Dice Maths Investigation

(5/18)3 = P(B winning in the 3rd round)

(625/104976)/(5/18) = 125/5832

(625/104976)/(5/18) = (5/18)3

(625/104976) = (5/18) x (5/18)3

(625/104976) = (5/18)4

(5/18)4 = P(B winning in the 4th round)

(5/18)/(5/18) = 1

(5/18)/(5/18) = (5/18)0

(5/18) = (5/18) x (5/18)0

(5/18) = (5/18)1

(5/18)1 = P(B winning in the 1st round)

There seems to be a relationship between the round no. and 5/18. You can see that the probability of B winning in the 2nd round is 5/182 this is the relationship that I think I can use to create a formula.

This is the formula: (5/18)n

n = No. of rounds

Test:

1st round - 5/18

(5/18)1 = 5/18 - Correct

2nd round - 25/324

(5/18)2 = 25/324 - Correct

3rd round - 125/5832

(5/18)3 = 125/5832 - Correct

4th round - 625/104976

(5/18)4 = 625/104976 - Correct

The formula appears to work.

As in all rounds, the probability of B winning is the same as C winning, the formula can be used for either. In addition to this, it can also be used to find the probability of a round passing.

P(B winning in the nth round) = (5/18)n

P(C winning in the nth round) = (5/18)n

P(Round passing) = (5/18)n

Formula for the probability of A winning in the nth round

The first thing we have to do as with B and C is to create a small table of results showing us the actual probability of A winning in the various rounds.

st round

2nd round

3rd round

4th round

P(A winning)

1/6

5/108

25/1944

25/34992

Here we have something different to the last formula. We still have a scale factor of 5/18 but the range does not start at 5/18. Instead it starts with 1/6.

I shall experiment by dividing the probabilities and then trying to relate the results to the scale factor (5/18)

(5/108)/(1/6) = 5/18

(5/108) = (1/6) x (5/18)1

(1/6) x (5/18)1 = P(A winning in the 2nd round)

(25/1944)/(1/6) = 25/324

(25/1944)/(1/6) = (5/18)2

(25/1944) = (1/6) x (5/18)2

(1/6) x (5/18)2 = P(A winning in the 3rd round)

(125/34992)/(1/6) = 125/5832

(125/34992)/(1/6) = (5/18)3

(125/34992) = (1/6) x (5/18)3

(1/6) x (5/18)3 = P(A winning in the 4th round)

(1/6)/(1/6) = 1

(1/6)/(1/6) = (5/18)0

(1/6) = (1/6) x (5/18)0

(1/6) x (5/18)0 = P(A winning in the 1st round)

We can see a very definite relationship in the work I have done above.

Firstly the fraction 1/6 - that must be included in the formula.

Secondly, 5/18 - that too must be included.

Finally and perhaps the most important thing is n-1.

We see that instead of the round number being the same as the power of 5/18 as with B and C, it is the round number minus 1.

The formula should be like this:

(1/6) x (5/18)n-1

Test:

1st round - 1/6

(1/6) x (5/18)0 = 1/6 - Correct

2nd round - 5/108

(1/6) x (5/18)1 = 5/108 - Correct

3rd round - 25/1944

(1/6) x (5/18)2 = 25/1944 - Correct

4th round - 125/34992

(1/6) x (5/18)3 = 125/34992 - Correct

The formula appears to work.

We now have the following formulas.

P(A winning in nth round) = (1/6) x (5/18)n-1

P(B winning in the nth round) = (5/18)n

P(C winning in the nth round) = (5/18)n

P(Round passing) = (5/18)n

The reasons for these formulae ratio are as follows:

The probability of each player winning decreases by a common ratio of 5/18 as each round passes. This is because 5/18 equals the original probabilities of each player not winning in the first round multiplied together - 5/6 x 4/6 x 3/6. This is equivalent to the probability of a round passing. Hence, the probability of a player winning must be multiplied by the probability of the game reaching the previous round multiplied by the probability of a round passing. - So the probability of (for example) player B winning in the 8th round must be the probability of the 7 rounds passing, multiplied by 5/18 (the probability of player B winning). The probability of say player B winning must be: 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 which in turn = 5/188

This explains the formula (5/18)n

The probability of Player A winning in the 8th round must equal the probability of the 7 rounds passing, multiplied by 1/6

The probability of player A winning in the 8th round must be:

5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 5/18 x 1/6.

The fact that only 7 five eighteenths are necessary explains the n-1 part and the fact that it must be multiplied by the first term explains the 1/6 part. So collectively I think that this explains the formula: 1/6 x (5/18)n-1

These formulas find the probability of the game lasting to a certain round and then a certain player winning. The overall aim of this investigation is to find the overall probability of a certain player winning. To do this I must begin working on the cumulative probabilities of a certain player winning - the probability of a player winning before or in the nth round.

st round

2nd round

3rd round

4th round

A

0.166666666

0.212963296

0.225823045

0.22939529

B

0.277777778

0.354938

0.37637

0.382325483

C

0.277777778

0.354938

0.37637

0.382325483

This shows the decimal cumulative probability of either A,B or C winning.

I have studied geometrical progressions in the book 'Pure mathematics 1' by J.K. Backhouse and S.P.T. Houldsworth to help me create a formula.

The geometrical progression I shall study first is for A

st round

2nd round

3rd round

4th round

P(A winning)

1/6

5/108

25/1944

25/34992

The first thin we need to identify is the common ratio. In our case it is 5/18.

The probabilities progress by 5/18 every round.

/6 x 5/18 = 5/108...5/108 x 5/18 = 25/1944...25/1944 x 5/18 = 125/34992.

Next we have to identify the first term: 1/6

Position of term

st

2nd

3rd

4th

Value

1/6

(1/6) x (5/18)1

(1/6) x (5/18)2

(1/6) x (5/18)3

Next, I studied infinite geometrical progressions from the same resource.

I found a formula for the sum of geometrical progressions:

Sn = a x ((1-rn)/(1-r))

Our details can be substituted in to the formula to create one accurate to our results:

P(A winning in or before nth round) = (1/6) x ((1 - (5/18)n)/(1 - (5/18))

st round

2nd round

3rd round

4th round

A

/6

23/108

439/1944

8027/34992

Test:

1st round - 1/6

(1/6) x ((1-(5/18)1)/(1-(5/18)) = 1/6 - Correct

2nd round - 23/108

(1/6) x ((1-(5/18)2)/(1-(5/18)) = 23/108 - Correct

3rd round - 439/1944

(1/6) x ((1-(5/18)3)/(1-(5/18)) = 439/1944 - Correct

4th round - 8027/34992

(1/6) x ((1-(5/18)4)/(1-(5/18)) = 8027/34992 - Correct

The formula appears to work: we now have the following cumulative probability formula:

P(A winning before or in nth round) = (1/6) x ((1 - (5/18)n)/(1 - (5/18))

The next step is to repeat this for B (and C and P(round passes))

I think the same formula can be used for them with a little modification: You will notice that in all rounds, the ratio of A:B:C is 3:5:5 I have created a table to illustrate this further. Player A's probabilities divided by three should be the same as play B or Cs' probabilities divided by five:

Player

Probability A

Probability B

P(A) Divided by 3 or 5

P(B) Divided by 3 or 5

Player A

1/6

5/108

1/18

1/18

Player B (or C)

5/18

25/324

1/18

1/18

This can be used in conjunction with the formula for A to find the cumulative probability of B (or C) winning in the nth round.

The formula would be:

((1/6) x ((1 - (5/18)n)/(1 - (5/18)))/3 x 5

= (1/6) x ((1 - (5/18)n)/(1 - (5/18))/3 x 5

P(B winning in or before nth round) = ((1/6) x ((1 - (5/18)n)/(1 - (5/18))/3) x 5

We now have the following formulas:

Conditional Probability:

P(A winning in nth round) = (1/6) x (5/18)n-1

P(B winning in the nth round) = (5/18)n

P(C winning in the nth round) = (5/18)n

P(Round passing) = (5/18)n

Cumulative Probability:

P(A winning before or in nth round) = (1/6) x ((1 - (5/18)n)/(1 - (5/18))

P(B winning in or before nth round) = ((1/6) x ((1 - (5/18)n)/(1 - (5/18))/3) x 5

P(C winning in or before nth round) = ((1/6) x ((1 - (5/18)n)/(1 - (5/18))/3) x 5

I shall now continue working on the formulae In order to simplify them:

P(B winning in or before nth round) = ((1/6) x ((1 - (5/18)n)/(1 - (5/18))/3) x 5

There is another way I could do this formula that requires much less work:

Instead of the ratio I could modify the formula for use with B and C:

I would have to change the first term but the common ratio would remain the same.

It should be like this:

P(B) = (5/18) x (1-(5/18)n)/(1-(5/18))

(5/18)/(13/18) = 5/13

Hence: P(B) = (1-(5/18)n) x 5/13

We can do the same for A:

(1/6) x ((1 - (5/18)n)/(1 - (5/18))

(1/6)/(13/18) = 3/13

Hence: P(A) = ((1 - (5/18)n) x 3/13

We now have the following simplified cumulative probability formulas:

P(A winning before or in nth round) = ((1 - (5/18)n) x 3/13

P(B winning before or in nth round) = ((1 - (5/18)n) x 5/13

P(C winning before or in nth round) = ((1 - (5/18)n) x 5/13

We can see the very interesting and important 3:5:5 ratio in these results in the form of 13ths.

Finding the overall probabilities of each player winning the game.

The cumulative probability formulae I have found should be extremely helpful in finding the overall probabilities.

The major problem with finding this is that there the game could, in theory, go on forever with no player winning. However, this is so unlikely that we are likely to get values so close to 1 that the game could not really go on.

The formula for finding the overall probability of player A winning is:

P(A winning before or in nth round) = (1 - (5/18)n) x 3/13

Only n= infinity.

Unfortunately there is no 'infinity' button on my calculator so I will have to substitute it with an alternative!!

The question is what degree of accuracy to do it to. I have chosen to do it to when the calculator thinks that the sum of the cumulative probabilities of players A. B and C is so close to 1 that it is 1. I feel that there is no need to calculate to any greater degree of accuracy because nobody using the formula is going to need the degree of accuracy to be any more than the capabilities of their calculator.

Therefore, infinity shall be replaced with the number 19.

The calculator thinks that there is no chance of the game continuing beyond the 19th round.

So the overall probability of A winning is:

(1 - (5/18)19) x 3/13

= 0.23076923

The overall probability of B winning is:

(1 - (5/18)19) x 5/13

= 0.38461538

The overall probability of C winning is:

(1 - (5/18)19) x 5/13

= 0.38461538

Finding the most likely winner

The most likely winner is the player with the highest overall cumulative probability.

The player with the highest overall cumulative probability is B or C

Players B and C are equally likely to win the game with an overall cumulative probability of 0.38461538

Player A is less likely to win with an overall cumulative probability of 0.23076923

Finding the most likely length of the game

The most likely length of the game in terms of dice throws is the point at which the probability (Not the overall cumulative probability) is highest.

The most likely length of the game is 2 or 3 dice throws as it is in the 1st round that players B and C have an equal probability of winning at 5/18.