# The Die Investigation.

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Introduction

James Chaffe. Die Investigation Coursework. 30/04/2007

3 People ‘A, B & C’ are playing a game of die. A wins the game if he throws a ‘1’, B wins the game if he throws a ‘2’ or a ‘3’ and C wins the game if he throws a ‘4’, ‘5’ or a ‘6’. Once someone has thrown their number that’s it, the game is over and they’ve won. A goes first, B second and C third. If none of those throw their number then A has another go and then if he doesn’t get it, it goes to B and so on until someone wins.

- What is the probability of each person winning?
- How many goes do you reckon it should take until someone wins?

I have decided to draw a tree diagram to see what I am dealing with and then put together in a chart the probability of each person winning on their first, second and third attempts. Then from that see if there are any patterns evolving.

Middle

25/1944

125/34992

A (LCD)

3/18

15/324

75/5832

375/104976

B (simplest form)

5/18

25/324

125/5832

625/104976

C (simplest form)

5/18

25/324

125/5832

625/104976

I then noticed some patterns emerging, such as:

P(B winning) = P(C winning).

I also noticed that 182 is 324, 183 is 5832 and 184 is 104976.

And 52 is 25, 53 is 125 and 54 is 625.

If n represents on which attempt they win on I can work out with the nth term that:

P(B Winning) = P( C winning) = (5/18)n

(5/18)1 = 5/18

(5/18)2 = 25/324

(5/18)3 = 125/5832

(5/18)4 = 625/104976

I then tested to see if A ∝ (B = C) in a similar pattern between each attempt.

## Attempt 1

A = K(B = C).

(3/18) = (5/18)K.

Therefore: K= 3/5.

## Attempt 2

A = K(B = C).

(15/324) = (25/324)K.

Therefore: K = 3/5.

## Attempt 3

A = K(B = C).

(75/5832) = (125/5832)K.

Therefore: K = 3/5.

## Attempt 4

A = K(B = C).

(375/104976) = (625/104976)K.

Therefore. K = 3/5.

That means that A is always directly proportional to B and C at the same proportion for throughout each attempt.

The probability of A winning is 3/5 of the probability of B winning and 3/5 of the probability of C winning.

So using the nth term I can determine that:

P(A winning) = 3/5.{(5/18)n}

P(B winning) = (5/18)n

P(C winning) = (5/18)n

Conclusion

From this we now only need to work out attempt 1s probabilities and to simply work out any other different types of scenario just change the denominator so it adds up to one when you add all the players changes together.

From this I can get:

P(A winning) = 3/13.

P(B winning) = 5/13.

P(C winning) = 5/13.

This probability investigation is entirely dependant on what position you go in. If for example C went first, then B second and A third.

Attempt 1 | Attempt 2 | |

C | 9/18 | 45/36 |

B | 3/18 | 15/324 |

A | 1/18 | 5/324 |

As shown from this we can see that in both attempts.

C is three times more likely to win than B.

And B is three times more likely to win than A.

We can work out using my early-described method:

P(A winning) = 1/13

P(B winning) = 3/13

P(C winning) = 9/13

So I can conclude that there is a set way to work out these problems as I have explained. You may also work out what the LCD is for these probability by making the total equal 1.

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

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