You might like to draw your own tree diagrams for these
solutions.
Several of you suggested that having the same number of the same
colour ribbons would make the game fair but this is not the case.
Take, for example, one that many of you chose, which was four red
and four blue ribbons. What happens?
\begin{eqnarray}\mbox{P (two the same colour)}&=& \mbox{P
(Red and Red)}+\mbox {P (Blue and Blue)}\\ &=&
\mbox{P(RR)}+\mbox{P(BB)}\\ &=&4/8 \times3/7 + 4/8
\times3/7 = 24/56 \end{eqnarray}
But this is less than a half so you have less chance of selecting
two of the same colour than choosing two ribbons of different
colours ($1 - 24/56 = 32/56$)
\begin{eqnarray}\mbox{P(two of different colours)} &=& 3/4
\times 2/3 \\ &=& 6/12 \; \mbox{OR}\;1/2\end{eqnarray}
Charles of Madras College sent in an excellent solution, which
arrived too late for full inclusion.
Laura and David of Cannock Chase High School gave the following
explanation of why the game is not fair (and the reason is not
because there are an uneven number of ribbons (see later)). It is
possible to make a fair game however and this is discussed
below.
The game is not fair because:
$$2/6 \times1/5 = 2/30$$
$$4/6 \times3/5 =12/30$$
Probability that the two ribbons are the same is $14/30$ $(2/30 +
12/30)$.
The game is not fair as it is not an even chance; if it was even it
would be $15/30$.
Nathan H. and Colby B. offered a solution to making the game
fair:
No, the game is not fair.
There only need to be 4 ribbons (3 same, 1 different) therefore
there can only be a 50/50 chance that one will win.
Working:
\begin{eqnarray}\mbox{P (Red and Red)} &=& \mbox{P (R)}
\times \mbox{P (R)} \\ &=& 3/4 \times 2/3 \\ &=&
6/12 \; \mbox{OR}\; 1/2 \end{eqnarray}
\begin{eqnarray}\mbox{P (Red and Blue)} &=& \mbox{P (B)}
\times \mbox{P (R)} \\ &=& 3/4 \times1/3 \\ &=&
3/12 \; \mbox{OR}\; 1/4 \end{eqnarray}
\begin{eqnarray}\mbox{P (Blue and Red)} &=& \mbox{P (R)}
\times \mbox{P (B)} \\
&=& 1/4 \times3/3\\
&=& 3/12 \; \mbox{OR}\; 1/4 \end{eqnarray}
$$\text{P (Same)} = 1/2$$
$$\text{P (Different)} = 1/2$$
There is another way to even the game: Put 2 ribbons of one colour
and 2 of another, and they each have to get the same colour (Red or
Blue).
***
Amy of Kingwinford Secondary School suggested the following
modification to the game to make it fair. Is she right?
There is more than one way to make a fair game - one way is:
The two players pick 1 ribbon, and say 1 player picks a red ribbon
and the other picks a blue ribbon, then 1 point would get added on
to you score, because you have both picked 2 different colours, but
if they both picked the same colour then they would have points
taken off their score.
There is another way of playing the game but differently, you could
have 4 different coloured ribbons (red, blue, green and yellow) 2
of each colour then the players have to try and pick 2
ribbons of the same colour, if they do they win!
***
Finally, Andrei of School 205, Bucharest started looking for
general solutions to a fair game and found some interesting
patterns (below). Would anyone like to suggest why triangular
numbers are appearing? Has he identified all the possible fair
games? To solve the problem, I used combinations. The formula of
combinations, i.e numbers of groups of k objects taken from n
objects, where the order does not matter is: \par $${n\choose{r}} =
\frac{n!}{r!(n - r)!} $$ I calculated the number of pairs of
ribbons of the same colour. There are two red ribbons and I take
two: $${2\choose2}$$ There are four blue ribbons and I take two:
$${4\choose2}$$ The total is: $${2\choose2} + {4\choose2}$$ In
total, there are 6 ribbons, and I take 2. So, the total number of
groups of 2 ribbons that could be formed from 6 is $${6\choose2}$$
The probability that the two ribbons are of the same colour (that
Jo wins) is: $$\frac{{2\choose2} + {4\choose2}}{{6\choose2}} =
\frac{1 + 6}{15} = \frac{7}{15}$$ Now, the probability that the two
ribbons are of different colours (that Chris wins) is: $$1 -
\frac{7}{15} = \frac{8}{15}$$ As the two probabilities are not
equal, the game is not fair. \par Now, I try to find possibilities
to make the game fair. I made a small program in which I introduced
the two numbers, and it calculated the probability. The numbers I
obtained formed an unexpected pattern. In the following table, the
numbers in the first column represent the ribbons of the first
colour, the numbers in the second column the ribbons of the second
colour, while in the third column their difference is
calculated:
@
| $1$ |
$3$ |
$3-1=2$ |
| $3$ |
$6$ |
$6-3=3$ |
| $6$ |
$10$ |
$10-6=4$ |
| $10$ |
$15$ |
$15-10=5$ |
| $15$ |
$21$ |
$21-15=6$ |
| $21$ |
$28$ |
$28-21=7$ |
All the numbers are found on the third
line of Pascal`s triangle, so they are triangular numbers.