I. Basic math.
 1 Conditional probability.
 A. Definition of conditional probability.
 B. A bomb on a plane.
 C. Dealing a pair in the "hold' em" poker.
 D. Monty-Hall problem.
 E. Two headed coin drawn from a bin of fair coins.
 F. Randomly unfair coin.
 G. Recursive Bayesian calculation.
 H. Birthday problem.
 I. Backward induction.
 J. Conditional expectation. Filtration. Flow of information. Stopping time.
 2 Normal distribution.
 3 Brownian motion.
 4 Poisson process.
 5 Ito integral.
 6 Ito calculus.
 7 Change of measure.
 8 Girsanov's theorem.
 9 Forward Kolmogorov's equation.
 10 Backward Kolmogorov's equation.
 11 Optimal control, Bellman equation, Dynamic programming.
 II. Pricing and Hedging.
 III. Explicit techniques.
 IV. Data Analysis.
 V. Implementation tools.
 VI. Basic Math II.
 VII. Implementation tools II.
 VIII. Bibliography
 Notation. Index. Contents.

## Monty-Hall problem.

n a TV show a prize is hidden behind one of three doors (the first, the second and the third). A player wants to guess that door. She randomly points at one of the doors (suppose this happens to be the third door). Then the host, who knows where the prize is, opens one of the other doors that does not have a prize behind it (let it be the second door). Then the player may stay with her choice (the third door) or switch to the other door (the first door) that remains closed. Should she switch?

The intuitive answer would be as follows. Suppose the player uses the switching strategy. She originally points to the right door with the probability and this is the only case when she loses. If the player uses the staying strategy then she loses with the probability if she originally points to the wrong door. Therefore, the switching strategy has twice better chances of success.

Let us consider the problem with more rigor. Suppose the third door is originally picked and the second door is opened. What is the conditional probability that the first door has the prize behind it? We introduce the notation We would like to calculate the probability Prob . Note that all probabilities of the form Prob are easily computable. Hence, we use the idea of the formula ( Inversion_remark ). According to the ( Bayes formula ) Using the ( Bayes formula ) again we calculate the numerator Using the independence of the prize location and the door picking we continue We calculate denominator using the ( Total probability rule ) with Prize: Again, by the independence of the prize location and the door picking we write

Note that we substituted Prob This means that the host opens a door at random when he has a choice.

Therefore, Hence, because the last two probabilities are complimentary. This brings us to the same answer as in the former non rigorous argument provided that the host acts randomly.

 Notation. Index. Contents.