The odds of a bet never reflect exactly the probability of the event: the difference is the house’s guaranteed profit. This example makes it quantitative.

Example — Odds 1,5:1 and the house margin

A betting house offers odds of 1,5:11,5:1 on the event BB (“Zverev beats Medvedev”): betting 11 EUR and winning, you receive 1,51,5 (of which 0,50,5 EUR is net winnings and 11 EUR is the return of the stake). Suppose P(B)=0,6P(B)=0,6.

I bet 100100 EUR. If I win I pocket 5050 EUR net, if I lose I lose 100100 EUR. Vˉ=0,650+0,4(100)=3040=10 EUR.\bar V = 0,6\cdot 50 + 0,4\cdot(-100) = 30 - 40 = -10 \text{ EUR}. The expected value is negative: on average I lose 1010 EUR for every 100100 EUR bet, that is, the house keeps a margin of 10%10\%.

Fair odds. For the game to be fair we would need Vˉ=0\bar V=0, that is 0,6x+0,4(100)=00,6\cdot x + 0,4\cdot(-100)=0, from which x=1000,4/0,666,6x = 100\cdot 0,4/0,6 \approx 66,\overline{6} EUR. The fair pricing would therefore be about 1,66:11,66:1 (and not 1,5:11,5:1).

Topics: Probability
Concepts: Fair game · Expected value
Methods: Fair game · Expected value
Skills: Probability calculation · Modelling