Analyzing the bwin Bonus: Can You Beat the House?

I’m not a big fan of gambling, but I do like probability theory. With the European Championship 2024, I noticed a surge in betting ads, including one from bwin offering a 100% bonus for new customers. This piqued my curiosity: What’s the probability of beating the house with this bonus? Or put another way, what are the expected earnings for a customer using this bonus?

Important notes

1. Disclaimer: I don’t want to encourage gambling. It can become an addictive behavior and can lead to financial ruin.
2. Background: I’m neither in the gambling industry nor do I have any experience with it. Hence, some assumptions I make along the way, could be wrong.
3. Correctness: I don’t guarantee the correctness of the calculations. If you really want to base financial decisions on this, recheck the calculations.
4. Qualifications: I’m not a mathematician, and my last statistics course was a few years ago, so my notation and methods might not be perfect. Constructive criticism is welcome!

Note on the figures

Throughout the blog post, I will use the following diagram to visualize how odds and probabilities are related.

For people completely new to the concept of odds: The odds tell you how much you will win if you bet 1€. For example, if the odds are 1.7, you will win 1.7€ for every 1€ you bet. Hence, when betting on events with increasing odds, you will win more money. But of course, the probability of those events will be lower. In a fair game, the odds are the inverse of the probability of the event happening. This is visualized in the diagram above.

In the real world, the odds are not fair, since the betting company wants to make a profit. But there are many sport betting companies, and they are in competition with each other. There are also websites that compare the odds of different betting companies. So each betting company has to offer odds that are not too unfair.

Hence, their goal is to estimate the probability as good as possible and offer odds better than the competition, but still in a safe margin, such that they make a profit.

Rules

The rules are as follows:

• Deposit up to 100€ and receive a 100% bonus, giving a total of up to 200€ to bet.
• You must bet the full amount (deposit + bonus) at least three times to withdraw the bonus and its earnings.
• Bet must be placed on events with odds of at least 1.7.

There are some additional rules¹, but they are not relevant for the calculation.

Strategy

To maximize the bonus, we deposit 100€ and receive 100€ bonus.

Since we only have one shot, and we are trying to earn money, we will act very conservatively and only bet on events with odds of 1.7.²

So we will bet all 200€ three times on events with odds of 1.7. Specifically, we will not continue gambling after fulfilling the requirements, and we will not bet with our earnings from the previous bets.

Calculation

Fair Game

Let’s first assume the odds are fair.

If the odds are \(1.7\), the fair probability of the event happening would be \(p_f := 1/1.7 = 0.5882\).

This assumes, the betting company knows the probability and is not trying to make a profit. In my opinion, the first assumption is a fair, since the betting company probably has a lot of data and can estimate the probability of an event happening quite well. We will loosen the second assumption later.

To calculate the expected value:

$$ E[X] = \sum_i p_i * x_i $$

where \(p_i\) is the probability of event \(i\) happening and \(x_i\) is the payout of event \(i\).

Let’s define the event space and the payouts. I already removed the deposit, so positive values are winnings and negative values are losses:

• \(A\): We win all three bets. We get the bonus plus the earnings of all three bets: $$ \begin{aligned} x_A &= 100 + 3 * (200 * 0.7) = 100 + 3 * 140 = 520 \\ p_A &= 0.5882^3 = 0.2035 \end{aligned} $$
• \(B\): We win the first and second bet, but then lose the last. We get the bonus plus the earnings of the first two bets, but loose 200€: $$ \begin{aligned} x_B &= 100 + 2 * (200 * 0.7) - 200 = 240 \\ p_B &= 0.5882^2 * (1-0.5882) = 0.1424 \end{aligned} $$
• \(C\): We win the first, then lose the second bet. In this case we now accept our losses and take out what is left. We can only take out the winnings and money from our deposit, not the bonus money. See Bonus 1, to learn whether we can improve this strategy. $$ \begin{aligned} x_C &= 100 + 100 * 0.7 - 200 = -30 \\ p_C &= 0.5882 * (1-0.5882) = 0.2422 \end{aligned} $$
• \(D\): We lose the first bet, game over. We lose our deposit: $$ \begin{aligned} x_D &= 100 - 200 = -100 \\ p_D &= (1-0.5882) = 0.4118 \end{aligned} $$

Check if the probabilities sum up to 1: \(0.2035 + 0.1424 + 0.2422 + 0.4118 = 0.9999\) (close enough).

Now we can calculate the expected value:

$$ \begin{aligned} E[X] &= p_A * x_A + p_B * x_B + p_C * x_C + p_D * x_D \\ &= 0.2035 * 520 + 0.1424 * 240 + 0.2422 * (-30) + 0.4118 * (-100) \\ &= 105.82 + 34.176 - 7.266 - 41.18 \\ &= 91.55 \end{aligned} $$

So the expected value with fair odds playing three times with the bonus is 91.55€.

Sounds like a very good deal, but of course, the odds are not fair.

So let’s rig the game in favour of the betting company, to see if we can still win.

Let’s rig the game

The next obvious question is: How unfair can the odds be, such that we still win on average.

If we formulate this mathematically, we want to write the expected value as a function of the probability \(p\) and then find the root \(p_0\) of this function:

$$ \begin{aligned} E_p[X] &= p^3 * 520 + p^2 * (1-p) * 240 + p * (1-p) * (-30) + (1-p) * (-100) \\ &= 280 p^3 + 270 p^2 + 70 p - 100 \end{aligned} $$

Solving \(0 = 280 p^3 + 270 p^2 + 70 p - 100\) for \(p\), we find \(p_0 = 0.424\). So if \(p \le 0.424\) the odds stay at 1.7, we will stop earning money on average.

Where does this left us?

I don’t know anything about sports. So I can’t know if they are rigged so badly, that the real probability could lower than \(0.424\).

Or maybe we can?

Lower bound for the probability

We don’t have any insights about the exact probabilities, which the betting company uses to determine the odds.

But we can calculate the lower bound of the probability, since the betting company allows us to also bet on the complementary event. If the real probability would be lower than the 1 - inverse of the odds of the complementary event, then betting on the complementary event would be a good deal. And it is probably safe to assume, that the betting company doesn’t offer good deals (except maybe this bonus?).

For example:
In Dart only one player can win. Let’s assume the offered odds for player A are \(1.7\). The betting company also lets us bet on player B. The betting company could also offer odds of \(1.7\) for player B. That would mean, with neither bet we could more than double our money, But there must be at least one player that has a probability of winning of at least \(0.5\).

I looked up some odds of complementary events. If the odds for outcome A are \(1.7\), then the odds for outcome B are usually around \(2.05\).

Hence, we know that the betting company estimates the \(p_A \le 0.5882\) and \(p_B \le 0.4878\). Since the sum of the probabilities is 1, we know \(p_A \ge 1 - p_B = 1 - 0.4878 = 0.5121\) and therefore \(p_A \in [0.5152, 0.5882]\).

The estimated probability by the sport betting company is probably somewhere in the middle, since otherwise it would offer good odds for one of the outcomes.

So the lower bound for the probability is \(p_l := 0.5121\). And the probability in the center is \(p_c := (0.5882 + 0.5121) / 2 = 0.5502\).

For our goal to earn money this is good, since it is larger than the minimum probability of 0.424, where we stop earning money.

Let’s calculate the expected value for the lower bound and the center:

$$ E_{p_l}[X] = 280 p_l^3 + 270 p_l^2 + 70 p_l - 100 = 44.31 $$$$ E_{p_c}[X] = 280 p_c^3 + 270 p_c^2 + 70 p_c - 100 \\ = 66.88 $$

Conclusion

The expected winnings are somewhere between 44.31€ and 91.55€, probably around 66.88€.

Again, while after doing this calculation, it would be rational to do this, I don’t want to encourage gambling.

Also looking at the probabilities above in 65.4% of the cases you will lose money.

Furthermore, with the time you spend on betting, you could probably earn more money by working.

Bonus 1

What if we would deposit another 60€ after winning the first but losing the second bet?

Since we already bet 200€ twice, we only have to do one more bet, to fulfill the condition of placing 3 bets. So there are two outcomes:

• \(C_1\): We win the last bet. We got the bonus, but had to deposit another 60€. $$ x_{C_1} = 100 - 60 + 200 * 0.7 = 180 $$
• \(C_2\): We lose the last bet. We lose our initial deposit and the 60€. $$ x_{C_2} = 100 - 60 - 200 = -160 $$

Let’s ignore everything that happened before and just compare the expected value of this strategy \(X'\) to the strategy, where we just cut our losses, which results in a loss of -30 €. If \(E[X'] > - 30\), then we should rebuy after losing the second bet.

$$ \begin{aligned} E[X'] &= p_{C_1} * x_{C_1} + p_{C_2} * x_{C_2} \\ &= p * 180 + (1-p) * (-160) \end{aligned} $$

If we insert various probabilities for \(p\), we get:

• \(p_f\): \(E[X'] = 39.99\)
• \(p_c\): \(E[X'] = 27.07\)
• \(p_l\): \(E[X'] = 14.15\)

So this strategy is clearly better and we can (on average) turn our loss into a profit.

To update the expected value for the whole game with the new strategy \(X'\), we simply have to replace the value for \(x_C\) with the expected value of the rebuy strategy.

$$ \begin{aligned} E[X'] &= p^3 * 520 + p^2 * (1-p) * 240 + p * (1-p) * E[X'] + (1-p) * (-100) \\ \end{aligned} $$

For \(p = p_c\) we get \(E[X_{rebuy}] = 81.00\).

So the expected earnings of the game with the rebuy strategy is 81.00€. You will still lose money in 56.1% of the cases.

Bonus 2

There is another “FreeBet” bonus, where you get a 10€ free bet. The only requirement is to bet it once and you will get the earnings (excluding the 10€) So if you bet 10€ on an event with odds of 1.7 and win, you get 7€. If you lose, you don’t lose anything, so this slightly improves the expected values.

$$ \begin{aligned} E[X_{FreeBet}] &= p_c * 7 + (1-p_c) * 0 \\ &= 0.5611 * 7 = 3.77 \end{aligned} $$

Combining the expected earnings with the rebuy strategy und assumption of the center probability:

$$ \begin{aligned} E[X_{total}] &= E[X_{rebuy} + X_{FreeBet}] \\ &= E[X_{rebuy}] + E[X_{FreeBet}] \\ &= 81.00 + 3.77 = 84.77 \end{aligned} $$

So under assumption of the center probability, the expected earnings of the game with the rebuy strategy and the FreeBet bonus is 84.77€.

I hope you enjoyed this post and maybe also got nostalgic about your probability theory course in university. If you have any questions or suggestions, feel free to contact me.