Probability and Counting
Probability and Life
- Genetics
- Finance
- History
- Games
- Fermat and Pascal Letters
- The problem of points, also called the problem of division of the stakes, is a classical problem in probability theory. One of the famous problems that motivated the beginnings of modern probability theory in the 17th century, it led Blaise Pascal to the first explicit reasoning about what today is known as an expected value.
- Fermat and Pascal Letters
- Game theory
- The study of mathematical models of strategic interaction among rational decision-makers.
What is a sample space S
?
The set of all possible outcomes of a (random) experiment.
What is an event?
A subset of the sample space.
What is probability?
- The very naive definition of $P_{(A)}$
- Number of favorable outcomes $A$ divided by all possible outcomes
- Assumptions:
- all outcomes are equally likely
- the sample space is finite
- The non-naive definition $P_{(A)}$ or Probability Axioms
- A function which takes an input, event $A$, a subset of sample space $S$,
- returns $P_{(A)} \in [0,1]$ as an output,
- such that
- the probability of the empty set $P_{\varnothing} = 0$
- An event that is impossible to happy
- the probability of the full set $P_{S} = 1$
- Certainty: a guaranteed event that always happens
- the probability of a countable infinite union equals to the sum of the probabilities of $A_1$,$A_2$,…, $A_n$ if they are disjoined (non-overlapping)
- $P_{(\bigcup_{i=1}^{\infty} A_{n})} = \sum_{n=1}^{\infty}P_{A_n}$
- the probability of the empty set $P_{\varnothing} = 0$
What are some basic principles of counting?
-
Multiplication rules:
- Independent events assuming
- $P_{(A and B)}=P_{(A)}⋅P_{(B)}$
- The general rule
- $P_{(A and B)}=P_{(A)}⋅P_{(B|A)}$
- If the events are independent, one happening doesn’t impact the probability of the other, and in that case $P_{(B|A)}=P_{(B)}$
- Independent events assuming
-
Binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ = $\frac{n(n-1)(n-2)(n-k+1)}{k!}$
- Example, what is the probability of having a full house card? (A full house has three cards of one kind and two of another)
- First you choose a type of card (13 choices), then you choose three out of four of those cards, then you choose a second type of card, and finally you choose two of those four cards.
- $P = \frac{\binom{13}{1} \cdot \binom{4}{3} \cdot \binom{12}{1 }\binom{4}{2}} {\binom{52}{5}} \approx 0.00144 $
How to choose $k$ objects out of $n$?
- Sampling table
Order matters | Order doesn’t matter | |
---|---|---|
Without replacement | S1: $n(n-1)…(n-k+1)$ | S3: $\binom{n}{k}$ |
With replacement | S2: $n^k$ | S4:$\binom{n+k-1}{k}$ |
- S1: The number of choices reduces every time
- S2: $n$ choices each time
- S3: $n$ choose $k$: think about how you choose the ice cream flavor and cones
- S4:
- Simple trivial cases
- $k = 0$:
- $P = \binom{n-1}{0} = 1$
- not choosing is also a choice
- $k = 1$:
- $P = \binom{n}{1} = n$
- you only choose once, there’s no difference whether there’s order or replacement
- $k = 0$:
- Simple non-trivial cases
- $n = 2$:
- $P = \binom{k+1}{k} = \binom{k+1}{1} = k + 1$
- put $k$ particles in two boxes
- $n = 2$:
- General cases
- how many ways are there to put $k$ indistinguishable particles in $n$ distinguishable boxes?
- convert the question to a “dot and separator” code
- $\binom{n+k-1}{k}$: $n+k-1$ positions, choose $k$ positions to put the dots
- the same as $\binom{n+k-1}{n-1}$: $n+k-1$ positions, choose $n-1$ positions to put the seperators
- e.g.
- $n= 4$, $k=6$
- $\cdot\cdot|\cdot|\cdot|\cdot\cdot$
- 6 dots and 3 separators = 9 positions
- choose 6 positions to put the dots
- the same as choose 3 positions to put the seperators
- Simple trivial cases
Ref
- Lecture 1: Probability and Counting | Statistics 110
- Financial Theory (ECON 251)
- An Introduction to Quantitative Finance by Stephen Blyth
Terms
In condensed matter physics, a Bose–Einstein condensate (BEC) is a state of matter (also called the fifth state of matter) which is typically formed when a gas of bosons at low densities is cooled to temperatures very close to absolute zero (-273.15 °C, -459.67 °F). Under such conditions, a large fraction of bosons occupy the lowest quantum state, at which point microscopic quantum mechanical phenomena, particularly wavefunction interference, become apparent macroscopically. A BEC is formed by cooling a gas of extremely low density (about one-hundred-thousandth (1/100,000) the density of normal air) to ultra-low temperatures. This state was first predicted, generally, in 1924–1925 by Albert Einstein following and crediting a pioneering paper by Satyendra Nath Bose on the new field now known as quantum statistics.
In quantum statistics, Bose–Einstein (B–E) statistics describe one of two possible ways in which a collection of non-interacting, indistinguishable particles may occupy a set of available discrete energy states at thermodynamic equilibrium.