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Introduction

Rules

10 counters are placed in the centre of the dodecahedron. Two dice are then rolled the amount of the two numbers on the dice are then recorded. A counter from the centre of the dodecahedron is then placed into “A” wins or “B” wins depending on the sum of the two numbers.

E.g.

3 + 2 = 5 therefore it goes into A wins

6 + 4 = 10 so obviously B wins

And so on. this continues until there are no counters remaining inside the dodecahedron.

Prediction

I predict that B will win because it is more likely to get the numbers that when added together make the sum of the numbers found in B.

Results

 Game Counters in A Counters in B Who won ? 1 5 5 draw 2 4 6 b won 3 3 7 b won 4 5 5 draw 5 5 5 draw 6 4 6 b won 7 3 7 b won 8 6 4 a won 9 2 8 b won 10 4 6 b

The results table

Middle

3

4

5

6

1

2

3

4

5

6

7

2

3

4

5

6

7

8

3

4

5

6

7

8

9

4

5

6

7

8

9

10

5

6

7

8

9

10

11

6

7

8

9

10

11

12 P(A Wins) = 12/36 = 1/3

P(A Wins) = 24/36 = 2/3

The possibilities found above are from using my possibility diagram.

All of the green coloured squares have a higher probability of being obtained, as shown in the diagram above. Therefore the game is unfair. For example, there are 5 different ways of getting 8 whereas there are only 2 ways of reaching 11.

The probability of A winning is 1/3 and B winning 2/3. Showing that the game is foul not fair.                                       Extension Work

The probability of A winning is 1/3, but this only applies to one counter. I am now going to find out the probability of A winning a game instead of winning a counter.

Conclusion

Pascal’s triangle

I have formed this triangle by arranging the results from the first 5 rows which have been taken from the various A and B. From this I could see a pattern developing. The pattern that I developed was the two numbers diagonally above the number were added together to get the bottom number.

E.G.

This triangle is called Pascal’s triangle.

Overall conclusion

From my findings it becomes evident to me that both the probability of B winning one counter and the game is considerably more likely than it is for A to win a counter and the match. The  probability of B winning a counter is 2/3 and the likely hood of B winning the game is 0.855. These statistics confirm to me that the game is foul because there is not an equal opportunity of a draw or A or B winning.

This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras section.

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