# Fair Bet

What is a fair bet? The fair bet is such a bet whereas the expected return (or value) of the bet is null (or neutral) for all parties involved. There are usually two parties: a player and a casino, a player and a betting company, alternatively two friends who make a bet between each other. The meaning of the fair bet is clearly illustrated by the examples below.

The expected return of the bet can be calculated on the basis of the so called Expected Value (EV). Instead of the word (expected) “return” we can use the synonyms such as “value”, “profit” or “loss” etc. However as the fair bet is neutral, which means that no party, in the long run, can gain profit or suffer loss, it may be more appropriate to speak of the zero expected return or zero expected value. That is just a brief introduction to the terminology and the meaning of the fair bet.

The fair bet is the one whereas the expected value equals zero (EV = 0). From the mathematical point of view the expected value is a weighted average where the profit and loss is weighted by the probability of winning and losing. The profit means a net payout of the wager, the loss is an amount wagered.

We can also put it in this way: the long-term average return of the fair bet is null. In the long run no party gains or losses. However, in the short run, that is a small number of games, rounds or attempts etc., there can be profit of one side and loss of another, but the longer the period the more it will settle down to the average outcome, which is zero.

Let us demonstrate the relationship between probability and payout and let us see how a low probability can be compensated for by a higher potential payout and vice versa.

Let us imagine we play the following four games together. If I manage to meet the conditions in the row “Win”, then you will pay me out by the amount shown in the column “Profit/Loss”. If I fail to do so, I will pay you \$1. The profit and loss is always from my point of view, however they are set to always be a fair bet.

## Game 1: Rolling a Dice – “Low” vs “High”

The first game uses an ordinary dice and in terms of the probability of winning and losing it is perfectly symmetric. If I roll 1-2-3 (any low number), you will give me `\$1`. If 4-5-6 (any high number) is rolled, then I will pay you \$1 (from my point of view it is a loss, thus `-\$1`).

For this game it is clear that we have the same odds of winning and losing. As for me the odds of winning, that is a low number of the dice 1-2-3, is `3/6 = 0.5` and the odds of losing are calculated in the same manner.

Here and in other examples too, if we have two possibilities only, that is either a win or a loss, we can also stipulate the probability of losing indirectly: one minus the probability of winning as the sum of the probabilities must equal one (or 100% if you like).

Rolling a Dice Outcome Probability Profit/Loss EV
Win 1-2-3 0.5 \$1 \$0.5
Loss 4-5-6 0.5 -\$1 -\$0.5
Total x 1 x \$0

`EV = 0.5 × \$1 + 0.5 × (-\$1) = \$0.5 - \$0.5 = \$0`.

The expected value formula means: the probability of winning times the possible win (the net payout) + the probability of losing times the eventual loss (therefore the minus sign).

We can see that the expected value or, in other words the average expected return, is null or neutral for both of us. It is possible that after a few rolls some of us will be profiting and the other losing, but in the long run it will tend to the average, thus we both will have a zero balance. The fair bet gives no edge to anyone. Let us try other games with asymmetric probabilities and payouts.

## Game 2: Still a Dice, But I Got to Roll a Six to Win

The second game still uses a single dice to determine the result. It does not matter who of us rolls the dice, but of course we assume that the dice is perfectly balanced and none of us cheats.

In this game I got to roll a six to succeed. If I fail to do so, that is I roll one of the numbers 1 to 5, I will have to pay you \$1 as always. The probability of rolling a six is `1/6`, the probability of the opposite result is `5/6` (or `1 - 1/6 = 5/6`, the indirect method). So far so clear. I have smaller odds of winning so the question remains how high the payout must be to compensate for.

Rolling a Dice Outcome Probability Profit/LossEV
Win 6 1/6 = 0.1667 \$5 \$0.8333
Loss 1-2-3-4-5 5/6 = 0.8333 -\$1 -\$0.8333
Total x6/6 = 1x \$0

Fair payout or fair odds (the same methodology e.g. applies for sportsbook) are a reciprocal of probability. The probability of rolling a six is `1/6`, the reciprocal is `1 ÷ (1/6) = 6`. Therefore I should be paid out at 6-times the bet (`\$6`) – that is the fair ratio corresponding to the probability. The probability of your winning is `5/6`, while the reciprocal is `1 ÷ (5/6) = 1.2`. That is why if I do not roll a six, then I should pay you 1 dollar and 20 cents.

These fair payouts may be put into ratio or reduced as follows: `6 ÷ 1.2 = 5`, which can be laid out in this way: `6 ÷ 1.2 = 5` and `1.2 ÷ 1.2 = 1`, and thus `5 to 1`. That is how we arrive at the condition above that if I fail to meet the conditions of the example games, then I will always pay you `\$1` (from my point of view it is a loss, thus `-\$1`).

In the table above we can see the prove that the payouts are fair for both parties as the average outcome (of both of us) is null.

`EV = 1/6 × \$5 + 5/6 × (-\$1) = \$0`.

In this game (and in the next games even more) the outcomes may vary a bit more in the short run. It is likely for me to lose the `\$1` more frequently (1-2-3-4-5 play against me), but there and then it will be compensated for by rolling a six and getting paid `\$5`. In the long run, or after many rolls, none of us would be neither profitable nor losing as the average return is null again (`EV = 0`).

## Game 3: Two Dice – I Got to Roll the Total of 2 or 12

Now let us play a game with two dice, there will be more possibilities. My goal is to reach the total of 2 or 12. If I fail to do so, then, as always, I will pay \$1 for each unsuccessful attempt.

Rolling the total of 2 or 12 is the hardest (the least probable). There is only one possibility to reach the total of 2 (1 & 1) and as well as only one way to roll the total of 12 (6 & 6). The total number of combinations with two dice is `6 × 6 = 36`. The probability of rolling 2 or 12 is `2/36 = 1/18`. The probability against that is `1 - (2/36) = 34/36 = 17/18`.

Rolling 2 Dice Outcome Probability Profit/LossEV
Win 2 or 12 2/36 = 1/18 = 0,0556 \$17 \$0,9444
Loss 3 to 11 1 - 2/36 = 34/36 = 17/18 = 0,9444 -\$1 -\$0,9444
Totalx1x \$0

What is now the fair payout to compensate for the lower probability so that the bet is fair?

The fair payout is again the reciprocal of the probability: `1 ÷ (1/18) = 18`. In case of winning I should be paid \$18. On the other side I should pay you `1 ÷ 17/18 = \$1.058823529` (calculated by Excel, not rounded). If we put the payout into a ratio then it is `18 ÷ 1.058823529 = 17` or `17:1`.

As compared to the previous game it is even more likely that I would be losing the \$1 even more frequently. It is obvious that rolling the total of 2 or 12 is less likely than rolling a six on a single dice. On the other hand my efforts would be compensated by the payout of \$17 and again noone would gain or lose in the long term (EV = 0).

`EV = 1/18 × \$17 + 17/18 × (-\$1) = \$0`.

## Game 4: Single Number Bet in Roulette

At last let us play a classical casino game – Roulette, more precisely French Roulette that has 37 numbers altogether (0 to 36). I will pick up a single number and I must catch it. If some other number is spun, then I will pay you \$1. The probability of hitting the single number is `1/37` only. On the contrary 36 out of 37 numbers play against me (or in your favor).

Single numbe bet in Roulette Outcome Probability Profit/LossEV
Win Hitting the number 1/37 = 0.0270 \$36 \$0.9730
Loss Hitting other numbers 36/37 = 0.9730 -\$1 -\$0.9730
Totalx1x \$0

In this game, or in terms of this single number wager in Roulette, my winning chances are even smaller than as in the previous three games. It is clear that the payout should be higher to compensate for this low probability and for the bet to be fair (EV = 0).

The procedure is already known to us, we will use the reciprocal of the probability: `1 ÷ (1/37) = 37`. If I hit the number, then I should be given `\$37`. If I don’t, then I should pay you `1 ÷ (36/37) = \$1.027777778`. Putting these into a ratio it is `37 ÷ 1.027777778 = 36` or `36:1`.

`EV = 1/37 × \$36 + 36/37 × (-\$1) = \$0`.

As for the single number bet in Roulette it is even more likely for me to keep losing \$1 more frequently than in the previous three games. But if I win, I will be given \$36 and it will cover some of the losses or create a safety cushion for the coming losses. However, in the long-term average, we both will be at zero. Even though we played a million rounds (spins), theoretically none of us should gain/lose.

Roulette is a real casino game, therefore we can calculate our real average return. Casinos are profit-driven and thus do not pay neutrally. In case of the single number bet you get paid at `35:1` only instead of 36:1, which gives house the edge (or margin) of 2.7%:

`EV = 1/37 × 35 + 36/37 × (-1) = -0.0270 = -2.7%` (after multiplying by 100).

## EV = 0 for All Games, But They Are Not The Same

We could have seen that all four games offer balanced chances for both parties, that is they are neutral from the long-term prospective. The expected average return is null and thus no party gains or losses in the long run. We have always put emphasis on the word “long-term” and also indicated that in the short run there can be smaller or bigger fluctuations. One party can be temporarily lucky, the other unlucky.

One does not need to be a specialist in mathematics or statistics to feel that the bigger the difference between the payouts (it increased gradually from game 1 to 4) the greater the short-term deviations from the expected profit/loss of both parties. This is measured by a statistics called the variance. It measures how far a set of numbers (profits and losses) are spread around the expected value (or mean value or long-term average).

We recommend a well-commented example of variance. The meaning and practical usage of variance will be clear from that. From the variance (VAR) we can derive a so called standard deviation (SD). The standard deviation can help us determine intervals (or borders) in which the numbers (or profits and losses in our case) can be found with a certain or demanded probability.

Suppose we play 100 rounds of each of the four games. The following table shows the spreads of the profits and losses with the approximate 99% probability (it is actually 99.73% which approaches the certainty 100%). Don’t be afraid of these crazy numbers or science. We will show how we arrive at these numbers step-by-step and explain their meaning. It is very useful.

Game GoalEV 99.73% Min 99.73% Max
Game 1 – A single dice roll Roll 1-2-3 \$0 -\$30 \$30
Game 2 – A single dice roll Roll a 6 \$0 -\$67 \$67
Game 3 – Two dice roll Roll 2 or 12 \$0 -\$124 \$124
Game 4 – Single bet in Roulette Hit a single number \$0 -\$180 \$180

All has been calculated by Excel and the procedure of calculations is described under the hyperlink above. But alright, let us calculate e.g. the second game where I roll a single dice and must get a six to win. Let us remind that if I roll a six, the probability is 1/6 and I get \$5, and if I roll any other number, the probability is 5/6 and I lose a dollar.

### Calculations for One Round Only

The expected return or value (EV) of one round is zero:

`EV = 1/6 × \$5 + 5/6 × -\$1 = \$0`.

Now let us calculate the variance (VAR):

`VAR = 1/6 × (\$5 - \$0)2 + 5/6 × (-\$1 - \$0)2 = \$5`.

The standard deviation (SD) is a square root of the VAR:

`SD = square root of 5 = \$2.236067977`.

### Calculations for 100 Rounds

The expected return or value of 100 rounds is the number of rounds times the expected value of one round. The result is zero again as any number times zero equals zero. The fair bet is always neutral, no matter how many rounds are played:

`EV = \$0 × 100 rounds = \$0`.

The variance of 100 rounds is calculated in a similar manner – the number of round times the variance of one round:

`VAR = \$5 × 100 rounds = \$500`.

The standard deviation of 100 rounds:

` SD = square root of 500 = \$22.36067977` (Excel, no rounding).

Now we can use the so called THREE SIGMA rule which says that about 99% (precisely 99.73%) of all values are found in the interval EV minus/plus three standard deviations. “Minus” determines the lower border of the interval, while “Plus” determines the upper border of the interval:

`Lower interval (min) = EV - 3 × SD = \$0 - 3 × \$22.36067977 = approx. -\$67`.

`Upper interval (max) = EV + 3 × SD = \$0 + 3 × \$22.36067977 = approx. +\$67`.

### And What is the Conclusion?

If we play each other 100 rounds (rolls) in the Game 2 – whereas I must roll a six to win \$5, otherwise I pay \$1 – then according to theoretical expectation none of use will either gain or lose. The balance of both of us shall be neutral (null). That is sort of the mean expected result, which the long-term playing shall convert to.

However, in the short run (= a small number of rounds played) and with the probability of 99.73% (that is almost reaching the certainty) it is possible that one's profits and other's losses or vice verca could fluctuate in the interval `<-\$67, +\$67>`. It means that if one of us were extremely lucky (the other one extremely unlucky) he could win (lose) up to \$67. This is what the probability admits and it could be considered normal, but of course, with a growing number of rounds played the balance shall approach \$0.

The bigger the variance, the greater fluctuations can occur. For the Game 1 we stated that it is symmetric and there we can expect the smallest short-term deviations, precisely -\$30 to +\$30.

The bigger the differences in probabilities and payouts, the greater the deviations, as measured by variance, can be. For the Game 3 with two dice and the need to roll the total of either 2 or 12 the short-term balance can range from -\$124 to +\$124.

In the Game 4 – the single number bet in Roulette – there is the biggest variance of possible short-term results that can be found in the interval -\$180 to +\$180. But again, let us highlight that the balance of each of the four games should approximate zero. The bets in all four games are fair, thus no party gains or losses a thing in the long run.

## Why a Casino Cannot Offer a Fair Bet

Casinos and other betting companies are profit-driven. In order to be profitable they must secure some edge right from the beginning. Their business model is based on the positive expected value. The players are paid out less than what corresponds to the probability (the fair bet). The expected value (or return) of the players is negative. In the long run they will lose a percentage of their bets (see the single number bet in Roulette above) and of course it holds true that a player's loss is a casino's gain (and vice versa, but the probability is on the casino's side).

The casino has the mathematical edge over the player, it expects a positive long-term average return. It is not precise to speak of a profit but a margin (the profit is a positive difference between revenues and costs). In the gambling industry there is a plenty of games and particular wagers, whereas the house edge can differ substantially.

If a casino paid a fair bet, it would work for a while as a charity before it would go bankrupt. The expected value of the fair bet is null (EV = 0 or 0%), so the margin is 0%. The casino would not earn a dime in the long run (no edge over the players), while it would be left with all the costs and expenditures such as rents, wages, security etc. The revenues from betting would be zero and financing of such operation could not be called otherwise than a charity.