• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

The Die Investigation.

Extracts from this document...


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.

...read more.









B (simplest form)





C (simplest form)





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

...read more.


        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










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.


...read more.

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

  1. Beyond Pythagoras

    I think these tests have proved that my formulae for all of the lengths of the triangles are correct, because in all three examples, the answers came out exactly right, down to the decimals. If there was any indiscretion, one would assume that it would have shown up in at least one of these three examples.

  2. Dice Game Maths Investigation

    Therefore when it says another letter, either A, B or C, it means that particular letter won on that specific throw. Table of Results After all this I decided it would be a good idea (as did my teacher) to collaborate my results with someone else.

  1. Beyond Pythagoras

    This proves that my nth term is correct. nth term for hypotenuse side I will find out this nth term using the same formula I used finding out the middle side, as it has a second difference.

  2. Beyond Pythagoras

    x (4n + n2 + 3) = 16n2 + 4n3 + 12n + 4n3 + n4 +3n2 + 12n + 3n2 + 9 = 9 + 24n + 22n2 + 8n3 + n4 c2 = (4n + n2 + 5)

  1. Beyond Pythagoras

    27 120 123 33 180 183 39 252 255 Finally to get the fourth part you have to times the first part by four. Shortest Side Middle Side Longest Side 12 16 20 20 48 52 28 96 100 36 160 164 44 240 244 52 336 340 So my theory is correct with the times.

  2. Research on Pythagoras and his work.

    However he did contribute to Pythagoras's interest in mathematics and astronomy, and advised him to travel to Egypt to learn more of these subjects. Thales's pupil, Anaximander, lectured on Miletus and Pythagoras attended these lectures. Anaximander certainly was interested in geometry and cosmology and many of his ideas would influence Pythagoras's own views.

  1. Beyond Pythagoras.

    (For instance in this case is 4), most likely 4n.However, because 4 is the difference, the formula must be n�. I now believe that the answer will have something to do with 4n�. So, I will now write the answers for 4n�.

  2. Investigate the probability of someone rolling a die and the probability of it landing ...

    wins = 5 x 2 x 1 x 1 = 5 6 3 2 6 108 P (B) wins = 5 x 2 x 1 x 5 x 1 = 25 6 3 2 6 2 324 P (C) wins = 5 x 2 x 1 x 5 x 2

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