creating a fair game with counters

this fascinating probability task originates with (as far as I am aware) and is described in detail
by Dr Natalya Vinogradova at Plymouth State University (New Hampshire, USA)
a pdf version of her article

consider various totals of counters
with just two colours: red and yellow (hers are red and blue)

can the (without replacement) probability of getting two different colours ever be the same as the probability of getting two of the same colours?

i.e. can P(YR or RY) ever = P(YY or RR)?

a suggested approach to this task starts with a total of 6 counters

a ppt is here

Natalya suggests that it is helpful for students to initially undertake a practical version of the task

maybe comparing the results for different number pairings (with a total of 6 altogether, initially at least)


an analysis of these results, for three of each colour (3R, 3Y), could involve a simple 'branching' diagram

from such a diagram, below,

are you more likely to obtain

• two of the same colour
• two different colours?

what makes you say that?

a more normal tree diagram shows the unfairness of this game (18 : 12)
with it being more likely to obtain one of each colour

what happens for other colour splits with a total of 6 counters?
consider three options:
(3 , 3) (4 , 2) and (5 , 1)
[their complements will have the same probabilities as these]

more likely to win with a 5 : 1 colour split

certain to win if they are all the same colour (6 , 0) ...

a (4 , 2) colour split is the fairest option (very nearly fair, 16 : 14) but is not completely fair

for any even number of counters with an equal split of colours, a proof that you are more likely to obtain two different colours than two colours that are the same:

try a different number of counters to see if a fair game can be created

this works!

so, it seems, do other pairs of consecutive
triangular numbers