What is the value of this promo

My local card casino offers a free sports pool promotion. All seated players get a free entry into the pool. It is similar to many sports pools I have seen elsewhere. Each ticket has two one digit numbers, one for the home team, one for the away team. Each number (for each team) is 0-9 and can be the same or different for each team. (apparently random) If you win in the correct order (based on the second number of the teams score i.e. 79 to 85 your numbers would be Home=9 and Away=5) you win $500. If you get the reverse, you win $250. For Basketball, You get a ticket at the beginning of the game, and one at halftime. The first is good for the first AND second half, the second ticket is only good for the second half. For football, you get one at the beginning of the game and one at each quarter. Only the first ticket is good all game, the others are only good during that quarter. What is the cash value (in EV terms) of these tickets. Please show your math.


Best regards,

Bob

Let's ignore the fact that, especially in football, some final digits are going to be more likely than others in the actual score.


A ticket is going to have one chance in ten of the first digit matching the home team score and one in ten of the second digit matching the away team score. Thus it has a probability of 1/10 * 1/10 = 1/100 of hitting the correct combination.


Likewise, it has a 1/100 chance of hitting the (away, home) combination as well.


So, *a priori*, a ticket is worth 1/100 of $500 plus 1/100 of $250 for each time it is good. That adds up to $7.50.


The first basketball ticket is worth $15 (half-time plus final score) and the second is worth $7.50. The first football ticket is worth $30, and the remaining three are worth $7.50 each.


This assumes that you get to keep the winning whole-game ticket if it wins at halftime or on a quarter break. If they take it away, its value declines slightly.


If tickets are transferable, i.e. if you can sell them or swap them with other players, you can get an overlay with the football tickets by swapping "bad" number combinations for "good" ones. (You can identify which is which by spending some time with a bunch of football scores. If you want to get really creative, you can work out most likely digits for running games, passing games, and strong or weak defenses. But don't take my word for this; I don't know doodly-squat about sports betting.)

Alan,


You can't win both ways ($500 and $250) at the same time (unless you have 6 6, etc. and then you win both ways.) So, wouldnt the value of the basketball ticket be a total of $7.50? I see it as you have a 1/100 chance of winning either $500 or $250*2 (once for the first half, and once for the final score)


Bob

If you evaluate just the value of a ticket from the (home, away) combination, it pays $500. The likelihood of this happening is 1/100, so the contribution to the value of the ticket from the (home, away) combination is $5.00. Likewise, the (away, home) combination pays $250 when hit, with a probability of 1/100, for a contribution to the ticket's value of $2.50.


There is one way for the ticket to pay $500, one way for it to pay $250, and 98 ways for it to pay nothing. The sum of all payments is $750, and the total number of possible outcomes is 100; so it's EV is $7.50.


That's $7.50 at halftime AND $7.50 for the final game, for a total of $15.00.

Alan,


I dont mean to appear brain-dead, but since you can only win either home OR away, shouldn't the value be the average of the potential winnings? $500 + $250 /2 = $375 ($3.75) for the half, and another $3.75 for the final for a combined value of $7.50 total for the game?


Bob