# Odds: Introduction

Lets say we have a bag con­tain­ing twice as many blue mar­bles as red mar­bles. Then, if you reach in with­out look­ing and pick out a mar­ble at ran­dom, the odds are 2 : 1 in fa­vor of draw­ing a blue mar­ble as op­posed to a red one.

Odds ex­press rel­a­tive quan­tities. 2 : 1 odds are the same as 4 : 2 odds are the same as 600 : 300 odds. For ex­am­ple, if the bag con­tains 1 red mar­ble and 2 blue mar­bles, or 2 red mar­bles and 4 blue mar­bles, then your chance of pul­ling out a red mar­ble is the same in both cases:

In other words, given odds of $$(x : y)$$ we can scale it by any pos­i­tive num­ber $$\alpha$$ to get equiv­a­lent odds of $$(\alpha x : \alpha y).$$

# Con­vert­ing odds to probabilities

If there were also green mar­bles, the rel­a­tive odds for red ver­sus blue would still be (1 : 2), but the prob­a­bil­ity of draw­ing a red mar­ble would be lower.

If red, blue, and green are the only kinds of mar­bles in the bag, then we can turn odds of $$(r : b : g)$$ into prob­a­bil­ities $$(p_r : p_b : p_g)$$ that say the prob­a­bil­ity of draw­ing each kind of mar­ble. Be­cause red, blue, and green are the only pos­si­bil­ities, $$p_r + p_g + p_b$$ must equal 1, so $$(p_r : p_b : p_g)$$ must be odds equiv­a­lent to $$(r : b : g)$$ but “nor­mal­ized” such that it sums to one. For ex­am­ple, $$(1 : 2 : 1)$$ would nor­mal­ize to $$\frac{1}{4} : \frac{2}{4} : \frac{1}{4},$$ which are the prob­a­bil­ities of draw­ing a red /​ blue /​ green mar­ble (re­spec­tively) from the bag on the right above.

Note that if red and blue are not the only pos­si­bil­ities, then it doesn’t make sense to con­vert the odds $$(r : b)$$ of red vs blue into a prob­a­bil­ity. For ex­am­ple, if there are 100 green mar­bles, one red mar­ble, and two blue mar­bles, then the odds of red vs blue are 1 : 2, but the prob­a­bil­ity of draw­ing a red mar­ble is much lower than 1/​3! Odds can only be con­verted into prob­a­bil­ities if its terms are mu­tu­ally ex­clu­sive and ex­haus­tive.

Imag­ine a for­est with some sick trees and some healthy trees, where the odds of a tree be­ing sick (as op­posed to heathy) are (2 : 3), and ev­ery tree is ei­ther sick or healthy (there are no in-be­tween states). Then the prob­a­bil­ity of ran­domly pick­ing a sick tree from among all trees is 2 /​ 5, be­cause 2 out of ev­ery (2 + 3) trees is sick.

In gen­eral, the op­er­a­tion we’re do­ing here is tak­ing rel­a­tive odds like $$(a : b : c \ldots)$$ and di­vid­ing each term by the sum $$(a + b + c \ldots)$$ to pro­duce $$\left(\frac{a}{a + b + c \ldots} : \frac{b}{a + b + c \ldots} : \frac{c}{a + b + c \ldots}\ldots\right)$$\$ Di­vid­ing each term by the sum of all terms gives us an equiv­a­lent set of odds (be­cause each el­e­ment is di­vided by the same amount) whose terms sum to 1.

This pro­cess of di­vid­ing a set of odds by the sum of its terms to get a set of prob­a­bil­ities that sum to 1 is called nor­mal­iza­tion.

# Con­vert­ing prob­a­bil­ities to odds

Let’s say we have two events R and B, which might be things like “I draw a red mar­ble” and “I draw a blue mar­ble.” Say $$\mathbb P(R) = \frac{1}{4}$$ and $$\mathbb P(B) = \frac{1}{2}.$$ What are the odds of R vs B?$$\mathbb P(R) : \mathbb P(B) = \left(\frac{1}{4} : \frac{1}{2}\right),$$ of course.

Equiv­a­lently, we can take the odds $$\left(\frac{\mathbb P(R)}{\mathbb P(B)} : 1\right)$$, be­cause $$\frac{\mathbb P(R)}{\mathbb P(B)}$$ is how many more times likely R is than B. In this ex­am­ple, $$\frac{\mathbb P(R)}{\mathbb P(B)} = \frac{1}{2},$$ be­cause R is half as likely as B. Some­times, the quan­tity $$\frac{\mathbb P(R)}{\mathbb P(B)}$$ is called the “odds ra­tio of R vs B,” in which case it is un­der­stood that the odds for R vs B are $$\left(\frac{\mathbb P(R)}{\mathbb P(B)} : 1\right).$$

# Odds to ratios

When there are only two terms $$x$$ and $$y$$ in a set of odds, the odds can be writ­ten as a ra­tio $$\frac{x}{y}.$$ The odds ra­tio $$\frac{x}{y}$$ refers to the odds $$(x : y),$$ or, equiv­a­lently, $$\left(\frac{x}{y} : 1\right).$$

Parents:

• Odds

Odds ex­press a rel­a­tive prob­a­bil­ity.

• The $$x/y$$ no­ta­tion is con­fus­ing—these ra­tios aren’t prob­a­bil­ities are they?