Singapore Mathematics Project Festival 2009

Transcription

1 Singapore Mathematics Project Festival 2009 Investigating a winning strategy for Big Two Done By: Peter Haw Seow Yongzhi Hwa hong Institution (High School)

2 ONTENTS 1 ABSTRAT INTRODUTION Terminologies: Rules and Instructions AIMS AND OBJETIVES RESEARH QUESTIONS LITERATURE REVIEW Dai Di Analysis (Teo, 2000) Probabilities of the same sets in other games Evaluation FINDINGS The probabilities of different sets hances of controlling of different combinations Deduction of opponents cards hances of promotion of different combinations ONLUSION BIBLIOGRAPHY AKNOWLEDGEMENTS APPENDIX I APPENDIX II

3 1 ABSTRAT This project was based on the popular 4-player card game, Big Two, otherwise known as hoi Dai-Di. The objective of the game is to be the first to finish playing all the 13 cards in one s hand. ards are played in sets, consisting of singles, pairs, triples and 5-card combinations (fivers). This project intended to find the most effective way to play any hand of Big Two. To do this, we conducted our research in 3 different areas: 1. Determining the probabilities of different sets 2. Determining the chances of controlling of different combinations 3. Investigating the possible combinations and strategies of opponents Summary of findings: 1) The probabilities we successfully determined are that of the pair, triple, quadruple, flush, full house, and straight flush. 2) We found a formula for the chances of controlling of single cards, pairs and triples. It is Q 8( R 3) + n, where Q is the chance of controlling, R is the rank value and n is dependent on the suit value. We separately calculated the chances of controlling for fivers. 3) We calculated the probability of opponents having a pair, triple and quadruple of a certain rank with 0, 1 or 2 cards of that rank in one s hand The probabilities of different sets would give players an edge since it helps in gauging the risk of a combination being beaten. The chances of controlling will also help players to know the probability of their combinations succeeding in controlling, allowing them to make more informed decisions about the game play. Lastly, the deduction of opponents cards can help players to determine whether their combinations can control. 3

4 2 INTRODUTION Big Two (commonly known as Dai-Di) is a popular card game with many different versions played around the world. However, we were unable to find much research done on how to effectively play the game. The objective of our project is to fill the missing gap in research by finding an effective way to play any hand of cards. We chose to embark on this project due to a fervent desire to learn more about the game. We hoped to increase our chances of winning in the game by expanding our knowledge of game plans and strategies. Given the lack of research on this subject matter, this undertaking is quite original. 2.1 Terminologies: In this report, we will use the following terminologies: Terminologies Hand Set Rank ombination ontrol Definite control Probable control Straggler Definition a group of 13 cards Any one of the following: Single ard Pair Triple Fiver the number value of a card a specific group of cards combinations which can make all other opponents pass, further grouped into: combinations which have a very high chance of making all other opponents pass combinations of sufficiently high rank, but not high enough to be considered definite combinations which have a low chance of being played 4

5 2.2 Rules and Instructions 1. Using the normal 52-card deck, four players are dealt 13 cards each. The objective of the game is to finish playing down all of one s cards first. In the game, 2 has the greatest value, followed by Ace, King, Queen... and ending with 3. The suit values, in ascending order of ranking, are: Diamonds, lubs, Hearts, and Spades. The number value takes precedence over the suit. 2. ards are played in sets, which consist of the following: singles, pairs, triples, and 5- card combinations, which we will informally refer to as fivers* throughout the report. The player beginning the game is the one holding the 3. He can play any set he wishes, but it must consist of this card. 3. Only higher-ranking combinations of the same set can be used to win the previously played combination (e.g. pairs cannot be used to beat singles). 4. If a player does not wish to play any cards, he can pass his turn. When all players pass, the person who played the last combination can then change the set being played. When one player has played all his cards, he wins, and the other players may choose to continue playing or start a new game. * Fivers include the following: Fiver Straight Flush Full House 4-of-a-Kind Straight Flush Definition 5 cards of consecutive numbers 5 cards of the same suit 5 cards combining a triple and a pair 5 cards combining a quadruple and a random fifth card 5 cards of consecutive numbers AND of the same suit 5

6 The fivers are ranked in this order: Straight < Flush < Full House < 4-of-a-kind < Straight Flush Note: 1. Straights of the same rank are compared by the largest card of each straight. 2. Flushes are compared based on their suit first, followed by their rank. 3. Full Houses and 4-of-a-kinds are compared based on the rank of the triple and quadruple in each combination respectively. 6

7 3 AIMS AND OBJETIVES Firstly, we intended to calculate the probabilities of the different sets occurring in the 13-card hand. This information would be advantageous to any player when considering the risk of a certain combination being beaten by an opponent. Secondly, we wanted to introduce benchmarks for classifying combinations into controls and stragglers, since we could only base our strategies on accurately defined terms. Thirdly, we identified various main processes in the game that we could analyze. These included passing, choosing between contrasting sets, using probable controls, and the deduction of opponents cards. To explore these areas, we decided to calculate the chance of controlling of different combinations, as well as to calculate the chance of promotion of different combinations to become definite controls before the end of the game i.e. the chance of all higher-ranking combinations being played out before the game ends. 4 RESEARH QUESTIONS We would like to investigate 1. The probability of obtaining each of the different sets of cards 2. The chances of controlling for different combinations 3. Deducing the possible combinations of opponents 7

8 5 LITERATURE REVIEW 5.1 Dai Di Analysis (Teo, 2000) This research paper is an analysis on the strategies used in Big Two. We have used some of the terminologies within this report, namely, controls and stragglers. The paper introduced a few common strategies important in playing the game, summarized below: 1. Probable controls should be played out early in the game. This is based on the likelihood that other players have stragglers of the same set and have insufficient controls to use to play them out. Thus they are likely to pass and wait for another chance to play their stragglers. 2. ertain cards can be withheld to guard against the possible strategies of opponents. For example, a hand with 4 Aces can be withheld to stop opponents from running pairs. However, for this to work, the player must have good controls in other sets. 3. Grouping combinations into 4 classes. a. lass A contains 100% controls at the point in time. b. lass B contains combinations with a very high chance of being played out and with a high chance of controlling. c. lass contains combinations which are sufficiently likely to be played out, but with little chance of controlling. d. lass D contains the stragglers that are unlikely to be played out without using a control prior to them Thus the main concern of the game is just to play out all the lass D cards. This class system can also be used to assess the strength of a hand. 8

9 4. In order to keep track of whether or not a certain combination has been promoted to a definite control, remembering high cards (such as Aces or 2 s) that have been played out can help one to eliminate the possible higher-ranking combinations. For fivers, one can note down which 4-of-a-kind plays remain available to opponents during the course of the opponents, since the probability of a straight flush is quite negligible (about 0.015). 5. One can infer the game plans of other opponents by noticing the sets which they continually gain control with. For example, if an opponent keeps gaining control with high pairs, then one could possibly split up a fiver to play a pair so as to minimize losses. 6. Generally lower-ranking combinations should be discarded as soon a possible, while middle-range and absolute controls could be held back in certain situations to yield greater advantages later in the game. 5.2 Probabilities of the same sets in other games We could only find one research paper discussing the strategies used in Big Two. Nevertheless, more work has been done on other games, such as Bridge, which also use the 13-card hand. Haigh(2003) calculated the probabilities of different suit distributions in the 13-card hand for bridge. From this we observed that the probabilities of suit distributions {4,4,4,1, {4,4,3,2 and {4,3,3,3 are , and respectively. It is apparent than all other hands which do not fall under these three contain a flush. Durango(2001) computed the number of poker hands containing various sets and their probabilities for 5-10 card hands. These sets were broadened to include royal flushes 9

10 (straight flushes of the rank AKQJ 10), 2/3/4/5 pairs and separate ranks of straights. However, for the hands containing 6 cards and above, there was no mathematical working for the calculations. 5.3 Evaluation The strategies found in Dai Di Analysis came in useful when trying to identify aspects of the game and obtaining methods for investigation. However, the paper was heavily slanted towards a qualitative approach to the problem. It lacked evidence of mathematical research. Hence, we performed mathematical calculations to provide academic matter to back up strategies cited. From the calculations of sets for other games, we can easily find the probability of the flush. However, the sets in Big Two are mostly irrelevant to the concepts in Bridge. Therefore we could not find any relevant information for the probabilities of other sets like the straight, quadruple, triple etc in the extensive research done on Bridge. As for Poker, although calculations were not done for the 13-card hand, the method by which the probabilities of sets in the 5-card hand were arrived at could provide inspiration for methods to calculate that for the 13-card hand, which would then be more relevant to Big Two. Due to the limited material for reference, much of the mathematical research still has to be undertaken by ourselves. 10

11 6 FINDINGS 6.1 The probabilities of different sets Number of 13-card hands = = 635,013,559,600 We define the probability of at least one combination of set x in 13 cards to be Px a. Pair Number of hands without a pair (i.e. A JQK) = 13 4 P pair = 1 = 99.99% b. Quadruple Let there be 3 quadruples X, Y and Z, of different unspecified ranks. 48 Number of hands with Quadruple X = Number of hands with Quadruple X and Y = Number of hands with Quadruple X, Y and Z = 3 1 c. Triple P quadruple = 3.42% (13 = 48 ) ( Number of ways to choose a triple = ) + ( ) P triple However, in the event that a quadruple is present in the hand, it will be calculated as a triple 4 times. Therefore we need to subtract 3 times the probability of a quadruple. 11

12 P triple 4 13 = 52 = 57.02% (3 3.42%) Fivers d. Flush There are 3 suit distributions with at most 4 cards of each suit i.e. they do not contain a flush. They are {1,4,4,4, {2,3,4,4, and {3,3,3,4. The calculation for their probabilities is shown below: Suit distribution Probability 1,4,4, ! ! = ,3,4, ! ! = ,3,3, ! ! = From Taking hances: Winning With Probability (2003) 4! 4! ( and represent the permutations of the suit distribution values) 3! 2! Probability of hands that do not contain a flush = 2.99% % % = 35.1% P flush = 64.9% = 100% 35.1% e. Full House We can split up a full house into a triple and a pair occurring in the same hand. Firstly, we need to calculate the number of 10-card hands that contain a pair, using the same method as that for the pair. 12

13 Number of ways to choose 10 different ranks out of 12 (-1 for the triple s rank) = Number of combinations for these 10 ranks = Number of 10-card hands without a pair = 4 10 Total number of remaining cards to choose 10 from = = 49 P pair in 10 cards = 100% 0.84% = 99.16% 4 = 100% However, calculated overlapping of the triple and the pair needs to be eliminated. We calculate number of hands with a triple and a pair of the same rank (impossible in reality). For each rank: Number of ways to choose pair = 4 2 = 6 Number of ways to choose triple = 4 3 = 4 Number of ranks = 13 Number of ways to choose 8 other random cards = P overlapping = 18.54% = ( ) P full house = 99.16% 57.02% 18.54% = 56.54% 18.54% = 38.0% 13

14 f. Straight For each rank of straight, number of combinations to form the straight = Number of ranks of straights (from A2345 to 10JQKA) = 10 Number of ways to choose 8 other random cards = P straight =? However, this gives rise to an impossibly huge value of 507%! The reason for this gross inaccuracy is that there is a lot of overlapping present in the calculation, and the chance for such overlapping to occur in real situations is far from negligible. For example, if a hand contains and seven other cards, using the calculation above, this hand would have 2 straights when in reality it only has 1. Another example would be and seven other cards. It would also be counted as having 2 straights when it actually has only 1. These are just simple examples. The problem becomes much worse if there are more overlaps present. This problem could not be easily solved using permutation with repetition since there were too many different cases to account for that could be present within the 13- card hand. Therefore, we attempted to use a computer program for the calculation (refer to Appendix I). It works by first generating all the possible 13-variable groups with 52 different variables, then checking each group for the existence of a straight (represented by 5 variables with consecutive number values). After counting the number of groups containing a straight, we can divide this by to obtain the probability of the straight. However, we discovered that a long time was needed for a normal computer to finish the calculation, and thus we have not obtained a result yet. 14

15 g. Straight Flush 47 Number of hands containing a straight flush 40 8 Within these hands there is also overlapping because of hands containing 6-9 cards having consecutive ranks and the same suit. For convenience, we call combinations of cards having consecutive ranks and the same suit nice. 46 Number of hands with 6-card nice combinations = Number of hands with 7-card nice combinations = Number of hands with 8-card nice combinations = Number of hands with 9-card nice combinations = 24 4 P 40 = ( straight flush = ) 1.63% Knowledge of these probabilities can give players an advantage over others. With these results, players can now more effectively determine the chance that an opponent has a combination of the same set as his, which leads to the chance that his combination can be beaten by an opponent. Thus one can decide whether to play the particular combination and risk being beaten or wait for a higher combination to be lured out, promoting his own. For example, when a player plays a low-ranking full house, he knows that the probability of an opponent having another full house is 38% and thus he has a good chance of controlling the round. 15

16 6.2 hances of controlling of different combinations a. Singles, Pairs, Triples Every card s relative percentile is 1/51, which is about 2%. E.g. 3 = 0 th percentile 9 = 50 th percentile 2 = 98 th percentile (5, 5 ) = 20 th percentile (10, 10 ) = 60 th percentile (A, A ) = 88 th percentile (8, 8, 8 ) = 40 th percentile Formula: Q 8 ( R 3) + n where Q is the percentile, R is the rank value (J, Q, K, A, 2 take the rank values of 11, 12, 13, 14, 15 respectively), n depends on the suit value for singles and pairs: Singles: For the 4 suits,,,, n = 0, 2, 4, 6 respectively Pairs: If the pair contains, n = 4, otherwise n = 0 16

17 b. Fivers Firstly, we calculated the total number of each fiver: i) Number of straights = = (-40 to exclude straight flushes) ii) Flush For each suit: Number of 7-led flushes i.e = 5 5 Number of 8-led flushes = 6 5 Number of 2-led flushes = Thus number of flushes = ) 4 40 = iii) Full House Number of ways to choose triple = 13 4 ( Number of ways to choose pair of different rank = (13-1) 6 = 12 6 Number of full houses = ( 13 4) (12 6) = 3744 iv) 4-of-a-kind Number of ways to choose a quadruple = 13 Number of ways to choose a random 5 th card = 48 Number of 4-of-a-kinds = = 624 v) Number of straight flushes = 10 4 = 40 Total =

18 Next, we converted these numbers into their relative percentiles: Fiver Straights Flushes Full Houses 4-of-a-kinds Straight Flushes Relative Percentile 10200/26580 = 0 to 38 th percentile 38 th to 83 rd percentile 83 rd to 98 th percentile 98 to 99.8 th percentile 99.8 to 100 th percentile We decided to assume that the bottom 50 th percentile of each set were to be considered as stragglers, as these combinations have a lower chance of being played than not. This includes all single cards, pairs and triples with rank less than 9 and all straights and flushes for fivers. For definite controls, we decided that it could be redefined to the 95 th percentile or more. This will include 2, 2, 2 for single cards, any 2 pair, 2 triple, and straight flushes, 4-of-a-kinds as well as Ace and 2 full houses. These definitions are important as players need to effectively classify their cards according to rank so that they can determine which combinations can be used as controls, and which ones are stragglers, to be discarded as soon as possible. More effective game plans can then be formulated with a better analysis of one s cards. Applications 1. Finding the chance of controlling of probable controls If a card is the n th percentile of its set, its chance of controlling is also n%. e.g. A 88 th percentile 88% chance of controlling Straight flush 99.8 th percentile 99.8% chance of controlling i.e. definite control 18

19 2. hoosing between contrasting sets E.g. A hand containing can be played in many ways, 2 of which are displayed below: a) b) 19

20 The first permutation groups all the possible pairs together, leaving the remaining cards while the second breaks up two pairs to form a flush. learly, the second permutation is a better choice. There are 3 straggler combinations in the first permutation (6 pair, 5, 8 ) which the player could find difficult to play as the game progresses. In the second, only the 6 pair is left. In other words, the 20 th and 44 th percentile cards (5 &8 ) can be combined to form a 70 th percentile fiver (Q -led flush), sacrificing only one 78 th percentile card (Q ) in the process. As shown in the example, our results can help players to choose the most efficient permutation with which to play their cards. This is an important process in the game as it affects the number of stragglers and controls in the hand. A wrong decision could result in too many stragglers left to clear with too few controls to do so. 20

21 6.3 Deduction of opponents cards We calculated the probability of a pair, triple, and quadruple of a certain rank existing while knowing the number of cards of that rank that the opponents cannot possibly have i.e. either the player is holding on to these cards or they have already been played out and thus the number of cards that are shared among the opponents (remaining cards). The results are shown in the table below: Remaining cards Pair 2 pairs Triple Quadruple 4 1 2/9 = /3 = /27 = /9 = /9 = /3 = The results were obtained firstly by listing all the possible ways in which the remaining cards could be divided among the 3 opponents. Next we counted the number of distributions where any opponent held at least a pair, triple and quadruple and divided this by the total number of possible distributions to obtain their probabilities. Refer to Appendix II. Besides the analysis of one s own cards, a major part of the game involves the deduction of the possible combinations existing in an opponent s hand. This helps players to determine the possibility of existing higher-ranking pairs, triples and quadruples. This is especially important as the game progresses and the chance of higher-ranking combinations existing decreases. However, our results are only limited to pairs, triples and quadruples, so this could be further extended on by including other sets. 21

22 6.4 hances of promotion of different combinations Initially, we wanted to include a fourth area of research the chances of promotion of different combinations to become definite controls before the end of the game. Regrettably, we have not found a way to calculate this. All higher-ranking combinations have to be played out before a particular combination can become a definite control, thus this question is synonymous with calculating the probability that all higher-ranking cards will be played before end game. Since different opponents will play their combinations differently, using strategies to lure out the others higher-ranking combinations, we find that there is no definite way to achieve this. We have decided to leave this as a possible extension of the project for others to investigate because we feel that it is an important part of the game which directly influences the decision of passing one s turn. 22

23 7 ONLUSION Our project has, to some extent, achieved its goal of finding methods to effectively play Big Two. Firstly, we were able to calculate the probabilities of most of the sets, the knowledge of which would provide an advantage for any player. Secondly, we introduced our own benchmarks for classifying controls and stragglers so that players can more effectively analyze their hand. Thirdly, we also succeeded in determining the chances of controlling for all combinations in general. This information will definitely help players in choosing their game plan and the best permutation with which to play their hand of cards. It also plays an important role in judging the probability of success of probable controls. Lastly, we conducted some preliminary research on the deduction of opponents cards based on analyzing one s cards. The calculations that we have done could be further extended to investigate other methods of deduction. Our last research question still remains unsolved. It could be used as a possible extension of the project. Its importance, as previously stated, lies in its implications on the decision to pass one s turn. This is an important aspect of the game, as holding cards back at the right times could result in an increase in the chance of controlling of one s combination(s) i.e. the promotion of one s combinations. In conclusion, the research that we have conducted will bring the strategy behind the game to a new level. With our findings, players can now strategize more substantially and thus play Big Two more effectively. 23

24 8 BIBLIOGRAPHY Printed Everitt, B.S.(1999) hance Rules: An Informal Guide to Probability, Risk and Statistics Haigh, J.(2003) Taking hances: Winning With Probability Online Teo, K.M., Kok, R., Ang, J. & Lim, I.(2000) Dai Di Analysis Retrieved Feb , from Bill, Durango Bridge Probabilities and ombinatorics Suit Distribution Retrieved Apr , from Durango, B. Poker Probabilities Retrieved Apr , from The Math Drexel Four of a Kind in a 13-ard Hand Retrieved June , from 9 AKNOWLEDGEMENTS We would like to thank the following people for their invaluable help: Our mentor for her guidance and encouragement Professor Steward Huang, for his help in the verification of our calculations. 24

25 APPENDIX I omputer Program for Straight Probability alculation #include<iostream> using namespace std; int pile1[13],pile2[5]; int compare(const void *a, const void *b) { return *(int*) a - *(int*) b; bool check(){ int a,b; qsort(pile2,5,sizeof(int),compare); for(a=0;a<4;a++){ if(pile2[a]!=pile2[a+1]-1) return false; return true; bool streak(){ int i,j,k,l,m; for(i=0;i<9;i++){ pile2[0]=pile1[i]; for(j=i+1;j<10;j++){ pile2[1]=pile1[j]; for(k=j+1;k<11;k++){ pile2[2]=pile1[k]; for(l=k+1;l<12;l++){ pile2[3]=pile1[l]; for(m=l+1;m<13;m++){ pile2[4]=pile1[m]; if(check()) return true; return false; int main(){ int card[52],i,j,k=0,l,m,n,o,p,q,r,s,t,u,count=0; for(j=0;j<13;j++){ for(i=0;i<4;i++) card[k++]=j; 25

26 for(i=0;i<40;i++){ pile1[0]=card[i]; for(j=i+1;j<41;j++){ pile1[1]=card[j]; for(k=j+1;k<42;k++){ pile1[2]=card[k]; for(l=k+1;l<43;l++){ pile1[3]=card[l]; for(m=l+1;m<44;m++){ pile1[4]=card[m]; for(n=m+1;n<45;n++){ pile1[5]=card[n]; for(o=n+1;o<46;o++){ pile1[6]=card[o]; for(p=o+1;p<47;p++){ pile1[7]=card[p]; for(q=p+1;q<48;q++){ pile1[8]=card[q]; for(r=q+1;r<49;r++){ pile1[9]=card[r]; for(s=r+1;s<50;s++){ pile1[10]=card[s]; for(t=s+1;t<51;t++){ pile1[11]=card[t]; for(u=t+1;u<52;u++){ pile1[12]=card[u]; if(streak()) count++; cout<<count<<endl; system("pause"); return 0; 26

27 1 st APPENDIX II Listing the ways in which the remaining cards can be distributed among the three opponents (referred to as A, B, ). a) Remaining cards = 4 A (/B/) card 2 nd A B 3 rd A B A B A B 4 th A B A B A B A B A B A B A B A B A B Key: 4 represents a quadruple existing 3 represents at most one triple existing 22 represents two opponents each having a pair 2 represents at most one pair existing Probability of: Quadruple = 1/27 = At least triple = 9/27 = 0.33 At least pair = 1 Two pairs = 6/27 =

28 b) Remaining cards = 3 1 st A (/B/) 2 nd A B 3 rd A B A B A B Probability of: Triple = 1/9 = 0.11 At least pair = 7/9 = 0.78 c) Remaining cards = 2 1 st A (/B/) 2 nd A B Probability of pair = 1/3 =