I would like a quick and simple way to understand how to work
out odds. I looked at the National Lottery but this is easier to
work out due to the fact that the balls have to come out! How do
the bookies work out what odds to put on a horse for example? Even
if something is 12-1, what does 5-4 mean?
Please clear up my maths blindness when it comes to gambling
odds!
Thanks.
Quick answer because I have to go off somewhere: 12-1 means that if you wager £1 they will give you £12 (in addition to your £1 back) if you win the bet. 5-4 means that if you wager £4 you get £5 back (and your £4 wager) if you win. If the odds were fair (which they're not because the bookies want to make a profit) this would mean that a gamble with odds of 12-1 would have a probability of 1/13=1/(12+1) of winning. Odds of 5-4 would have a probability of 4/9=4/(4+5) of winning.
I believe that quoting odds of "12-1" is
really a shorthand for "12-1 odds-against". So you wager 1 and get
12 if you win (plus original stake of 1).
Now it could be (if the event were very probable) that you were
quoted odds of "1-12" (or equivalently "1-12 odds-against"). Then
you would wager 12 and get 1 (plus original stake of 12). Under
such circumstances, it would be more usual to quote it as "12-1
odds-on".
To summarise:
"12-1" : Wager 1, get 12+1 if you win
"12-1 odds against" : Wager 1, get 12+1 if you win
"12-1 odds on" : wager 12, get 1+12 if you win.
I hope this is what you wanted. If instead, you are interested in
how the bookie decides what odds to quote in the face of his
extensive knowledge of a horse and two punters wanting to place
opposing bets, then write back and I'll give you an example.
Andre
12-1 on (or 12-1 odds on) is occassionally
called 1-12. This makes life logically a lot easier.
ie. if the odds are m-n and you bet £n, if you win, you get
£(m+n) back - £m profit plus your original
stake.
I was daydreaming about this question
yesterday, reflecting on how bookmakers alter the odds that they
quote as they receive more bets for or against an event.
I’m not a gambler, but I don’t think you see strange
looking quotes like “101 to 10, odds against”. Suppose
we postulate that a bookie has a rule in his mind that the sum of
the two numbers in his quote (101 and 10 in the case) cannot exceed
a certain number (so as not to confuse the punter). What do the
resulting probabilities look like?
First, if we have a quote of “a to b” this implies a
probability of the event of b/(a+b)
Now lets take an example: suppose the limit is set at, say, 6.
Possible quotes are:
0 to 6; 1 to 5; 2 to 4; 3 to 3 etc.
0 to 5; 1 to 4; 2 to 3 etc
0 to 4; etc
etc.
Note that I am allowing zero in either a or b for completeness.
These equate to probability quotes of:
6/6, 5/6, 4/6, 3/6, 2/6, 1/6, 0/6
5/5, 4/5, 3/5, 2/5, 1/5, 0/5
4/4, 3/4, 2/4, 1/4, 0/4
etc
Now gather these all up, remove duplicates, cancel common factors
and sort them. We get
0/1, 1/6, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 5/6,
1/1
Believe it or not, these series go by a name: Farey Series.
They have some interesting properties. For example, pick any three
consecutive terms, e.g. “3/5, 2/3, 3/4". Form a fraction
composed of the sum of the numerators divided by the sum of the
denominators of the first and third term ( (3+3)/(5+4) in this
case). This is the same as the term in the middle. Take another
triplet: 4/5,5/6,1/1. As before, (4+1)/(5+1)=5/6, the middle
term!
Just to be clear, Farey series are properties of fractions (whose
numerator cannot exceed a given number) but the bookies is possibly
the nearest to a practical example of their use.
Andre
Of course, bookmakers regularly offer odds
of 100-30, so large numbers sometimes occur (but only for a few
examples).
Incidentally, in France, a bookmaker's job is a lot easier - French
odds are always offered as x to 1, where x is a decimal to at most
one decimal place, x>0.