You meet a nice man on the street offering you a game of dice. For a wager of just 2 $, you can win 8 $ when the dice shows a six. Sounds good? Let’s say you join in and play 30 rounds. What will be your expected balance after that?

You roll a six with the probability p = 1/6. So of the 30 rounds, you can expect to win 1/6 · 30 = 5, resulting in a pay-out of 40 $. But winning 5 rounds of course also means that you lost the remaining 25 rounds, resulting in a loss of 50 $. Your expected balance after 30 rounds is thus -10 $. Or in other words: for the player this game results in a loss of 1/3 $ per round.

Let’s make a general formula for just this case. We are offered a game which we win with a probability of p. The pay-out in case of victory is P, the wager is W. We play this game for a number of n rounds.

The expected number of wins is p·n, so the total pay-out will be: p·n·P. The expected number of losses is (1-p)·n, so we will most likely lose this amount of money: (1-p)·n·W.

Now we can set up the formula for the balance. We simply subtract the losses from the pay-out. But while we’re at it, let’s divide both sides by n to get the balance per round. It already includes all the information we need and requires one less variable.

B = p · P – (1-p) · W

This is what we can expect to win (or lose) per round. Let’s check it by using the above example. We had the winning chance p = 1/6, the pay-out P = 8 $ and the wager W = 2 $. So from the formula we get this balance per round:

B = 1/6 · 8 $ – 5/6 · 2 $ = – 1/3 $ per round

Just as we expected. Let’s try another example. I’ll offer you a dice game. If you roll two six in a row, you get P = 175 $. The wager is W = 5 $. Quite the deal, isn’t it? Let’s see. Rolling two six in a row occurs with a probability of p = 1/36. So the expected balance per round is:

B = 1/36 · 175 $ – 35/36 · 5 $ = 0 $ per round

I offered you a truly fair game. No one can be expected to lose in the long run. Of course if we only play a few rounds, somebody will win and somebody will lose.

It’s helpful to understand this balance as being sound for a large number of rounds but rather fragile in case of playing only a few rounds. Casinos are host to thousands of rounds per day and thus can predict their gains quite accurately from the balance per round. After a lot of rounds, all the random streaks and significant one-time events hardly impact the total balance anymore. The real balance will converge to the theoretical balance more and more as the number of rounds grows. This is mathematically proven by the Law of Large Numbers. Assuming finite variance, the proof can be done elegantly using Chebyshev’s Inequality.

The convergence can be easily demonstrated using a computer simulation. We will let the computer, equipped with random numbers, run our dice game for 2000 rounds. After each round the computer calculates the balance per round so far. The below picture shows the difference between the simulated balance per round and our theoretical result of – 1/3 $ per round.

(Liked the excerpt? Get the book “Statistical Snacks” by Metin Bektas here: http://www.amazon.com/Statistical-Snacks-ebook/dp/B00DWJZ9Z2)

This should be corrected or taken down. This appears on the first SEO page for google on the topic & incorrectly calculates the expected value. This site has probably caused a lot of confusion on the topic.

The expected loss per round is -2/3, not -1/3.

“You roll a six with the probability p = 1/6. So of the 30 rounds, you can expect to win 1/6 · 30 = 5, resulting in a pay-out of 40 $. But winning 5 rounds of course also means that you lost the remaining 25 rounds, resulting in a loss of 50 $.”

The author needs to decide whether to use gross or net payouts. If we consider the pay-out of 30 rounds as $40 (aka assuming gross payouts), then the loss is NOT $50. Instead, we lost $2 per round regardless of whether or not we won, thus the total loss is $60. Thus, your expected balance is -$20 after 30 rounds; said differently, -2/3 loss per game.