0

Brazil are currently playing Mexico, and at the start of the game Brazil were 2/5 to win. As it's the 38th minute and still 0-0, their odds have changed to 8/15.

Now, if I'm not wrong that represents implied probabilities of 71.4% and 65.2% respectively.

What I'm wondering, is whether there's an easy way of working out which odds are better without having to convert them into percentages?

Dan
  • 293

4 Answers4

2

Odds against of a to b is the same as a/b to 1 So 2-5 is 0.4 and 8-15 is 0.53 or just notice it is > 0.5 so it is larger than 0.4 . Higher "odds against" implies a lower prob of winning.

Mr.Spot
  • 2,797
1

To compare two fractions $\frac ab$ and $\frac cd$, you can compute $ad-bc$. If it is greater than zero, $\frac ab \gt \frac cd$. If it is less, $\frac ab \lt \frac cd$ You are really just putting them over a common denominator.

Ross Millikan
  • 374,822
  • Perfect! Knew I was missing something simple... Slowly getting my high school maths back in my head! Thanks Ross! – Dan Jun 17 '14 at 19:49
  • So, in this example: (2x15)-(5x8) = 30 - 40 = -10 Therefore our second fraction is larger, thus implying a lower probability of winning. – Dan Jun 17 '14 at 19:54
  • Your conversion to percents implies that Brazil being $2/5$ to win means their chance to win is $5/(2+5)$. At least that gives $71.4%$ That is separate from comparing fractions. It looks like you are quoting that Brazil is $5/2$ times as likely as its opponent to win. – Ross Millikan Jun 17 '14 at 19:58
  • Okay, but would it imply that the odds have indeed lengthened as the resulting -10 lets us know the second odds are greater than the first? – Dan Jun 17 '14 at 20:00
  • It tells you that $\frac 25 \lt \frac 8{15}$. What you want to compare are $\frac 57$ and $\frac {15}{23}$ and this works fine for that, too. If the conversion is as I think, that odds of $a/b$ means $b/(a+b)$ chance of winning, the lower odds (viewed as a fraction) are always the higher chance of winning. – Ross Millikan Jun 17 '14 at 20:18
  • It sucks that you can't mark two answers correct... – Dan Jun 17 '14 at 20:20
  • To prove that, say we are comparing quoted odds of $a/b$ and $c/d$. We are given $ad -bc \lt 0$ As fractions we are comparing $\frac b{a+b}$ and $\frac d{c+d}$, so we compute $b(c+d)$ and $d(a+b)$ and find $bc + bd \gt ad + bd$ – Ross Millikan Jun 17 '14 at 20:21
0

I'm not sure if this is what you mean, but if you convert them to fractions, which does not involve a calculator, you get $5/7$ and $15/23$ respectively. If you multiply $5/7$ by $3/3$, you get $15/21$, which is greater than $15/23$, and so better odds.

Avi
  • 1,780
  • 11
  • 21
0

Or, try a "double division": $$\frac{\frac{2}{5}}{\frac{8}{15}}=\frac{15(2)}{8(5)}=\frac{30}{40}<1 $$

user99680
  • 6,708