Problem of the Week Day 3: Week of 8/20/12 - 8/24/12

Today is day three of the problem of the week. Unlike normal Wednesdays, I have decided to give a few probability questions. The easy one you should be able to figure out with some simple math, and the hard one is explained in a previous post on the math behind a card trick.

Easy: You and your friend decide to play a game where you take a b card deck and deal down a card. On each turn, you both bet a dollar out of the two twenties that you brought with you. If the card dealt is a face card, you win the money and if it is a number card, your friend wins the money. Whoever runs out of money first loses.

Before the game, you rigged the deck so that y% of the cards are face cards and x% of the cards are number cards, making it much easier for you to win. Based on this cheat, determine how many turns it will take before your friend runs out of money.

t =

If you can't figure out a way to solve this problem, you probably have the values of x and y wrong. If you do, go back and make sure you have checked for dominant strategies, and then a mixed strategy equilibrium.

Hard: First off, solve the following problem:

2g - s = k
k =

Your friend bets you f dollars that you can't win a game he made up, which you accept. The game is that he takes a k card deck with numbers from 1 to s written on them. You will randomly guess a number from 1 to s, and then see if the next card has that number on it. If you fall into a conscious pattern, your friend automatically gets the money. Your goal is to get through the whole deck and never guess a card correctly. Determine the odds that your friend will get your money and you will therefore lose. Round to the nearest full percent.

l = ___%

This problem seems pretty daunting, but we did go over it in a blog post (not a problem of the week). If you can find this post and your calculator, the problem won't be too difficult. Good luck.