 1 more card, Player D will get his huge bonus and is impossible to take over So what is the best interest for Player A. Many will say just skipping the turn and hope for your best i.e. accepting losing the game My argument will be going for 6 cards, at least you have 30-40% chance And again this is not just a made up scenario, You can see this post for a real game example http://www.dominating12.com//forum/?cmd=topic&id=2177 And another real example I just played recently http://www.dominating12.com//?cmd=game&sec=play&id=481139 (Andy is Player A, aeronautic is Player B and lazer36 is Player D) Any way, when we (I or you) do more maths for the case n=2, 3 or 4. (more realistic cases), we will see if the probability increasing more or not.

I think it is about 40% but I am not sure. I would love to solve this problem but it seems more complex than expected. I cannot see a great difference based on the cards you and your opponent hold (except in case of a wild - obvious!). Wilds cannot be ignored as they make a Great difference. If we assume the other player could not turn in in the previous round, otherwise he would have done it, then you only have to consider the % of his last card to be a wild (assuming all the others are not). But in some situations you cannot be sure that he didn't intentionally decided to go for 5 cards for many possible reasons. The paradox is that one of the possible reasons is that he actually holds a wild card, but he doesn't want to waste it, preserving the wild for another moment. So you should make two separate calculations: one assuming he could not turn in the previous round and one assuming nothing; and then you'll evaluate in each game what of 2 corresponds better to the situation you are in.

I think the actual calculation is more complex, involving lots of probability and differs according to the number of players in the game. The odds will increase if there are less players in the game tho. The thing is Player A moves will significantly change the game. If Player A kills Player B, double turn in -> Player A gains advantage, possibly wins, by killing Player D If Player A kills Player B, single turn in -> Player D can now kills player A and wins. If Player A don't kill Player B, get a card -> Player B then may or may not kill Player A, since he cannot trade after the kill. (unless in advanced card) **Unless Player D, is blocked from Player B, then Player B should kill Player A and gain card advantage.** If Player A don't kill Player B, don't get a card -> Player B will not kill Player A, and this will then gives the game to Player D. Condition, assuming Player D now hold <3 cards, since player D traded after killing Player C. However if Player A, can't turn in when holding 3 cards then chances are, Player A will be dead holding 4 cards. So, assuming this is not a fog game, you can see the cards played in the game log. If probability of having a turn in at 3 cards < probability of double trade with 6 cards. Then process to kill player B. If probability of having a turn in at 3 cards > probability of double trade with 6 cards. Then don't kill player B. Do I make any sense here? Sound like I am just going in a loop. Well, if my reason is not good, feel free to gives your, I am just still a newbie. ps: I am old school, I did the calculation other cards combination on paper, but it is long to type. If I am free later today, I will try to post it here.