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  • Level: GCSE
  • Subject: Maths
  • Word count: 2900

Dice Maths Investigation

Extracts from this document...

Introduction

The instructions above tell us how to play a very simple dice game that relies on probability. From the offset, we can see that the game is an unfair one as the probability of one of the players winning varies. Below is a table showing us how different dice roles affect each player. Dice Number 1 2 3 4 5 6 A Wins B Wins C Wins We can see that the distribution of numbers is not even. However, the probability of each player winning is conditional because if A throws a one, the game finishes and B does not get a chance to throw. To illustrate this more clearly I have created a conditional probability tree diagram: From this simple diagram, the following results are easily obtainable: 1st Round 2nd Round 3rd Round 4th Round A 1/6 5/108 25/1944 125/34992 B 5/18 25/324 125/5832 625/104976 C 5/18 25/324 125/5832 625/104976 The most important thing I found that each probability had in common with one another, except Player A's first term, was that all the numbers on top were divisible by five and that all the bottom numbers were divisible by 18: 25/5 = 5...324/18 = 18...625/5 = 125...104976/18 = 5832. You will notice that when any of the probabilities are divided by 5/18 the result is the previous probability. This is because the probability of a player winning is multiplied by the probability of all the previous rounds passing. ...read more.

Middle

This is equivalent to the probability of a round passing. Hence, the probability of a player winning must be multiplied by the probability of the game reaching the previous round multiplied by the probability of a round passing. - So the probability of (for example) player B winning in the 8th round must be the probability of the 7 rounds passing, multiplied by 5/18 (the probability of player B winning). The probability of say player B winning must be: 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 which in turn = 5/188 This explains the formula (5/18)n The probability of Player A winning in the 8th round must equal the probability of the 7 rounds passing, multiplied by 1/6 The probability of player A winning in the 8th round must be: 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 1/6. The fact that only 7 five eighteenths are necessary explains the n-1 part and the fact that it must be multiplied by the first term explains the 1/6 part. So collectively I think that this explains the formula: 1/6 x (5/18)n-1 These formulas find the probability of the game lasting to a certain round and then a certain player winning. The overall aim of this investigation is to find the overall probability of a certain player winning. ...read more.

Conclusion

I have chosen to do it to when the calculator thinks that the sum of the cumulative probabilities of players A. B and C is so close to 1 that it is 1. I feel that there is no need to calculate to any greater degree of accuracy because nobody using the formula is going to need the degree of accuracy to be any more than the capabilities of their calculator. Therefore, infinity shall be replaced with the number 19. The calculator thinks that there is no chance of the game continuing beyond the 19th round. So the overall probability of A winning is: (1 - (5/18)19) x 3/13 = 0.23076923 The overall probability of B winning is: (1 - (5/18)19) x 5/13 = 0.38461538 The overall probability of C winning is: (1 - (5/18)19) x 5/13 = 0.38461538 Finding the most likely winner The most likely winner is the player with the highest overall cumulative probability. The player with the highest overall cumulative probability is B or C Players B and C are equally likely to win the game with an overall cumulative probability of 0.38461538 Player A is less likely to win with an overall cumulative probability of 0.23076923 Finding the most likely length of the game The most likely length of the game in terms of dice throws is the point at which the probability (Not the overall cumulative probability) is highest. The most likely length of the game is 2 or 3 dice throws as it is in the 1st round that players B and C have an equal probability of winning at 5/18. ...read more.

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