Probability: how many blocks needed to make odds even?

Two friends X and Y are fighting over doing laundry. X being smarter than Y put two red blocks and two green blocks in a bag. Both will take turn and choose the blocks out of the bag. X put the condition that if there comes out same coloured blocks, X will do the laundry and if comes out different colored Y will do the laundry.
The questions is :

1. How many blocks should be in a bag to make odds even.

2. What are the combinations that make the game fair ?

I tried this and figure out that odds are 2:1 but my both questions are still unanswered? Please help

Re: Probability problem

Hello, Ashiya!

Two friends X and Y are fighting over doing laundry.
X, being smarter than Y, put two red blocks and two green blocks in a bag.
Each will randomly draw a block from the bag.
X put the condition that: if they are the same color, X will do the laundry
and they have different colors, Y will do the laundry.

Click to expand...

This is a well-known hustle . . . but it seems that Y hasn't heard of it.

X can convince Y of the "fairness" of the arrangement like this:

[A] There are only two cases: (1) the color match, (2) the colors do not match.
. . .\(\displaystyle P(\text{diff.colors}) \:=\: \tfrac{1}{2}\)

There are four outcomes: RR, RG, GR, GG.
. . .The color are different in two of them.
. . .\(\displaystyle P(\text{diff.colors}) \:=\:\tfrac{2}{4}\:=\:\tfrac{1}{2}\)


The Truth:

There are: .\(\displaystyle _4C_2 \:=\:6\) possible outcomes.

Same color
Either both Reds are drawn or both Greens are drawn.
. . \(\displaystyle (_2C_2) + (_2C_2) \:=\:1 + 1 \:=\:2\text{ ways.}\)

Different colors
One of the Reds and one of the Greens is drawn.
. . \(\displaystyle (_2C_1)(_2C_1) \:=\:2\cdot2 \:=\:4\text{ ways.}\)

\(\displaystyle \text{Therefore: }\;\begin{array}{ccc}P(\text{same color}) &=& \frac{2}{6} \:=\:\frac{1}{3} \\ \\[-3mm] P(\text{diff.colors}) &=& \frac{4}{6} \:=\:\frac{2}{3} \end{array}\)

1. How many blocks should be in a bag to make odds even.

Click to expand...

Surprisingly, there must be more of one color than the other.
For example: 3 Reds and 1 Green.

There are: .\(\displaystyle _4C_2 \:=\:6\) possible outcomes.

Same color: both red
. . \(\displaystyle _3C_2 \:=\:3\text{ ways.}\)

Diff. colors: one Red, one Green
. . \(\displaystyle (_3C_1)(_1C_1) \:=\:3\cdot1 \:=\:3\text{ ways.}\)

2. What are the combinations that make the game fair ?

Click to expand...

This was trickier to work out ... and it took some creative guesswork at the end.

Suppose there are \(\displaystyle r\) Red blocks and \(\displaystyle g\) Green blocks.

Same color: both Red or both Green
. . \(\displaystyle (_rC_2) + (_gC_2) \:=\:\frac{r(r-1)}{2} + \frac{g(g-1)}{2}\) .[1]

Diff. colors: one Red and one Green
.\(\displaystyle (_rC_1)(_gC_1) \:=\:r\cdot g\) .[2]


We want [1] and [2] to be equal.

. . \(\displaystyle \frac{r(r-1)}{2} + \frac{g(g-1)}{2} \:=\:rg \quad\Rightarrow\quad r^2 - r + g^2 - g \:=\:2rg\)

. . \(\displaystyle r^2 - 2rg + g^2 \:=\:r + g \quad\Rightarrow\quad (r-g)^2 \:=\:r+g\)


By inspection, we find these pairs: .\(\displaystyle (3,1),\;(6,3),\;(10,6),\;(15,10),\;\hdots\)
. . These are consecutive Triangular Numbers.


\(\displaystyle \text{Therefore: }\;\begin{array}{ccc}r &=& \frac{n(n+1)}{2} \\ g &=& \frac{n(n-1)}{2} \end{array}\quad\text{for any integer }n \ge 2\)