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.
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
5/6 not a 1 2/6
4/6 not 2,3 3/6
3/6 not 4,5,6 1/6
5/6 not a 1 2/6
4/6 not 2,3 3/6
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
So if 1/2 X = 32 then X = 64
The following geometric series is for A winning the game overall in any
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
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