So let me give you three example games.

1st game is the mega millions. You pay $2 for a chance to win $250 million. You have a 1 in 250 million chance of winning.

2nd game is state lotto. You pay $1 for a chance to win 5 million. You have a 1 in 13 million chance of winning.

3rd game is the pick 4. You pick 4 numbers for $1 and get $5000 if you win. You have a 1 in 10,000 chance of winning. Obviously this is by far the most probable game to win but with the low payout is it the most +EV?

Is it correct to say game 1 and game 3 are equal in EV and game 2 is the lowest EV?


Which game is the most plus EV? Notice you pay $2 for the first game but $1 for the other games. Also what if the grand prize of the mega millions was 5 times more? Would that make it the most +EV?

They're all -EV of course, but Game 3 is the least -EV. Game 1 is twice as bad as Game 3 because you're paying twice as much with the same odds ratio. In general you can use the EV formula, namely take the weighted average of the possible net profit amounts. To be thorough, you'd have to factor in taxes, 2nd/3rd prizes etc, and estimated probabilities of splitting with other winners (at least in the case of Mega Millions).

To follow up on heehaw's response:

No. The EVs are -1, -0.6154, -0.5 (with very. very minor rounding)

The third game with EV = -0.5 has the best EV but it is not +EV as that symbol is used to represent a positive gain. If the megamillions prize was 5 times more, then, yes, it would be the best EV of +3.

The questions are fine for EV analysis interest but IMO of no practical value for determining which lottery you should play, if any.

For example, suppose the mega million $ win is 5x but you must bet $200 with 200/2=100 times the chance of winning. Then your EV = 300. Betting and losing $2 may have little effect on providing for family but betting and almost certainly losing $200 may be more significant, so the increase in EV is not compensatory.

Also, EV is over the long run. Mega Millions has a very long run indeed.