# The Dice Game - Probabilities.

Extracts from this document...

Introduction

The Dice Game

Three players A, B, C play a game with a single dice.

The rules of the game are:

Player A ALWAYS goes first.

A rolls the dice. If the dice lands showing a 1 then A wins the game. If A does not throw a 1 then B has a turn.

B rolls the dice. If the dice lands showing a 2 or a 3 then B wins the game. If B does not throw a 2 or 3, then C has a turn.

C rolls the dice. If the dice lands showing a 4 or a 5 or a 6 then C wins the game. If C does not throw a 4 or a 5 or a 6 then A starts again.

The procedure continues until there is a winner.

## INVESTIGATE ANY OR ALL OF

- The probabilities of each A, B, or C winning the game.

- Who will be the most likely winner?

- The most likely length of the game in terms of the number of rolls of the dice to produce a winner.

I’ve decided to investigate question 1 out of the three.

Player A has to throw a 1 to win the game. This means that in terms of probability, the probability that A will win is going to be (1/6). The probability that A will loose is (5/6) because:

1 – (1/6) = (5/6)

Player B has to throw a 2 or a 3 to win the game. This means that the probability that B will win would be (2/6), because:

(1/6) + (1/6) = (2/6)

The probability that B will loose would be (4/6) because that is the remainder of 1:

1 – (2/6) = (4/6)

Player C has to throw a 4 or a 5 or a 6 to win the game. This means that the probability that C will win would be (3/6) because:

(1/6) + (1/6) + (1/6) = (3/6)

The probability that C will loose the game would be (3/6)

Middle

P(B) 2nd go = (5/6) × (4/6) × (3/6) × (5/6) × (2/6)

For B to have a third go, it is an extended version of the 2nd go: A would have lost, B would have lost, C would have lost, then back to A but he lost, B would have lost again, C would have lost again, A would lose again and then (finally!) B would have a 3rd go. This phrase would look like this:

P(B) 3rd go = [(5/6) × (4/6) × (3/6)]² × (5/6) × (2/6)

Now, if I put all the values into a long series for P(B), this is what it would look like:

P(B) = [(5/6) × (2/6)] + [(5/6)² × (4/6) × (3/6) × (2/6)] + {[(5/6) × (4/6) × (3/6)]² × (5/6) × (2/6)}…

The problem is the same as before in that I don’t know when B will win, so I have to cancel out the times when B will not win to find out the probability that B will win. To do this, I have to multiply the probability that everyone will lose with the probability that B will win. This would lead to a series like so:

P(B) × (5/6) × (4/6) × (3/6) = [(5/6)² × (4/6) × (3/6) × (2/6)] + {[(5/6) × (4/6) × (3/6)]² × (5/6) × (2/6)} + {[(5/6) × (4/6) × (3/6)]³ × (5/6) × (2/6)}…

With this, I now have to take that series away from the series of P(B), and I will be left with the following:

P(B) – [P(B) × (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6)] + {[(5/6) × (4/6) × (3/6)]³ × (5/6) × (2/6)}

The term that is printed in bold is the infinitely small term, and can actually be ignored in this circumstance because it will make no difference to the answer since it is so close to zero. Leaving it out will leave this:

P(B) – [P(B) × (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6)]

Now, if I simplify it then I can get this:

[P(B)][1 – (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6)]

As before, I shall now take over the [1 – (5/6) × (4/6) × (3/6)] and will be able to obtain an answer:

P(B) = (5/6) × (2/6) = (5/18) = (5/13)

(1 – (5/6) × (4/6) × (3/6)) (13/18)

I shall now find out what the probability is for C winning the game.

Conclusion

B – Bk = d – dkª

B(1 – k) = d(1 – kª)

B = d(1 – kª)

(1 – k)

This is the formula for the investigation. It is called “the sum of a geometric series”. This investigation was based on geometric series, but in this case I was able to obtain an answer. In this investigation, k stands for (5/6) × (4/6) × (3/6), which sums up to a total that is smaller than 1. When you give the fraction a power, you make the fraction smaller the bigger the power number is. If you make the index as large as infinity, then the fraction will get smaller and smaller. This value is so close to zero that it can almost be counted as zero, because the value is so infinitely small. That is why I ignored the infinitely small term; it would make no difference to the answers. kª in this case is 0. 1 – 0 = 1. d × 1 = d. With these sums, the formula would look like this:

B = d

(1 – k)

This is the full algebraic solution to the investigation; the formulae that I found for each of the probabilities look somewhat like this. But what would happen if k was not a fraction? Well, any number that is larger than 1 will grow very large when you give the number a power. If the power were infinitely big, then the number would be infinitely big too. With an infinitely big number, you cannot obtain an answer, because infinity is an endless number. What I mean is 1 – (k to the power of infinity) is immeasurable. The reason why in this investigation I was able to obtain an answer to the probability series was because of k being smaller than 1, i.e. a fraction, I was able to obtain an answer, and the number is a fraction because you use fractions in probability.

This concludes my investigation.

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