• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28
  29. 29
    29
  30. 30
    30
  31. 31
    31
  32. 32
    32
  33. 33
    33
  34. 34
    34
  35. 35
    35

This essay will examine theoretical and experimental probability in relation to the Korean card game called Sut-Da. First, a definition of probability and how it is used in general life will be examined. Each hand of Sut-Da provides the theore

Extracts from this document...

Introduction

ABSTRACT

This essay will examine theoretical and experimental probability in relation to the Korean card game called ‘Sut-Da’. First, a definition of probability and how it is used in general life will be examined. Each hand of ‘Sut-Da’ provides the theoretical probability for a player to win the game. It is clear however, that the theoretical value of winning in ‘Sut-Da’ does not always apply in real life games. Secondly, the experimental probability of winning for each hand is examined. To find out the probability of winning with each hand, I am using permutation & combination, theoretical probability and experimental probability. Experimental probability data was gained from my friend and I playing the game. Finally, within my evaluation, I looked at experimental probability using excel spread sheets and also using calculations that were compared with the experimental probability data gained from actually playing the game. The theory of probability has been covered in a number of textbooks and I used these textbooks to help me get used to the formula. I have then worked out all the possibilities of hands and their probabilities for winning, performing all the calculations myself and using my own numbers in presenting my data for the experimental probability data.

Introduction

        It all began when I started to watch a Korean drama called “Ta-JJa”. I decided to watch this drama as one of my favourite actor was the main character in the drama. This drama was about this man getting into the world of gambling in order to be successful in his life. This drama was only based on the tricks that could have been used while playing the game “Sut-Da”.

...read more.

Middle

        Probability of winning with April (Ddaeng-4) is split into eight different                         possibilities

Graph 7. Tree Diagram of Ddang-4

image68.png

image70.pngimage70.png

                    = 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

image71.png

        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.

image67.pngimage67.png

                  = 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

image72.png

image73.pngimage73.png

                  = 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

image74.png

        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.

image75.pngimage75.png

                  = 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)        = image77.pngimage77.png

                                        = 9.76952185 x 10-3

Dok-Sa

        Probability for player 1 to get Dok-Sa is 2/20C2.

...read more.

Conclusion

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’image33.jpgimage34.jpg

image35.jpgimage37.jpg

Bibliography

Chaudhri, Vivek. "Game Theory and Business Applications." Economic Record (2005): 88+.

"Game Theory." The Wilson Quarterly (2005): 92+.

Gendenko, B. V. and B. D. Seckler. The Theory of Probability. New York: Chelsea Publishing Company, 1962.

Hanafuda The Japanese Flowercard Game. 2010. 10 October 2009 <http://www.hanafuda.com/>.

Henry E. Kyburg, Jr. Probability Theory. Englewood Cliffs: Prentice-Hall, 1969.

Kneale, William. Probability and Induction. Oxford: Clarendon Press, 1949.

Leonard, Graham. Sutda. 26 04 2004. 12 December 2009 <http://hanafubuki.org/sutda.html>.

Urban, Paul, et al. Mathematics for the international students Mathematics HL (Core). Adelaide: Haese & Harris, 2008.

Wilson, Matthew C. "Uncertainty and Probability in Institutional Economics." Journal of Economic Issues (2007): 1087+.

 | Page

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Extended Essay- Math

    �&$h���(Ú]C a���$Hn��du�{C3/4 ��g�}6�Rd���$Û·oO���"'c�ÌE��(KL��L�g��'�m0J�f...j# �Y�"w�o~�H'f��h6i�d�3/4} �+l��Nj��M���(c)-BÖ¬Y��;�i�&o,'�ST �L"�~�qI�&'R"u�m�z+�:N�6������,-8�� "��=.ImT�mä·L�3/4�$�gÙ²e��QdoÅ9���V0P4�;�%�"�()u;���c�=��G�d�P...-"0oh���(c)�Ñ�-$��1/2�Gy�h�[5!���_}���A?�x�U"Vï¿½Ô ï¿½e-�LDǸ$�' J�L�1/41/2��=x�`�E� �×8I�"�;y�ƿO�)�h�-�L� v���+yC�"��� ص(c)�-�´@A�d�vB��u�Kâ·¡V �*(c)����-wM��]1/4dbIXD�,3/4(tm)�k)�Ѩ� -$,[�UW]m�z+?�-' ���H6R �n��{����N�k��pI�\!E�Q?N#��[$A?� ��'�K�6-�Hã°1/2��'"W��ӧO"� �G{�����(`�7q'H:�d"�Ó<�7�x�U"V�7o�-j�U(c)M �E����v-.]�,(c)�0'{/L{?Ú21��DM7dI�m�3/4H[is��(�x�C�H�!� Wx(r)�''���'hFR�#P7'�w ,��c�(c)uOq(c)� d-�e?�i�P�. QU-1/4-��.A?�^{-,2�g�G�-B�=�o��WG&� /1/4P|G� "�-�P �� -s�S -mX$��EX~�f�pv/X� �"�3/4�ƭ�Zig"��0�-Z��Q����" M�,,���)�m�"�ÍϪ@�h��' 8m�"�Q��`'x��d'�C�RG~��(c):� ��1/2��l�'"I^z饱/[:��a'�/�`�,Y8"i�%��-���O9pj%��\'"ï¿½× V���cP/P� �"^=�nâ°ï¿½x Z���ڴiS�] #`5-��dU" �_Þ"E'�M�x�Rd:�9T,�{\��Źç C,���K&(c)���ÙN�p\�1/2��ͣ'�Q�m1/4y$7w�����M�?�"�1/2{�b�j%�-�I@�ٳ',2.E"%�cZ�l�k�.��}1/2>��#�� O\{eI�m~b'�i�qK�^1/2z%�"� ��1/4���@y-���^��'��#F������Y�� � � 3/4bX�.��S,�$�OK�$$ä� V�GN#�N��M1/2Ȥ7�]��K/9 W��c�"�;$V4�"KR���m"-��^�"E�h�}��$\'1/2@b�...0�� ���-"�dxB��%�\�(9jԨ��PK�@�3/4�'K.\��4�(tm)c-�8�,(tm)(c)9-�...�\'�5\��NbY�:q�A?��$<����v�<Y�:K�F-3/4 �h]�}��Q�kZ��"�69�x�a[)�"w�Þ��� (tm)

  2. Math IA type 2. In this task I will be investigating Probabilities and investigating ...

    Therefore although it can be used relatively accurately to predict the most probable outcome, still it is not set in stone. Also unless, the total numbers of points played are multiples of three, the last 2 points or any 2 points will have to be distributed such that for example

  1. THE DICE GAME - calculating probabilities

    The probability is: P(N) * P(A)= 15/36 * 15/36 -------------------------------------- TASK 3 & 4 -------------------------------------- Solution: There are two cases when both players can roll their dice two times: Ann is winning: P(N) * P(A) = 15/36 * 15/36 Bob is winning: P(N)

  2. Infinity Essay

    A German mathematician named George Cantor showed that there are different Gibson, Lourdes page 2 orders of infinity, the infinity of points on a line being of a greater order than of prime numbers (positive integer that has exactly two positive factors, one and itself).

  1. Mathematics (EE): Alhazen's Problem

    solutions (unless these chords are in fact the diameter, as we will see in the following example). Analysis of specific scenarios: Let us analyze another more specific scenario.

  2. Modelling Probabilities in Tennis. In this investigation I shall examine the possibilities for ...

    but in which, if both players have three points, deuce is called and the next point determines the winner. This scenario means that no game may go beyond 7 points. We can model games as arrangements of 4 A's and 4 B's in 7 spaces, thus: Number of possible games Some of these games finish earlier than 7 points, however.

  1. The purpose of this investigation is to create and model a dice-based casino game ...

    Therefore the probability that players A and B both roll the same number is , or . Because the probability that both players roll the same number is , it stands to reason that the probability in which both players roll different numbers is . Let the ordered pair (m,n)

  2. Modelling Probabilities on games of tennis

    The table below shows the values of the probabilities: x P (X=x) [Fraction form] P (X=x) [Decimal form] 0 0.000017 1 0.000339 2 0.003048 3 0.016258 4 0.056902 5 0.136565 6 0.227608 7 0.260123 8 0.195092 9 0.086708 10 0.017342 To check whether the fractions found are correct, they all are added up, as their sum should be 1.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work