Probability: how many blocks needed to make odds even?

[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. 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

 

[soroban]

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.

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.

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 ?

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.


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