Stats, Probability & Games – A Presentation for Draper University

This is a recent math workshop I did for the students of Draper University in San Mateo, California.  It presents basic probability theory in a fun and accessible light.

Draper University is a two-month long intensive program for aspiring entrepreneurs.  Throughout the course, students work on small teams to complete challenges designed to build creativity, problem-solving skills, and business sense.

When I came, students were preparing for a carnival, where each team was designing a booth.  Teams were competing against each other in a winner-take-all model: whichever team made the most profit would get to keep all profit from all teams.

I wanted students to understand basic probability and outcome analysis in order to optimize their pricing and payout strategies at the carnival.

I interspersed Challenge Problems throughout the presentation.  Teams who submitted solutions to these problems were eligible for extra points on the program-long team scoreboard.

We reviewed the difference between probability and odds — two words which are often used interchangeably, but have different mathematical meanings.

I have always wondered if the concept of odds was invented by casinos and lotteries to make their games more attractive.  Instead of comparing the winning event to all possible outcomes, they can compare it only to the losing outcomes, making it seem more likely.  Doesn’t “2 to 1 odds” sound better than “0.67 probability”?

We brought mathematical precision to two commonly used phrases:

And then played a huge rock-paper-scissor tournament.

Some people thought the probability of winning a rock-paper-scissors game would be 1/3, since there are 3 choices in every game.  However, there are only 2 choices that end the game.  This can be illustrated with a tree of all possible outcomes:

Note that there are 9 possible outcomes, but only 6 possible outcomes that end the game, three of which are losing outcomes, and three of which are winning outcomes.

I recently read Delivering Happiness, by Zappos founder and CEO Tony Hsieh.  Hsieh writes about the story of Zappos, as well as the research he did on human happiness.  He notes that one of the main factors contributing to happiness is perceived control.  Not necessarily control, but perceived control.  Rock-paper-scissors is a perfect illustration of this.  Why else would we prefer it over a coin toss?  We believe that by choosing between rock, paper, and scissors, we somehow have control over the outcome.

Another factor that contributes to happiness, according to Hsieh’s research, is human connection.  Since rock-paper-scissors involves synchronization, some eye contact, and body language, students remarked that it creates more opportunities for laughter, creativity, and human connection.  For instance, many players like to act out the outcome: if I played paper and you played rock, I might wrap my hand around yours to assert my victory.

The bottom line is, there is more than mere numerical probability in these activities.  By maximizing the psychological “payouts” of your game, you can increase its appeal without lowering your profits.

One team ended up utilizing this at the carnival by making a dunk tank: zero monetary payout, but a huge experiential reward — watching someone get soaking wet due to your throw.

The winner of our tournament won five games in a row.  We found that the probability of winning five games in a row is (1/2)^5, or 1/32.  That’s about 3%.

Do you agree?

Can you make a general formula for the probability of winning n games in a row?

(Hint: make a tree showing all the possible outcomes after 2 games, 3 games, 4 games, etc)

And now a new game to consider:

At first people said, “Charge 5 million dollars!”

Yes, if we charged 5 million dollars, we would make huge a profit every time someone played.  But no one would want to play.

So what is the least we could charge, while still making a profit?

The class was divided.  Some people said \$3.01; others said \$3.51.  Others were unsure.

This is was our doorway into discussing expected value.

In the example with a fair dice, the expected value is the same as the average, because all the outcomes are equally likely: all have a probability of 1/6.  If they had different probabilities we’d have to multiply the value of each outcome by its unique probability.  We’ll see an example of this later on.

Expected value has endless applications but we focused on its application to carnival games and casino-style games.

We used a numerical example to familiarize ourselves with process, and then generalized:

Here’s a simulation of how the expected profit per player is affected by the price to play, amount of payout, and probability of winning.  Click here to edit this spreadsheet and see how the results change.

Now, what if you had multiple levels of payouts?

No one knew how to calculate the expected profit per player.  Why?  We don’t know any of the probabilities!

Suppose we did know the probabilities of hitting each zone.

Now we are able to calculate the expected profit per player (assuming these probabilities are legitimate) (are they?  where did they come from?).

To find the expected payout per player, we multiplied each payout by its likelihood, and added them together.  We expect to payout \$3 one-third of the time.  We expect to payout \$6 one-tenth of the time.  We expect to payout \$9 one-twentieth of the time.  Thus,

expected payout per player = \$3*(1/3) + \$6*(1/10) + \$9*(1/20).

This game ends up being a loss for the house.  The expected payout is 5 cents more than the price to play.  This means that on average, we are losing 5 cents for every person who plays.

How can we improve this?

We did another simulation.  To get more accurate probabilities, we built a real-life model of the game and each of us took a shot.  Surprisingly, no one was able to hit the green zone.  Hitting the blue zone was pretty easy but many people aimed for the green and yellow zones and instead got nothing at all.

Here are the results of our simulation.  We were emboldened to increase the green payout to \$50 because no one was hitting it.

Here is a spreadsheet where you can simulate it yourself.  The red numbers are the ones you have access to change.  The black numbers are calculated automatically from the red numbers. Click here to access and modify the spreadsheet.

I spent the next 5 minutes letting teams work on the task below and answering questions.

That’s when we ran out of time.

Check out the bonus questions!

You’ll find that, without autocorrect, the probability of accidentally texting “I love you” is very, very low.  The probability of getting just the “I” would be around 1/50, because there are roughly 50 characters (26 letters + punctuation) on a typical texting keyboard.  By the time we get to “I” with a space afterwards, we’re already down to 1/2500.  That’s less than 0.04%!

However, with autocorrect, it’s a different story.  You could push something like “Umkocdeyi” and suddenly you’ve notified your boss that you love her.

#MathProblemOfTheDay: Interesting Combination Lock

I saw this combination lock at a Duane Reade.  It has only four options: you move the knob up, down, left, or right. They advertise that it’s easier to operate, especially if you need to open it in the dark.

However, it also seems easier for someone else to crack the code.

Assuming a 4-step combination code, how many tries would it take a stranger–at most–to crack the code? (I didn’t read exactly how many steps the code is, I just blindly assumed it would be 4). Assuming it would take 5 seconds to try a given combination, how many minutes, at most, would it take someone to bust open your lock?

You could also consider the following problem: suppose there is no fixed length of the combination. You can make any combination between 1 step long and 4 steps long. In that case, how many possible combos would exist?

Also: see what happens if the combination is 5 steps long.  Or 6 steps long.  The number of possible combos is shooting up.

Can you make a formula for the number of possible combos, if the combination is n steps long?