Bayesian Probability

Question

Dave flips a coin with a 60% (40%) probability of landing heads (tails). If the coin lands heads, Kevin chooses a random variable $X$ from a uniform probability distribution $u(x)$ in the range $-1 \leq X \leq 1$. If the coin lands tails, Kevin chooses $X$ from the same range but from a probability density of $p(x)=\frac{1}{2}+\frac{x}{5}$. Kevin tells you $X=\frac{1}{2}$. 


What probability can you assign to Dave's coin having landed heads?

Discrete Probablity

Question

Dave flips a fair coin until he gets two consecutive heads and defines $X$ to be the number of flips it takes; Kevin flips another fair coin until he gets three consecutive Tails and defines $Y$ to be the number of flips it takes. What is $P(X>Y)$?

Diamonds

I recently got engaged, i.e., I gave in to the tradition of spending an absurd amount of money on a rare rock. As much as I don't like the concept, I love my fiancee more and spent the last 2 months shopping and building a ring. Let's just say she's happy with the final product.


Along the way, I learned a lot about this process and figured I'd post some of the interesting things I found.

Calculating Expectation

Question

You are running a casino with the following game:


The player rolls a fair die until a 4, 5 or 6 is thrown. For every number 1, 2 or 3 that is thrown the player's score increases by 1. If the game stops with a 4 or 5 the player is paid the accumulated score. If the game stops with a 6 the player is paid nothing.


How much do you charge patrons to play this game?

Mixed Equilibrium

Question

Consider a simple card game between you and a single opponent using only six cards from a standard 52 card deck {Ac, Ad, Ah, As, 5c, 5d}: 


Each player is provided one red and one black ace. You are given a red five and your opponent is given a black five. Each player picks a card ignorant of the other's choice and have a showdown. You win if colors match; you lose if colors differ. The amount won is the numerical value of the winner (where aces are equal to one). If the two 5s are shown, the result is a draw.


How would you play this game?

Continuous Probability

Question

The Cumulative Density Function of a random variable $X$ is $CDF(x)=c*e^{-2*x}$ if $0 \leq X < \infty$ and is $0$ otherwise. What is $P(X \geq 2)$?

Threes

How to play

Threes is played with five dice where the objective is to get the lowest score. Someone starts by rolling five dice. On each roll, you are required to keep at least one of your dice, and you re-roll the remaining ones. You do this until all five dice have been kept. So at most you will roll five times.  For example, you can choose to keep anywhere from one to five dice on the first roll. If you kept two of them, then you'd re-roll the remaining three dice, and you can choose to keep one to three of them. And, so on...

The sum of the five selected dice becomes your score, with one caviat: threes are worth zero. Therefore a minimum score is zero  by rolling 3-3-3-3-3, and a maximum score is thirty  by rolling 6-6-6-6-6.

When beginning, the order is determined randomly. After a round has been played, the highest score goes first and the lowest score goes last. Clearly, going last has the advantage due to the ability to know when to simply just stop trying for a lower score while taking on the risk of getting a higher one. Say the first person scores a 8. And, as the 2nd player, your very first roll is 3-3-2-2-2, you can simply stop right there and claim a win with a score of 6.

For a three, four, or more player game, only a single distinct winner can win the pot. So, say in a four player game, two people score 6, one guy scores 10, and another scores a 9. Since there is no distinct winner due to two best scores tying, there is no winner this round. All players ante up again, building the pot. All players are still in the game, and therefore, are eligible to win. You play until there is a single distinct winner, who scoops.

Is there an optimal strategy for this game? If so, what is it?