We have been asked to investigate the probability of someone rolling a die and the probability of it landing on particular number for a player to win the game. For A to win he/she must roll a 1 and if he/she does this they have won the game. For B to win, first of all A must lose and they must roll 2 or a 3 and then they have won the game. For C to win they must roll a 4,5 or 6 and of course B must have lost. I have to investigate these tasks:

1. The probability of A, B or C winning.

2. Who will be the most likely winner?

3. Most likely length of the game.

I have first of all drawn a tree diagram so it is easier to interpret and it is easier to see things visually:

From this I tried to find the probability that no one wins in Round 1 and this is how I did it:

P (LLL) = 1- (5 x 2 x 1)

6 3 2

P (LLL) = 1 – 5

18

P (LLL) = 13

18

I also found the probability of A, B and C winning in Round 1:

P (A) wins = 1

6

P (B) wins = 5 x 1 = 5

6 2 18

P (C) wins = 5 x 2 x 1 = 5

6 3 2 18

In the second round the probabilities of winning will be different, as you must say that no one won in the last round. This is how I found out the probability of A, B and C winning in the second round:

P (A) 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 x 1 = 25

6 3 2 6 3 2 648

Here I found that there was a pattern and this was that each time there is another round the difference increases by 5. I will now test my theory to see if it is correct.

18

P (A) winning in first round = 1

6

P (A) winning in second round = 5

108

When you divide the probability of winning in the second round by the probability of winning in the first round you get 5.

18

I will now test my theory again:

P (A) winning in 2nd round = 5

108

P (A) winning in 3rd round = 25

1944

Again my theory is correct as if you times the probability of A winning in the second round by 5 you get the correct answer 25.

18 1944

This also works for B as well as A:

P (B) winning in 1st round = 5

18

P (B) winning in 2nd round = 25

324

Again the difference between these two is 5. I will now test this for C.

18

P (C) winning in 1st round = 5

18

P (C) winning in 2nd round = 25

648

This time it does not work as the difference between the two is 5

36

I then found that the difference of 5 was again apparent when finding out the difference

18

between the probability of having no winner.

P (No Winner) in Round 1 = 5

18

P (No Winner) in Round 2 = 25

324

You also know that the game can go on for infinity if no one wins and this can be written as follows:

5

18

After working out the difference between the two rounds to be 5 I came up with a formulae: 18

A winning in a particular round = 1 x 5

6 18

n = the round number

I will now test this:

What is the probability of A winning in the 4th round?

1 x 5

6 18

= 0.00357 (The answer from the formulae)

= 125 (The correct answer)

34992

I now divide 125 and I get the same answer. We now know that the formulae works for

34992

the fourth round but now I am going to test it for the third round.

1 x 5

6 18

= 0.0129 (Formulae answer)

= 25 = 0.0129

1944

Again my formulae works but I will give it one more test. The chance of A winning in the tenth round:

1 x 5

6 18

= 0.00000164 (Formulae answer)

= 0.00000164 (Actual answer)

I have now tested my formulae 3 times and each time has come out with the correct answer. I have now come to the conclusion that this is correct. From this I think that I might have found a formula for B:

5 x 5

18 18

Again n is the round number and this is the formula for B winning in a particular round. I will now test my formulae to find the probability of winning in round 4:

5 x 5

18 18

= 0.00595

Actual answer: 0.00595

P (B) winning in Round 3:

5 x 5

18 18

= 0.0214

Actual answer: 125 = 0.0214

5832

P (B) winning in Round 10:

5 x 5

18 18

= 0.00000274

Actual answer: 0.00000274

I have now proved that this formulae works by testing it 3 times and has come out with the correct answer each time. I will now test my formulae for C to win in a particular round:

5 x 5

18 18

The formula is the same as B and therefore will come out with the same answers and as B and C have the same probabilities I know that this formulae is correct. From these answers I can see that again B and C have the same probabilities and that A doesn’t have much of a chance of winning because the probability of A winning in a particular round is less than B and C.

I am now going to find the probability of A winning in the 1st or 2nd or 3rd or 4th round (etc.):

P (A) winning in 1st or 2nd round:

1 + 5 = 23

6 108 108

As it is easier to see a trend in a decimal I am going to convert all fractions into decimals:

1 + 5 = 23 = 0.213

6 108 108

P (A) winning in 1st, 2nd or 3rd round:

1 + 5 + 75 = 0.223

6 108 5832

As you can see there is a very slight increase of 0.01.

P (A) winning in 1st, 2nd, 3rd or 4th round:

1 + 5 + 75 + 125 = 0.229

6 108 5832 34992

This time there is even a smaller difference of 0.006.

P (A) winning in 1st, 2nd, 3rd, 4th or 5th round:

1 + 5 + 75 + 125 + 625 = 0.230

6 108 5832 34992 629856

Again there is even a smaller difference between the 2 of 0.001.

As you can see from above the probability of A winning is getting slightly bigger each time another round is played. If you carry this on, there will be almost no change in the end between the two probabilities, So, as another round is played the probability of winning only increases by a very small amount.

I know needed to find the actual probabilities of winning to tell me who would be the most likely winner. I did this by finding this equation:

y = 1__

(1-x)

I have taken x to equal 5 as it is not variable which means that it will not change. I’ll

18

now test this formula:

A = 1

1- 5

18

A = 1 x 1

6 18

13

A = 3

13

B = 5 x 1

18 13

18

B = 5

13

C = the same as B

C = 5

13

I will now check this:

3 + 5 + 5 = 1

13 13 13

I know now that this is correct as they all add up to 1.

From this we can see that B and C are the most likely winners in any round.

In question 3 we must decide the most likely number of rolls of the dice before someone wins. This is how I worked it out:

%Chance that A will win 1st round = 16.6

%Chance that B will win in 1st round = 27.7

I found that 16.6% + 27.7% is 44.4%. Therefore it is quite unlikely that you will have someone who has won as it is less than 50%. So, I added C:

%Chance that C will win in 1st round = 27.7

This time it is adds up to be more than 50% (72.2) so you have a very good chance that it will be 3 rolls. From this I can also work out that there is a 27.7% chance that you will have no winner.

So, the mostly likely number of rolls is 3.