Superbowl Grid Pools
$100 per box - $10,000 prize pool
A prize of $325 pays out for every scoring change and whatever remains of the prize pool goes to the winning box of the final score.
An extra point or 2-point conversion will be scored as separate scoring changes.
Example...let's say the Patriots play the Giants in the Super Bowl. If the Giants score a touchdown and go up 6-0, whoever has the box "Giants 6, Patriots 0" will win $325. After the Giants kick the extra point and go up 7-0, whoever has the box "Giants 7, Patriots 0" will win $325 as well.
The score 0-0 will win a "round" of $325 before the game even starts for having the beginning score of the game.
The final score box gets final pay out (money left in pool) plus the score change payout.
If somehow there are so many score changes occur that the prize pool runs out before the end of the game, the final box will win $1250 and the score change payouts will be adjusted accordingly.
Should we each get our own boxes or does it make more sense to pool our money to split several boxes and does it matter how many we split?
Quote: Juggler77
Should we each get our own boxes or does it make more sense to pool our money to split several boxes and does it matter how many we split?
It doesn't matter at all since the numbers are random. If I buy more than one box I keep them away from each other since I once had two boxes one box apart and the winner was said one box apart.
I wonder if 5-5 gets a rebate?
Quote: DRichInteresting. I have had Super Bowls squares for the past 30 years and never seen one run like this. I like the concept but not the fact that the payouts will change if the number of scores depletes the pool.
I never have either, and I found it interesting as well. I also found several websites that administrate a Football Squares Pool pretty easily online, and I wish it had occurred to me earlier, because I think it's too late to get one up and running here and get everybody's money ahead of time. Maybe next year.
I do think it's worth pooling your money to get partials on more numbers, especially if it's 100/square and each score pays 3x + that amount. 3 people partnered up seems ideal for those payouts. It would take 30 scores, roughly, to zero out the final win below a useful amount, which would take a combined score over 80 at least; seems unlikely that would be a problem.
Quote: AZDuffmanIt doesn't matter at all since the numbers are random. If I buy more than one box I keep them away from each other since I once had two boxes one box apart and the winner was said one box apart.
Depending on the rules, it may be advantageous to have the boxes in an array. I was in one where the adjoining edges of each winning box got a small amount. Surprisingly, folks still signed up for the edges and corners.
Pooling something with an EV of 0 is still an EV of 0, just with less variance.
If you want to actually make money off this, you have to trade/buy/sell boxes with other people after the numbers are drawn in a +EV fashion.
Quote: AyecarumbaDepending on the rules, it may be advantageous to have the boxes in an array. I was in one where the adjoining edges of each winning box got a small amount. Surprisingly, folks still signed up for the edges and corners.
True, rules differ, but in the end all the numbers are random, no?
Quote: DRichInteresting. I have had Super Bowls squares for the past 30 years and never seen one run like this. I like the concept but not the fact that the payouts will change if the number of scores depletes the pool.
There would have to be 26 score changes, not including 0-0, for the prize pool to drop below $1250. That's 13 touchdowns and conversions. In a college game, maybe, but in the Super Bowl?
Quote: ThatDonGuyThere would have to be 26 score changes, not including 0-0, for the prize pool to drop below $1250. That's 13 touchdowns and conversions. In a college game, maybe, but in the Super Bowl?
I understand it is unlikely but generally the ones we run we pay the winner every quarter. I think I would still want to pay the winners after each score and I would probably have to just end up paying the difference out of my pocket if there were more scores.
Quote: Juggler77My friends and I are joining a Superbowl Grid Pool. These are the rules.
$100 per box - $10,000 prize pool
A prize of $325 pays out for every scoring change and whatever remains of the prize pool goes to the winning box of the final score.
An extra point or 2-point conversion will be scored as separate scoring changes.
Example...let's say the Patriots play the Giants in the Super Bowl. If the Giants score a touchdown and go up 6-0, whoever has the box "Giants 6, Patriots 0" will win $325. After the Giants kick the extra point and go up 7-0, whoever has the box "Giants 7, Patriots 0" will win $325 as well.
The score 0-0 will win a "round" of $325 before the game even starts for having the beginning score of the game.
The final score box gets final pay out (money left in pool) plus the score change payout.
If somehow there are so many score changes occur that the prize pool runs out before the end of the game, the final box will win $1250 and the score change payouts will be adjusted accordingly.
Should we each get our own boxes or does it make more sense to pool our money to split several boxes and does it matter how many we split?
I was in a pool like this last year, but it was only $1 or $2 a square. I did pretty good.
The 2nd $20 is still open, prolly 20 remaining. Might go 2 more squares. Same rules.
Quote: AyecarumbaIf not 51, how many would you have to buy to have an edge?
You can't have an edge.
Every box is 0 EV (again, assuming a payout of 100% and no rigging).
Quote: sc15You can't have an edge.
Every box is 0 EV (again, assuming a payout of 100% and no rigging).
Since I have more than half the boxes, don't I have a better chance of winning more of the prizes? To simplify things, assume the price for each box was $1 and there were only four prizes, one for each quarter. (10, 20, 10, 60). Since I have more than half the squares, don't I have a 51% probability of winning the big prize and coming out $9 ahead?
Does the fact that the same box can win multiple times make a difference?
Quote: mipletI was in a pool like this last year, but it was only $1 or $2 a square. I did pretty good.
My mom is in a score changing pool like I was in last year. $1 a square pays $7 per score change, up to 14 changes. $2 for 0/0 start, and any left over goes to the final score. She has the following squares (2 separate pools) hawks tens digit pats ones digit.
63 82 17 09 40 34 41 58 29 63
Quote: AyecarumbaSince I have more than half the boxes, don't I have a better chance of winning more of the prizes? To simplify things, assume the price for each box was $1 and there were only four prizes, one for each quarter. (10, 20, 10, 60). Since I have more than half the squares, don't I have a 51% probability of winning the big prize and coming out $9 ahead?
Does the fact that the same box can win multiple times make a difference?
No. This is exactly like roulette (if the wheel had no zeroes). You can bet on as many numbers as you want and get an extremely high chance of winning, but it wouldn't change your EV.
You could buy all 100 squares of this pool and have a 100% chance of winning $0 -- basically playing with yourself.
If it's a fair drawing each square has $0 EV and you can't add up a bunch of $0 EV to get ahead.
Quote: AyecarumbaSince I have more than half the boxes, don't I have a better chance of winning more of the prizes? To simplify things, assume the price for each box was $1 and there were only four prizes, one for each quarter. (10, 20, 10, 60). Since I have more than half the squares, don't I have a 51% probability of winning the big prize and coming out $9 ahead?
Yes, but "expected value" and "probability of winning" are two different things.
Let's do this backwards - we'll play the game and then draw the squares.
The expected return of the first quarter prize is 51% x 10 + 49% x 0 = 5.1
The expected return of the second quarter prize is 51% x 20 + 49% x 0 = 10.2
The expected return of the third quarter prize is 51% x 10 + 49% x 0 = 5.1
The expected return of the final prize is 51% x 60 + 49% x 0 = 30.6
The total expected return = 5.1 + 10.2 + 5.1 + 30.6 = 51.
Expected value = expected return minus cost = 51 - (51 squares x 1 per square) = 0.
Quote: AyecarumbaSince I have more than half the boxes, don't I have a better chance of winning more of the prizes? To simplify things, assume the price for each box was $1 and there were only four prizes, one for each quarter. (10, 20, 10, 60). Since I have more than half the squares, don't I have a 51% probability of winning the big prize and coming out $9 ahead?
Does the fact that the same box can win multiple times make a difference?
Gambling isn't your strong suit..
Hehe.. Advantage play, perhaps not...Quote: sc15Gambling isn't your strong suit..
