Prediction Markets

Prediction markets trade in contracts that pay off according to whether some future event occurs. If the event occurs the contract pays off 1 unit, otherwise the contract becomes worthless. As a result, the price of a contract can be interpreted as the market's estimate of the probability that the future event will occur. Assume frictionless markets: one can buy (risking $r$ units to win $1-r$ units) and sell (risking $1-r$ units to win $r$ units) contracts at no cost.

QUESTIONS

1.) Individuals $A$ and $B$ think the probability that a future event will occur is $p$ and $q$ respectively. $A$ and $B$ have risk capital $X$ and $Y$. $A$ and $B$ invest so as to maximize the expected log of their capital. What price $r$ for the contract will assure that the sum of the demand for the contract from $A$ and $B$ is 0?

2.) A future event has $n$ possible outcomes each with associated contract $c_{i}$ which will pay off 1 unit if the $i$th possibility occurs while the other $n-1$ contracts will become worthless. Each contract has a market price $m_{i}$ and the market is efficient: $\sum_{i}^{n} m_{i}=1$.

You believe the true probability of the $i$th possibility occurring is $q_{i}$ for $i=1,...,n$ (where the $q_{i}$ also sum to 1). You have $X$ units to invest and you wish to do this in such a way that the expected value of the logarithm of your capital is maximized. How many of each of the $n$ contracts should you buy?

Poker Dice

In Poker Dice, the aim is to build the best possible poker hand, where straights and flushes don't count. The first person to roll has up to three throws of the dice and after each throw can put aside any dice they wish to use for their hand; also, they can stop after the first of second roll if they wish. After the first person establishes a number of rolls, each successive player may only throw the dice at most that many times. The player with the highest ranking hand after everyone has had a turn wins the game. Is there any way to gain an edge?

1-4-24 (Midnight)

How to play

Midnight is played with six dice where the objective is to get the highest score. Someone starts by rolling six 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 six dice have been kept. So at most you will roll six times.  For example, you can choose to keep anywhere from one to six dice on the first roll. If you kept two of them, then you'd re-roll the remaining four dice, and you can choose to keep one to four of them. And, so on...

The sum of the six selected dice becomes your score, with one caviat: you must qualify the hand by rolling a one and a four. Therefore a maximum score is twenty-four by rolling 1-4-6-6-6-6, and a minimum score is zero by not qualifying.

When beginning, the order is determined randomly. After a round has been played, the lowest score goes first and the highest score goes last. Clearly, going last has the advantage due to the ability to know when to simply just stop trying for a  higher score while taking on the risk of getting a lower one. Say the first person scores an 8. And, as the 2nd player, your very first roll is 1-4-6-5-1, you can simply stop right there and claim a win with a score of 12.

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 18, one guy scores 15, and another scores a 13. 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?

Rock Paper Scissors

The best game ever? If you don't know RPS, I feel bad for you. What is the optimal way to play? Is there a way to get an advantage? I think so...

Optimal Proportional Play

Question

You are at a craps table next to a gambler looking for some action. Both the table and the die are fair. She offers you the following "even money" proposition: starting with the next roll (where the sum of the die's faces are tallied) bet whether two consecutive 7s or a 12 is rolled first. You can pick either side and the size of the bet.


What side of the proposition would you take and how would you wager?

Optimization

Question

You are bidding for an asset that you can immediately sell for 1.5 times what you pay for it but don't know exactly what the seller believes it is worth. You are told the seller's value can be accurately modeled as a random variable uniformly distributed between 375 and 1000. You can only make one bid, and the seller will only trade if your bid is higher than what they believe the asset is worth. What is your bid?

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?

Russian Roulette (2)

...please see the post Russian Roulette to catch up...


Obviously, the further down the chain your turn is, the better. But how much better? An interesting exercise is to create tables showing your edge against the field for two to five people taking turns in 1st to 5th place when everyone is 1) on equal footing and are forced to spin and 2) where one player uses the strategic advantage against all the others...

Russian Roulette

I recently acquired the iRevolver app for my iPhone. It has done a wonderful job settling disputes and providing gambling action. Most recently, I used the app to settle a round of drinks with friends. Usually we play credit-card roulette, but this was a bit more novel. Here was the specifics of our game:

The revolver has six chambers. I loaded one bullet. Four of us randomly selected    the order of the game and each person has a choice when it is was his/her turn: 1) pull the trigger or 2) spin the cylinder and then pull the trigger.

I was the second player, "P2", and lost on turn number 6. Everybody spun the cylinder when it was his/her turn. Naturally, since our game allowed a strategic decision to be made (whether to spin or not spin the cylinder) I was curious whether one could use this decision to their advantage for future rounds.

Common Birthdays

This data from the New York Times caught my attention the other day regarding "a ranking based on how many babies were born in the United States on that date between 1973 and 1999. Jan. 1".

The only problem is that it is not easily summarizeable. Hence this is a great example where data visualization techniques can help transform a complex data set into a simple story. I don't use Windows often, but here is where Microsoft Excel does a fantastic job creating a fast and easy heat map via conditional formatting:

The simple story: parents in the US from 1973 to 1999 made babies in the winter.