Expected Value Calculation of Lotto Max
Posted in Commentary on June 3rd, 2010 by Sacha PeterI notice that the new national lottery, Lotto Max, has a June 4, 2010 draw for a $50M jackpot and five separate $1M draws.
You have to choose 7 numbers from 1 to 49. The actual odds of getting all 7 numbers is (7!/(49!/42!)) or approximately 1 chance in 85,900,584. (For those of you that don’t know the mathematical notation, “!” means “factorial”, which means 7! is 7*6*5*4*3*2*1 = 5040).
Each $5 ticket enables you to pick three draws of 7 numbers, so the precise odds of winning with one $5 ticket is the above odds multiplied by three, which turns out to be one in 28,633,528.
So your expected value out of a $5 ticket is $50M / 28,633,528 + 5*($1M / 28,633,528) = $1.92/ticket, at least with the main prizes and most importantly, not assuming any splits of prizes (i.e. somebody else having a winning ticket). The expected value of the 6 plus bonus, 6 and 5 number draws, given past performance, is approximately $1.34, so the value of an entire ticket is about $3.26 with a large main prize. In other words, you are sacrificing about 35% of your equity for a chance of winning a million dollars (or more).
Your chance of getting any prize worth $1M or above (which is generally what people want, rather than a high expected value) is one in 4,772,254, again assuming no splits.
The Lotto 6/49, by comparison, has far superior expected value, but slightly less probability of winning a $1M+ prize, assuming you buy 2.5 tickets (which would given a chance of winning of 5,593,526 to one).
Finally, it should be stated that 1 in 5,000,000 is an exceedingly unlikely probability. To give you an idea how unlikely it is, if you played a 5,000,000-to-one lottery continuously once per second, it would take you 38.2 days in order to have a 50% chance of winning at least one big prize.
It’s probably not worth $5 to discover this.