Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If you bet on the result of a soccer match, you usually have three possibilities. You bet

  • 1 - if you think the home team will win
  • X - if you think the match ends in a draw
  • 2 - if you think the away team will win

Lets say we have the following soccer match with the following betting quotes:

Kansas City vs. Portland - 1 = 1.92, X = 3.57, 2 = 5.00

This means: If you think Portland (In the opinion of the bookie, the underdog) will win the match, you bet on 2

Example (I bet \$100): In case Portland wins I win \$400 $$100*5.00-100 = 400$$ $$stake*quote-stake = net win$$ (When Portland loses the match, or it ends in a draw, I'll lose my stake)

Now, some bookies offer a so-called double chance bet. This kind of bet takes one possibility out. That leaves you to following bets. You bet

  • 1/X - if you think the home team will win or the match ends in a draw
  • X/2 - if you think the away team will win or the match ends in a draw

This variant is perfect if you think Portland will win the match, or at least it will end up in a draw. To calculate this quotes I use the following formula: (Q1 = 1st quote 1, Q2 = 2nd quote)

$$ 1/(1/Q1+1/Q2) $$

For the 1/X bet $$ 1/(1/1.92+1/3.57) = 1.25 $$

For the X/2 bet $$ 1/(1/3.57+1/5) = 2.08 $$

Now comes my math problem: When the bookie does not offer a double chance bet, I want to create it my self: With two single bets. For the Kansas City vs. Portland bet I'd like to place a X/2 bet. The quote for the bet is as I showed before 2.08. I want to place \$100 on it. When I win the bet, I'll get \$108 net win:

$$100*2.08-100 = 108$$

How do I have to split the money on two (X and 2) single bets, to win \$108, when Portland wins or the match ends in a draw?

I got to the solution for this case by trying out. But with the result in my hand, I still don't get the formula to calculate it.

I bet \$58.35 on X and \$41.65 on 2

$$ 58.35*3.57-58.53-41.65 ≈ 108$$ and $$ 41.65*5.00-41.65-58.53 ≈ 108$$

Notice the last subtraction. You have to subtract the stake of the other bet. Because when Portland wins, I win only the 2 bet and lose the stake for the X bet.

share|improve this question
add comment

4 Answers 4

up vote 3 down vote accepted

Let's examine the X/2 case. Denote by $Q_X=3.57$ the bookie's quote for X and $Q_2=5.00$ for 2. Denote by $Q=1/(1/Q_X+1/Q_2)=(Q_X Q_2)/(Q_X+Q_2)\approx 2.0828$ the quote you have calculated for the X/2 bet. You want to split the total bet $B=\$100$ into two bets $B_X$ (for X) and $B_2$ (for 2) so that $B_X Q_X=BQ=B_2 Q_2$. From the first equation you get $B_X=B(Q/Q_X)$. Similarly from the second equation you get $B_2=B(Q/Q_2)$, or alternatively $B_2=B-B_X$ (as the value of $B_1$ is already known). Let's substitute the values: $$B_X=100(2.0828/3.57)\approx 58.34\quad\text{and}\quad B_2=100-58.34=41.66.$$ In fact, you can do this more easily without calculating $Q$ at all, since $$B_X=B\frac{Q}{Q_X}=B\frac{Q_2}{Q_X+Q_2}.$$

share|improve this answer
add comment

The defining feature of a double chance bet is that, if either of the events you bet on happens, you win the same amount regardless of which event it was.

To simulate a double chance bet with single bets, you need to divide the stake so that the same will happen.

So, let $Q_1$ and $Q_2$ be the quotes offered for the two events. We seek a value $0 \le \alpha \le 1$ such that, if we bet a fraction $\alpha$ of our total stake on event 1 and the rest on event 2, the payout in either case will be the same, i.e.

$$\alpha Q_1 = (1 - \alpha) Q_2.$$

To solve this, expand the right hand side, collect the $\alpha$ terms together on one side and divide to get

$$\alpha = \frac{Q_2}{Q_1 + Q_2}.$$

Then, to ensure equal payout in either case, you should bet $\alpha$ times you total stake on event 1, and the rest on event 2.

share|improve this answer
add comment

Planning on betting that either team wins?

$S_1*Q_1-S_1-S_2=S_2*Q_2-S_1-S_2$

$S_1=S_2*\frac{Q_2}{Q_1}, S_2=S_1*\frac{Q_1}{Q_2}$

$S_1+S_2=1$ (to find $S_1$ and $S_2$ as percentages)

$S_1+S_1\frac{Q_1}{Q_2}=1, S_2+S_2\frac{Q_2}{Q_1}$

$S_1=\frac1{Q_2/Q_1+1}, S_2=\frac1{Q_1/Q_2+1}$

For X/2, S_1 = 58.3% just as above.

share|improve this answer
    
If you think it isn't a tie with these odds, put 72.2% on 1 and 27.7% on 2 for +38.7% on a win. Remember that you never win a double-chance if $\frac1{1/Q_1+1/Q_2}<=1$. –  user474632 Aug 17 '11 at 23:50
add comment

You bet $\$100$ $Q2 / (Q1 + Q2)$ on X and $\$100$ $Q1 / (Q1 + Q2)$ on 2. To the nearest cent, these work out at $\$58.34$ and $\$41.66$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.