One of the things I do for my clients is I help them design games, typically gambling games. Usually what will happen is that a client has a game idea in mind and is looking to figure out what the payouts and probabilities should be so that it’s a game where the “house wins”, but is still fun enough for players to try their luck. In an overly simplistic way, these games are nothing more than a variation on flipping coins. Some games have memory features, some games don’t. Some involve dice, some involve cards, some involve wheels, some involve coins, but typically all of them are in the world of discrete probability. Coming up with good game design is not easy — the ideas may be easy, but getting the right balance of pays and probabilities is a bit of art and a bit of math.

Here’s a fun little game I came up with a long time ago as a segue I sometimes use to introduce probability theory and to address some persistent misconceptions of what 50-50 odds mean, independent vs dependent events, and conditional probability.

Here’s how the game works. You can use this with or without the gambling backdrop — just change dollars to points and ask who can get the most points [no wager]. You can play the game here or you can see some examples below.

For this entire game I assume that you have the ability, either virtually or physically, to flip a fair coin [a coin where there is a 50% chance of getting “heads”, and a 50% chance of getting “tails”].

### Gambling Version

A player pays $1 to flip two coins. Upon flipping both coins, he records the number of heads and the number of tails observed and adds this total, respectively, to the number of heads and the number of tails observed on previous flips in the game. Thus, if the player has paid their first dollar to initiate the game, then the previous number of heads and tails observed are both zero. After flipping both coins and recording the result, the player can elect to do exactly one of the following: (a) get paid out and end the game or, (b) flip two more coins or, (c) end the game.

A player can get paid out if the total number of heads observed

equalsthe total number of tails observed. In this case, the payout equals (in dollars) the total number of heads (\(N\)), squared. Otherwise, the payout is zero.If the player elects to flip two more coins, then the game is said to have continued and previous possible payouts are no longer valid.

Here are some fun things. If we have an infinite balance, then we’ll visit a 50-50 split infinitely often. But what if we have a finite balance? Will we eventually go bankrupt or is this game generally in favor of the player?

There are a bunch of good strategy questions. When do you want to hit the reset button? When do you want to keep the count rolling up?

And there are some good probability questions. What’s the likelihood that the next set of flips will produce an equal number of heads and tails given that you already have an equal number of heads and tails? What if the difference were 2? What’s the probability that the difference is odd?

Here are a few examples of games.

- Flip two coins (cost $1) and observe, HT. Payout, if elected, is $1 since \(N^{2} = 1\).
- The player chooses to get paid out.
- The player received $1 and paid $1. The game is over. The net return is zero.
- This scenario happens 50% of the time.

- Flip two coins (cost $1) and observe, HT. Payout, if elected, is $1 since \(N^{2} = 1\).
- The player chooses to forgo the payout and continue play.
- Flip two more coins (cost $1) and observe, HT, again. Payout, if elected is $4 since \(N^{2} = 4\) [the one H from the first set of flips and the second H from the second set of flips and the number of heads observed thus far equals the number of tails]. The total spent so far is $2.
- The player chooses to get paid out and receives $4 and while having paid $2. The game is over and the player made $2.

Clearly, your payout is zero if you have an odd number of coin flips. But the interesting thing about this game is that if you have an even split of heads and tails, then the probability that you will maintain an even split in two more coin flips is 50%. But this is materially different from finding the probability of having an even split after some even number of coin flips.

In other words, **if** I have flipped a coin 8 times and have observed 4 heads and 4 tails, **then** the probability that I will observe 5 heads and 5 tails is 50%. The operative word here is “if”. That’s conditional probability. But what is the probability before any number of coins have been flipped that I would observe 5 heads and 5 tails in 10 coin flips? That’s a whole ‘nother problem involving the binomial distribution.

As a classroom exercise, if the class truly believes [and they often do before any formal introductions to the topic] that getting 5 heads and 5 tails happens 50% of the time, then it’s a matter of asking each student, individually to conduct this experiment. If true, they should observe roughly half the students achieving this result and half the students not achieving this result. But this is unlikely to happen!

And what of the game? Well, when do you want to quit? Would you play this game? Why or why not?

Thank you for reading! I want to keep in touch with my readers. If you are interested, click here to sign up!

Your Thoughts Are Welcome. Leave A Comment!