Dice Game

I have worked out all of the probabilities B will have the largest chance of

winning the game by far and A will have the smallest chance.

ROUNDTHROWSROUND

THROWS1122126212224123242338441124245132526626261371327128112

8259372913102530381126311112514322513383338144113439152535121611

364141712371318133861819253951420124013

This frequency graph shows that in practice the game is most likely to be

won in the first round.

This frequency graph shows each round will normally take between 2 and 3

rolls of the dice to produce a winner. This also coincides with what I have

said about the being most likely to be won in the first round.

A B C A B C

1

/6

2,3

5/6 not a 1 2/6

4,5,6

4/6 not 2,3 3/6

1

3/6 not 4,5,6 1/6

2,3

5/6 not a 1 2/6

4,5,6

4/6 not 2,3 3/6

3/6

Not 4,5,6

Probability of A winning first throw = 1/6= 6/36

Probability of B winning first throw = 5/6 x 2/6= 10/36

Probability of C winning first throw = 5/6 x 4/6 x 3/6 = 60/216 = 10/36

Probability of A winning second throw = 5/6 x 4/6 x 3/6 x 1/6 = 60/1296 = 15/324

Probability of B winning second throw = 5/6 x 4/6 x 3/6 x 5/6 x 2/6 = 600/7776 = 25/324

Probability of C winning second throw = 5/6 x 4/6 x 3/6 x 5/6 x 2/6 x 3/6 = 3600/46656 = 25/324

The probabilities of B and C winning the game will always be the same

however many games you play, as the numbers don't change (5/6 x 2/6 for B

and 5/6 x 4/6 x 3/6 for C) they just repeat. The answers are all multiples of

the first one.

Formula of A winning = 1/6 (5/6 x 4/6 x 3/6)^N-1

N = the number of rounds minus the first round where A did not win

otherwise there would not be a probability there.

This is the probability to calculate the overall probability of A winning in any

round. For this formula to work you must substitute the N with the round

number for which you are trying to calculate the probability.

Formula of B or C winning = 5/18^(N)

This is the formula for B and C winning in any round apart from the first.

For this formula to work, substitute the N for the round that you wish to

calculate the probability for.

To find the overall probability of each player A, B or C winning the game

you need to use a geometric series.

This is a simple example of a geometric series:

X = 32 + 16 + 8 + 4 + 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 ......

In this line you can see that each number is half the previous number

/2 X = 16 + 8 + 4 + 2 + 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 ......

In this line I have divided everything by half so that I the 32 is taken out.

Therefore if 1/2 X = 16 + 8 + 4 + 2 + 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32

......

then the other half of X must equal 32 as they were both taken out in the

previous line.

So if 1/2 X = 32 then X = 64

The following geometric series is for A winning the game overall in any

round:

A wins P= 1/6 x (5/18 x 1/6) + ((5/18)^2 ) x 1/6) +((5/18)^3) x 1/6) + .......

5/18 P = (5/18 x 1/6) + ((5/18)^2 ) x 1/6) +((5/18)^3) x 1/6) + .......

3/18 P = 1/6

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

P = 1/6 x 18/13

P = 18/78 = 3/13

The probability of A winning the whole game in any round is 3/13

This is the geometric series for both B or C winning:

B or C wins P = 5/18 + ((5/18)^2) + ((5/18)^3) + ((5/18)^4) ........

5/18 P = ((5/18)^2) + ((5/18)^3) + ((5/18)^4) ........

3/18 P = 5/18

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

P = 5/18 x 18/13

P = 90/234 = 10/26 = 5/13

This means that the probability of B winning is 5/18 and the probability of C

winning is also 5/18.

To prove that the probabilities of A, B and C winning you can add them up

and if the equal 1 then they are correct. If they do not equal 1 then they will

be wrong.

Probability of A winning = 3/13

Probability of B winning = 5/13

Probability of C winning = 5/13

so 3/13 + 5/15 + 5/15 = 1

These probabilities are correct because they have been reached using a

geometric series which are reliable and I also know that B and C's

probabilities for winning in each round are the same so the overall

probabilities must be the same aswell.

From looking at the overall probabilities of A, B and C winning the overall

game you can see that A by far has the worse chance of winning the game.

The probabilities of of B and C winning the game are identical so they both

have the same chance of winning the game, so there is not 1 clear player who

is most likely to win the game.

At first I thought that player B was the most likely player to win the game

because I thought the overall probabilities were in fact the probabilities of

each player rolling one of their allocated numbers. but then I realised that this

was wrong because B only has 2 numbers (2 and 3) and C has 3 numbers

(4,5 and 6) so if what I thought was correct then C would have the better

probability and they would not be the same. The probabilities are really the

probability of the player rolling one of their numbers and the order that they

roll in put together. Because although B only has 2 numbers it rolls before C

so it has chance to win before C gets to have a go. This order increases the

probability of B winning and decreases the probability of C winning the game

overall.