Diagram 2. Example of 3-8 Guang-Ddaeng
Guang-Ddaeng is when a player gets January and March or January and August card with 光.
Diagram 3,4. Examples of Guang-Ddaeng
Ddaeng is when a player gets any same months, October is highest and January is lowest.
Diagram 5. Examples of Each Ddaeng
Ali is when a player gets two cards formed from January and February.
Diagram 6. Examples of Ali
Dok-sa is when a player gets two cards formed from January and April.
Diagram 7. Examples of Dok-Sa
Gu-Bbing is when a player gets two cards formed from January and September.
Diagram 8. Examples of Gu-Bbing
Jang-Bbing is when a player gets two cards formed from January and October.
Diagram 9. Examples of Jang-Bbing
Jang-Sa is when a player gets two cards formed from April and October.
Diagram 10. Examples of Jang-Sa
Sae-Luk is when a player gets two cards formed from April and June.
Diagram 11. Examples of Sae-Luk
Gab-Oh is when a player gets two cards' numbers get added up and the 1st digit is 9.
Diagram 12. Examples of Gab-Oh
Gget is when a player gets two cards' numbers get added up and become 1 ~ 8 (8 is highest, 1 is lowest).
Diagram 13. Examples of Each Gget
Mang-Tong is when a player gets two cards' numbers get added up and become 0.
Diagram 14. Example of Mang-Tong
Calculations
3-8 Guang-Ddaeng
3-8 Guang-Ddaeng is when a player gets March and August card with 光.
For player 1 to get this hand, they firstly need to receive one of these two cards from the deck, which is 2/20. Then player 2 needs to receive any cards that are not those two cards, which is 18/19. As player 1 already has one of those two cards, for player 1 to get that hand, player 1 must get the other card to match the hand which is 1/18. As player 1 has the highest hand in the whole game, the card that player 2 has does not matter anymore, hence the probability of him getting any card is 17/17.
Hence probability of winning with this hand is;
P(1) = (2/20) x (18/19) x (1/18) x (17/17).
=1/190
= 0.0052631579
Guang-Ddaeng
Guang-Ddaeng is when a player gets January and March or January and August card with 光.
There are two different situations of getting these hands, getting January with光 first and getting either of March or August with光.
P(a) = 2 x P(player 1 getting January with 光) x P(player 2 getting any cards excluding March or August with光) x P(player 1 getting either March or August with光) x P(player 2 getting any card)
= 2 x (1/20) x (18/19) x (2/18) x (17/17)
= 1/95
P(b) = P(player 1 getting either March or August with 光) x P(player 2 getting any cards excluding January with光) x P(player 1 getting January with光) x P(player 2 getting any card)
= (2/20) x (18/19) x (1/18) x (17/17)
= 1/190
P(all) = P(a) + 2P(b)
= 1/95+ 1/190
= 0.0157894737
Ddaeng
Ddaeng is when a player gets any same months, October is highest and January is lowest.
Starting with October, probability of winning with this hand is split into two different ways. First one is when player 2 has one card that has 光 and second one is when player 2 does not have card with 光.
1st and 3rd bracket is the probability of cards that player 1 receive and 2nd and 4th bracket is the probability of cards that player 2 receive.
In P(1) and P(2), 1st bracket represents two cards that is in the deck of 20, which are October. Second bracket is three cards that is 光 in the deck of 20, third bracket is one card left in the deck of 20 which is October and the last bracket is representing the amount of cards left that is not 光 for player 2 to get, which makes the hand lower than Ddaeng-10.
Graph 1. Tree Diagram of Ddang-10
P(1) = (2/20) x (3/19) x (1/18) x (15/17)
= 1/1292
P(2) = (2/20) x (15/19) x (1/18) x 17/17)
= 1/228
P(all) = (1/1292) + (1/228)
= 5/969
Probability of winning with September (Ddaeng-9) is split into three different possibilities.
First one is when player 2 has one card that has光, second one is when player 2 does not have光 and does have October and last one is when player 2 does not have both光 and October.
Graph 2. Tree Diagram of Ddang-9
P(1) = (2/20) x (3/19) x (1/18) x (1/17)
= 1/1292
P(2) = (2/20) x (2/19) x (1/18) x (16/17)
= 16/29070
P(3) = (2/20) x (13/19) x (1/18) x (17/17)
= 13/3420
P(all) = (1/1292) + (16/29070) + (13/3420)
= 0.005125559
Probability of winning with August (Ddaeng-8) is split into four different possibilities.
First one is when player 2 has one card that has 光, second one is when player 2 does not have 光 but has one 10, third one is when player 2 does not have 光and 10 but has 9 and last one when player 2 does not have 光, 10 and 9.
Graph 3. Tree Diagram of Ddang-8
P(1) = (2/20) x (2/19) x (1/18) x (16/17)
= 16/29070
P(2) = (2/20) x (2/19) x (1/18) x (16/17)
= 16/29070
P(3) = (2/20) x (2/19) x (1/18) x (16/17)
= 16/29070
P(4) = (2/20) x (12/19) x (1/18) x (17/17)
= 1/285
P(all) = (16/29070) + (16/29070) + (16/29070) + (1/285)
= 0.0051599587
Probability of winning with July (Ddaeng-7) is split into six different possibilities. However, from this hand, since the numbers of possibilities are increasing, the mathematical equation with the ‘combination’ notation will be used, instead of using the expanded fraction method.
C is representing ‘combination’, where represents. The equation above represents, simply taking away the probability of opponent winning from 1 and multiplying by the probability for player 1 to receive a specific hand.
Graph 4. Tree Diagram of Ddang-7
= 5.0567595 x 10-3
Probability of winning with June (Ddaeng-6) is split into seven different possibilities.
Graph 5. Tree Diagram of Ddang-6
= 5.0225698 x 10-3
Probability of winning with May (Ddaeng-5) is split into seven different possibilities.
Graph 6. Tree Diagram of Ddang-5
= 4.9879601 x 10-3
Probability of winning with April (Ddaeng-4) is split into eight different possibilities
Graph 7. Tree Diagram of Ddang-4
= 4.9545604 x 10-3
Probability of winning with March (Ddaeng-3) is split into nine different possibilities.
Graph 8. Tree Diagram of Ddang-3
For Ddaeng-3, it is slightly different than other hands. This is because Ddaeng-3 includes a card that has 光. As this hand requires光card to complete the hand, 3-8 Guang-Ddaeng and 1-3 Guang-Ddaeng cannot be created. Hence these two hands were eliminated from the probability of opponent winning.
= 4.9879601 x 10-3
Probability of winning with February (Ddaeng-2) is split into ten different possibilities.
Graph 9. Tree Diagram of Ddang-2
= 4.8847609 x 10-3
Probability of winning with January (Ddaeng-1) is split into eleven different possibilities.
Graph 10. Tree Diagram of Ddang-1
For this hand, it is similar with Ddaeng-3. Because this hand requires 光card, 1-3 Guang-Ddaeng and 1-8 Guang-Ddaeng cannot be created. Hence these two hands were eliminated from the probability of opponent winning.
= 4.9191606 x 10-3
Ali
Probability for player 1 to get Ali is 2/20C2. Ali could be formed using one January and one February card; hence there are four combinations and 20C2 represents getting two cards from a deck of 20 cards. However, this needs to be split into two different ways where the January card may be 光card or not. Hence, the probability for player 1 to win with Ali is (1/2) x (2/20C2) x (1 – (10/18C2) – (12/18C2). (10/18C2) is the sum of probabilities where player 2 will win when player 1 has Ali with January 光card. (12/18C2) is the sum of probabilities where player 2 will win then player 1 has Ali with normal January card.
Hence, by subtracting from 1, 2nd bracket represents the sum of probabilities of player 2 receiving a lower hand than player 1.
P(player 1 win with Ali) =
= 9.76952185 x 10-3
Dok-Sa
Probability for player 1 to get Dok-Sa is 2/20C2. Dok-Sa could be formed using one January and one April card; hence there are two combinations and 20C2 represents getting two cards from a deck of 20 cards. Again split into player 1 having 光 card or not.
P(player 1 win with Dok-Sa) =
= 9.5638477 x 10-3
Gu-Bbing
Probability for player 1 to get Gu-Bbing is 2/20C2. Gu-Bbing could be formed using one January and one September card; hence there are two combinations and 20C2 represents getting two cards from a deck of 20 cards. Again split into player 1 having 光 card or not.
P(player 1 win with Gu-Bbing) =
= 9.36250355 x 10-3
Jang-Bbing
Probability for player 1 to get Jang-Bbing is 2/20C2. Jang-Bbing could be formed using one January and one October card; hence there are four combinations and 20C2 represents getting two cards from a deck of 20 cards. Again split into player 1 having 光 card or not.
P(player 1 win with Jang-Bbing)=
= 9.1653982 x 10-3
Jang-Sa
Probability for player 1 to get Jang-Bbing is 4/20C2. Jang-Bbing could be formed using one April and one October card; hence there are four combinations and 20C2 represents getting two cards from a deck of 20 cards.
The equation below represents, simply taking away the probability of opponent winning from 1 and multiplying by the probability for player 1 to receive a specific hand. Because to create Jang-Sa, no 光cards are needed, firstly taking out all the Guang-Ddaeng and taking Ali, Dok-sa and Jang-Bbing.
P(win) =
= 8.903643 x 10-3
Sae-Luk _
Probability for player 1 to get Jang-Bbing is 4/20C2. Jang-Bbing could be formed using one April and one June card; hence there are four combinations and 20C2 represents getting two cards from a deck of 20 cards.
The equation below represents, simply taking away the probability of opponent winning from 1 and multiplying by the probability for player 1 to receive a specific hand. Because to create Sae-Luk, no 光cards are needed, firstly taking out all the Guang-Ddaeng and taking Ali, Dok-sa, Jang-Bbing and Jang-Sa.
P(win) =
= 8.7161979 x 10-3
Gab-Oh
Gab-Oh could be created with five different combinations; 1&8, 2&7, 5&4, 10&9 and 3&6. For 1&8 and 3&6, they are slightly different from others because there is 光 in the hand, then it must be split into have光 or no.
P(1&8) =
= 1.68902649 x 10-2
P(2&7) =
= 1.58238734 x 10-2
P(3&6) =
= 1.59614723 x 10-2
P(4&5) =
= 1.6374269 x 10-2
P(10&9) =
= 1.60990712 x 10-2
P(all) = 1.68902649 x 10-2+ 1.58238734 x 10-2 + 1.59614723 x 10-2 + 1.6374269 x 10-2 + 1.60990712 x 10-2
= 8.11489508 x 10-2
Gget-8
Gget-8 could be created with four different combinations; 1&7, 2&6, 3&5 and 8&10. Again on 1&7, 3&5 and 8&10, each contains a card with 光, therefore, it must be split into have光or no.
P(1&7) =
= 1.47230822 x 10-2
P(2&6) =
= 1.4035088 x 10-2
P(3&5) =
= 1.42052758 x 10-2
P(8&10) =
= 1.4172687 x 10-2
P(all) = 1.47230822 x 10-2 + 1.4035088 x 10-2 + 1.42052758 x 10-2+ 1.4172687 x 10-2
= 5.9219698 x 10-2
Gget-7
Gget-7could be created with five different combinations; 1&6, 2&5, 3&4, 8&9 and 7&10. Again on 1&6, 3&4 and 8&9, each contains a card with 光, therefore, it must be split into have光or no.
P(1&6) =
= 1.3484692 x 10-2
P(2&5) =
= 1.1971104 x 10-2
P(3&4) =
= 1.1833505 x 10-2
P(8&9) =
= 1.2108703 x 10-2
P(7&10) =
= 1.2246302 x 10-2
P(all) = 1.3484692 x 10-2 + 1.1971104 x 10-2 + 1.1833505 x 10-2 + 1.2108703 x 10-2 + 1.2246302 x 10-2
= 6.1644306 x 10-2
Gget-6
Gget-6 could be created with four different combinations; 1&5, 2&4, 6&10 and 7&9. Again on 1&5, each contains a card with 光, therefore, it must be split into have光or no.
P(1&5) =
= 1.0870313 x 10-2
P(2&4) =
= 1.0457516 x 10-2
P(6&10) =
= 1.0182319 x 10-2
P(7&9) =
= 0.9356725 x 10-2
P(all) = 1.0870313 x 10-2 + 1.0457516 x 10-2 + 1.0182319 x 10-2 + 0.9356725 x 10-2
= 4.0866873 x 10-2
Gget-5
Gget-5 could be created with four different combinations; 2&3, 5&10, 6&9 and 7&8. Again on 2&3 and 7&8, each contains a card with 光, therefore, it must be split into have光or no.
P(2&3) =
= 1.0044720 x 10-2
P(5&10) =
= 1.0182319 x 10-2
P(6&9) =
= 0.9907121 x 10-2
P(7&8) =
= 0.9631923 x 10-2
P(all) = 1.0044720 x 10-2 + 1.0182319 x 10-2 + 0.9907121 x 10-2 + 0.9631923 x 10-2
= 3.9766083 x 10-2
Gget-4
Gget-4 could be created with three different combinations; 1&3, 5&9 and 6&8. Again on 1&3 and 6&8, each contains a card with 光, therefore, it must be split into have光or no.
P(1&3) =
= 0.7258342 x 10-2
P(5&9) =
= 0.6467148 x 10-2
P(6&8) =
= 0.6329549 x 10-2
P(all) = 0.7258342 x 10-2 + 0.6467148 x 10-2 + 0.6329549 x 10-2 = 2.0355039 x 10-2
Gget-3
Gget-3 could be created with four different combinations; 3&10, 4&9, 5&8 and 6&7. Again on 3&10 and 5&8, each contains a card with 光, therefore, it must be split into have光or no.
P(3&10) =
= 5.3663571 x 10-3
P(4&9) =
= 5.2287582 x 10-3
P(5&8) =
= 4.8159615 x 10-3
P(6&7) =
= 5.3663571 x 10-3
P(all) = 5.3663571 x 10-3 + 5.2287582 x 10-3 + 4.8159615 x 10-3 + 5.3663571 x 10-3
= 2.10746478 x 10-2
Gget-2
Gget-2 could be created with three different combinations; 3&9, 4&8 and 5&7. Again on 3&9 and 4&8, each contains a card with 光, therefore, it must be split into have光or no.
P(3&9) =
= 3.3023736 x 10-3
P(4&8) =
= 3.0271758 x 10-3
P(5&7) =
= 3.4399725 x 10-3
P(all) = 3.3023736 x 10-3 + 3.0271758 x 10-3 + 3.4399725 x 10-3
= 9.7694769 x 10-3
Gget-1
Gget-1 could be created with four different combinations; 2&9, 3&8, 4&7 and 5&6. Again on 3&8, each contains a card with 光, therefore, it must be split into have光or no.
P(2&9) =
= 1.9263846 x 10-3
P(3&8) =
= 2.1671827 x 10-3
P(4&7) =
= 2.2015824 x 10-3
P(5&6) =
= 2.4767802 x 10-3
P(all) = 1.9263846 x 10-3 + 2.1671827 x 10-3 + 2.2015824 x 10-3 + 2.4767802 x 10-3
= 8.7719299 x 10-3
Mang-Tong
Mang-Tong could be created with two different combinations; 2&8 and 3&7. Again on 2&8 and 3&7, each contains a card with 光, therefore, it must be split into have光or no. However, probability of winning with Mang-Tong is 0. Because, since this hand is the lowest hand of all the possible hand in the game, hence when a player receives Mang-Tong, he automatically loses the game.
Table 1. Summary Table for the Theoretical Probability of each Hand
Practical Probability
These probability in table 2 was gained through number of experiments under the two conditioned mentioned in the beginning, where there are two players playing the game and the deck is shuffled ten times before the game starts to create a fair deck. The total number of games played is 1000 and there were 484 win and 516 lose.
Table 2. Summary Table for the Practical Probability of each Hand
Table 3. Percentage Difference between Theoretical and Practical Probability
Conclusion and Evaluation
The theoretical probabilities for each hand were calculated under the assumption that an event occurs under certain probability. Also, the experimental probabilities for each hand were recorded under the trial and error experiments.
Comparing these two probabilities, it was expected that experimental probability would not be similar to the theoretical probability when there are small frequencies of trials but it would get closer to the theoretical probability as the number of trials increased. For example, when the die with six sides was thrown six times, the data may not be spread equally. However, when it was thrown 600 times, then the data would then be evenly spread. Considering the same concept, this game, Sut-Da, was played 1000 times.
The theoretical probability of winning was around 0.473. However experimentally, player 1 won 484 times out of 1000 which ended up being 0.484. There was only 2.33 percentage difference between the theoretical and experimental probability. Also, for hands such as; Guang Ddaeng, Ddaeng-10, Ddaeng-9, Ddaeng-7, Ddaeng-5, Ddaeng-3, Jang-Sa, Sae-Luk, Gab-Oh, Gget-8, Gget-7, Gget-6, Gget-5, Gget-4, Gget-3, Gget-2 and Gget-1, the experimental probabilities were closer to the theoretical values. These hands had less than 20 percentage difference between the theoretical and experimental probability.
Also, there were hands that had a percentage difference which were greater than 20 percent. Hands such as, 3-8 Guang Ddaeng, Ddaeng-6 and Jang-Bbing did not come as frequently as expected. Also, hands such as, Ddaeng-8, Ddaeng-4, Ddaeng-2, Ddaeng-1, Ali, Dok-Sa and Gu-Bbing, turned out to be more frequent than expected. Ddaeng-8 and Gu-Bbing had extremely high frequencies when compared to the theoretical values.
By looking at the comparison between theoretical probability and experimental probability, two conclusions can be made. The first conclusion is that the game setup was not balanced. In the introduction, it states that the deck will be shuffled ten times to create a fair deck. However, by repeating shuffling cards, it was observed that the cards did not mix as well and same hand appeared twice in a row. With this observation, if the shuffling of cards were performed in a more random way, the deck will become fairer.
The second conclusion is that the number of trials was too low to illustrate the overall spread of data. As the lowest probability is in the range, it was expected that all the probabilities would be spread out when 1000 trials were performed. However, as stated above, some of the hands evenly appeared but there were also hands that were extremely higher or lower than the theoretical probability. Hence, it could be concluded that if the number of trials were increased, then the data will be evenly spread.
This game’s rules are constructed into two main parts; selection and distribution and win or lose. Firstly, selection and distribution is about the cards that the players receive which always have a limitation. Selection and distribution deal with expectation which makes this game a game where the player has limited information. Since there are only 20 cards and the cards are being dealt by the dealer, there is no choice of cards that the players can pick, and the cards are just given randomly. Also, no matter what a player receives, those two cards belong to only one hierarchy of hands. This becomes a limiting factor and the players are only allowed to participate in the game with the received cards; therefore, there is no choice for the players in what hands they receive.
The second part of the game deals with winning or loses. For this game ‘Sut-Da’, winning or loses is split into three parts; scarcity, rule and addition of the cards. Scarcity means that the hand with lowest probability of occurring will have a higher chance of winning. The rule describes situations where the probability is same, but one will win because it is in a higher place in the hierarchy. This applies to hands such as Ali, Dok-Sa, Gu-Bbing, Jang-Bbing, Jang-Sa and Sae-Luk. For example, to receive 1 and 4, has same probability as receiving 2 and 3. However, 1 and 4 is Ali and 2 and 3 is Gget-5. Since Ali is a higher hand in the hierarchy, Ali will win over Gget-5. The last part, the addition of cards acts as a balance throughout the game. Hands from Gab-Oh to Mang-Tong represent single digits after the cards are added. These hands balance the validity of the game as they redistrict the probability of winning to be just below 50 percent.
However, considering that the game is limited, a game where the player has limited information, there are not many situations in different parts of the world that undergoes similar concepts. However, this could be seen similarly in combats between two groups. For example, when there is a debate between two parties, then both sides are going through the same concept compared to ‘Sut-Da’. Since they lack in information, they are handicapped in preparing for the discussion, however, they do have choice in what they will do in order to win the discussion.
To conclude, it was obtained by the calculations that the percentage difference between the theoretical probability and the experimental probability is similar for some hands but also have huge difference in other hands. To fix this problem, when the number of trials were increased, then the probability would be more spread out than by just having 1000 trials. Also, this concept of ‘Sut-Da’ could be seen in the combat between two groups such as debating.
Appendices
Appendix 1. Photos of My Friend and I Playing ‘Sut-Da’
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