We Got A Problem #13: The Birthday Problem

At my son’s recent 5th birthday party, he got 15 birthday cards.  Sure enough, he got more than one of the same card.

Suppose everyone buys their cards at Foyerjohn, and randomly picks one of the cards.  How many different cards would need to be on sale for there to be a 50% chance that all 15 people pick a distinct card?

Harder: how many different cards would need to be on sale for there to be a 50% chance that my son doesn’t get three of the same card?

We Got A Problem #13: Deal or No Deal

This short game is played with ten playing cards: an ace, two, three, …, nine, and a joker.  I shuffle the cards and lay them face down.

Your decision: grab as many cards as you want, still face down. When you’re ready, flip them all over at once and win: $1 multiplied by the sum of the card values.

Except the joker. If you flip over the joker, you win nothing. Good day.

So, how many cards should you take to maximize your expected return? What would change if all nine non-joker cards were aces? Eight aces and a ten?

We Got A Problem #12: Tile Shufflin’

Online word games, such as Scrabble, now let you “shuffle” your tiles with the touch of a button.  As far as I can tell, when you click the “shuffle” button the 7 tiles are randomly rearranged into any of the 7! = 5040 different orderings.

And there’s an animation of the tiles moving.  Well… sort of.  Because, sometimes one or more of the tiles doesn’t move, because they are in the same position before and after the shuffling.

How does the probability of having one stuck tile compare to the probability of having no stuck tiles? How does the probability of having one stuck tile compare to the probability of having two stuck tiles? Three? Four?

A Scrabble variant called Lexulous gives players eight tiles instead of seven.  What happens then?

What is the approximate probability of having no stuck tiles?

We Got A Problem #11: Rock Rock Paper Scissors

Most people know Rock-Paper-Scissors (or its more fun cousin, Ninja-Cowboy-Bear).  Kate and Ogden are locked in a long battle of online Rock-Paper-Scissors, with a winner scoring 1 point, and a tie scoring nothing for either player.  Game’s to 100.

Ogden won the first series 100-77 by constantly looking for patterns in Kate’s choices. For the second series, Kate tries something interesting: she has no pattern.  She randomly selects rock, paper, or scissors with equal probability.  She even tells Ogden that she is doing this, and he can do nothing about it but sigh and accept that he cannot beat this “strategy”.  Kate wins the second series 100-95 by luck.

For the third series, Ogden changes a rule: if rock beats scissors, the winner scores 2 points instead of 1.  Other victories (scissors over paper, paper over rock) are still worth 1.

Now Kate’s panicked: if she still randomly selects, is she beatable in the long run, or not?  Can she alter the probabilities to make a new “strategy” that Ogden can’t beat?  It seems like it would be bad to play scissors so often, and she should probably play more rocks… or not?

Let us know whether you find any ways that Kate can play to force Ogden into equilibrium.

We Got A Problem #10: Cliff Hangers!

My wife was recently playing the Cliff Hangers slot machine, based on a great game from The Price Is Right.  Here’s how the bonus game works:

The goal is to climb exactly to the 25th and final step on the mountain (while yodeling, of course).  You keep taking turns, picking from 5 doors, one each with the numbers 3, 5, 10, 15, and 20 on them.  Whatever pick you make, that’s the number of steps taken on that turn.  The doors are randomized, so you don’t know what you’re going to get, and you can get the same number more than once.

If after any number of turns your total is exactly 25 steps, you win.  If you go over 25, you lose.

There are clearly some ways to win the game with lucky picking, such as 5 + 20 or 10 + 10 + 5.  My wife won even though her first pick was a 3.

What is the probability of hitting 25 and winning this game?  Which of these three goals gives the greatest chance of winning: 18, 55, or 58?  If there were no stopping point, what is the long-term probability of hitting a high number like 507?

We’ll provide a solution next week that involves a single mathematical calculation.  It’s pretty awesome.

We Got A Problem #9: Yearly Multiplication

This Saturday (December 1, 2012) gets written as a short date as 12/1/12 (or, outside the US, generally 1/12/12).  Either way, it’s interesting because 12 x 1 = 12.  That’s happened a lot this year, including June 2.

1. How many times in a century will this happen? It happened once in 2001, twice in 2002.

2. In what year(s) will this happen the most times?  Are there years when it won’t happen?

3. Days like 12/12/44 or 8/25/00 sort of work: 12 x 12 = 144, which you’d write as 44 in a two-digit year.  If you include these days too, how many times in a century does it happen?  Cooool.

We Got A Problem #8: Super Ultra Mega Man!

The new toy craze is Mega Men, where kids buy a Mega Man in a box. The box won’t tell you which one it is, and kids open it up when they get home in hopes of getting one they don’t already have. (Sadly, eBay is not an option, since the only cool Mega Men are the ones still in their original packaging.)

There are ten toys in all, each equally likely when you buy a box. If you collect all ten, you can make Super Ultra Mega Man!

On average, how many Mega Men will you have to buy for your kid before he can finally collect them all?

(Stolen from PCMI 2007 Day 1 Problem #11: http://mathforum.org/pcmi/hstp/sum2007/morning/bowen/day1handout.pdf)