Investigate the probability of someone rolling a die and the probability of it landing on particular number for a player to win the game

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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:

. 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
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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 ...

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