Super Bowl Squares
Posted by Chase Stuart on February 4, 2010
This is a re-run of Doug's post from two years ago.
Three Super Bowls ago, I wrote this post over at Sabernomics. In it, I looked at your probability of winning a squares pool with any given square. For example, I found that in a one-unit-per-square pool, either of the '0/7' squares would have an expected value of about 3.8 units. Compare that with, say, a '5/6' square, which has an expected value of 0.22, or the lowly `2/2' square and its expected value of .04. Because it was all the data I had at the time, I only considered the last digits of the final scores of games, but someone correctly pointed out in the comments that most pools also give prizes for (the last digits of) the cumulative scores at the end of each quarter.
Well, now I have score-by-quarter data for the entirety of the NFL's 2-point-conversion era (1994--present), so it's time for an update.
I'm sure there are lots of ways to do this, but a bit of googling indicates that a standard payout structure is something like 10% of the pot after each of the first three quarters, and 70% for the final. This doesn't alter things too drastically, but it does have a couple of effects.
- The '0/7' squares enjoy an even larger advantage over an average square. The '0/7' squares have an expected value of about 4.9 under this scheme.
- The '0/0' square starts climbing the charts.
More than 20% of all games are in a '0/0' situation (remember, that includes 10-0 and 10-10 as well as 0-0) after one quarter. At halftime, about 7.5% of all games are a '0/0.' So the more weight you put on the intermediate stages, the better the '0/0' square looks. Here is a chart that shows the expected value of a given square after each quarter, along with a final column that shows the expected value under a 10/10/10/70 system:
+-----+-----+------+------+------+------+------+ | | | q1ev | q2ev | q3ev | q4ev | ev | +-----+-----+------+------+------+------+------+ | 7 | 0 | 11.8 | 5.6 | 4.7 | 3.9 | 4.9 | | 0 | 0 | 20.5 | 7.5 | 4.4 | 1.9 | 4.5 | | 3 | 0 | 9.2 | 5.1 | 3.4 | 3.3 | 4.1 | | 7 | 7 | 6.9 | 6.3 | 4.2 | 2.2 | 3.3 | | 7 | 4 | 1.3 | 3.0 | 3.3 | 3.4 | 3.1 | | 7 | 3 | 4.7 | 4.5 | 3.3 | 2.0 | 2.7 | | 4 | 0 | 3.5 | 3.6 | 2.6 | 2.1 | 2.4 | | 4 | 1 | 0.0 | 0.5 | 1.6 | 2.3 | 1.8 | | 3 | 3 | 3.1 | 3.2 | 3.3 | 1.2 | 1.8 | | 4 | 3 | 0.9 | 2.3 | 2.3 | 1.5 | 1.6 | | 7 | 1 | 0.1 | 1.5 | 2.0 | 1.8 | 1.6 | | 6 | 0 | 1.1 | 2.2 | 1.7 | 1.5 | 1.6 | | 4 | 4 | 0.2 | 1.8 | 2.3 | 1.5 | 1.5 | | 6 | 3 | 0.3 | 1.5 | 1.5 | 1.7 | 1.5 | | 1 | 0 | 0.3 | 1.2 | 1.3 | 1.5 | 1.3 | | 7 | 6 | 0.5 | 1.7 | 1.6 | 1.0 | 1.1 | | 3 | 1 | 0.1 | 0.9 | 1.0 | 1.0 | 0.9 | | 8 | 1 | 0.0 | 0.0 | 0.0 | 1.3 | 0.9 | | 8 | 0 | 0.0 | 0.4 | 0.8 | 1.0 | 0.8 | | 6 | 4 | 0.0 | 1.1 | 1.2 | 0.8 | 0.8 | | 9 | 7 | 0.1 | 0.5 | 0.7 | 0.8 | 0.7 | | 6 | 1 | 0.0 | 0.4 | 0.5 | 0.9 | 0.7 | | 9 | 3 | 0.1 | 0.4 | 0.5 | 0.7 | 0.6 | | 9 | 0 | 0.2 | 0.7 | 0.5 | 0.7 | 0.6 | | 7 | 5 | 0.0 | 0.2 | 0.4 | 0.8 | 0.6 | | 8 | 7 | 0.0 | 0.0 | 0.0 | 0.8 | 0.6 | | 1 | 1 | 0.0 | 0.0 | 0.0 | 0.8 | 0.6 | | 5 | 0 | 0.1 | 0.2 | 0.4 | 0.7 | 0.6 | | 8 | 3 | 0.0 | 0.0 | 0.0 | 0.7 | 0.5 | | 9 | 4 | 0.0 | 0.3 | 0.6 | 0.6 | 0.5 | | 7 | 2 | 0.1 | 0.3 | 0.5 | 0.5 | 0.5 | | 6 | 6 | 0.0 | 0.6 | 0.5 | 0.5 | 0.5 | | 8 | 4 | 0.0 | 0.0 | 0.0 | 0.8 | 0.5 | | 4 | 2 | 0.0 | 0.2 | 0.4 | 0.6 | 0.5 | | 2 | 0 | 0.1 | 0.4 | 0.6 | 0.6 | 0.5 | | 9 | 6 | 0.0 | 0.0 | 0.0 | 0.7 | 0.5 | | 9 | 1 | 0.0 | 0.0 | 0.0 | 0.6 | 0.4 | | 3 | 2 | 0.0 | 0.1 | 0.3 | 0.5 | 0.4 | | 8 | 5 | 0.0 | 0.0 | 0.0 | 0.6 | 0.4 | | 5 | 4 | 0.0 | 0.0 | 0.0 | 0.5 | 0.4 | | 8 | 6 | 0.0 | 0.0 | 0.0 | 0.5 | 0.3 | | 6 | 2 | 0.0 | 0.1 | 0.1 | 0.4 | 0.3 | | 5 | 3 | 0.0 | 0.0 | 0.0 | 0.5 | 0.3 | | 9 | 2 | 0.0 | 0.0 | 0.0 | 0.4 | 0.3 | | 8 | 8 | 0.0 | 0.0 | 0.0 | 0.4 | 0.3 | | 5 | 1 | 0.0 | 0.0 | 0.0 | 0.3 | 0.2 | | 6 | 5 | 0.0 | 0.0 | 0.0 | 0.2 | 0.2 | | 2 | 1 | 0.0 | 0.0 | 0.0 | 0.3 | 0.2 | | 5 | 2 | 0.0 | 0.0 | 0.0 | 0.3 | 0.2 | | 9 | 8 | 0.0 | 0.0 | 0.0 | 0.3 | 0.2 | | 5 | 5 | 0.0 | 0.0 | 0.0 | 0.2 | 0.1 | | 9 | 5 | 0.0 | 0.0 | 0.0 | 0.2 | 0.1 | | 8 | 2 | 0.0 | 0.0 | 0.0 | 0.2 | 0.1 | | 9 | 9 | 0.0 | 0.0 | 0.0 | 0.2 | 0.1 | | 2 | 2 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | +-----+-----+------+------+------+------+------+
Here's how to read that. Take the top line for example. If you have one of the two '0/7' squares, then your expected value is 11.8% of the first-quarter pot, 5.6% of the second-quarter pot, and so on. With the 10/10/10/70 system, your overall expected value would be about 4.9.
As I was poking around the web looking for info on standard payout schemes for these kinds of pools, I came across this page. One of the commenters there suggests using not the final digit of each team's score, but the final digit of the sum of each team's score. So a 17 would be an 8, a 22 would be a 4, and a 38 would be a 1.
I was too lazy to check this one out quarter-by-quarter, but just looking at final scores, this scheme produces a much flatter expected value curve.
+------+------+------+ | | | ev | +------+------+------+ | 7 | 4 | 2.0 | | 8 | 2 | 1.6 | | 4 | 1 | 1.5 | | 6 | 3 | 1.5 | | 8 | 6 | 1.4 | | 9 | 6 | 1.4 | | 6 | 4 | 1.4 | | 8 | 5 | 1.4 | | 7 | 0 | 1.3 | | 4 | 0 | 1.3 | | 6 | 5 | 1.3 | | 8 | 7 | 1.3 | | 4 | 3 | 1.3 | | 3 | 0 | 1.2 | | 5 | 3 | 1.2 | | 6 | 1 | 1.2 | | 7 | 6 | 1.1 | | 8 | 3 | 1.1 | | 8 | 1 | 1.1 | | 8 | 4 | 1.1 | | 9 | 3 | 1.1 | | 5 | 4 | 1.1 | | 5 | 2 | 1.1 | | 7 | 3 | 1.1 | | 5 | 1 | 0.9 | | 9 | 8 | 0.9 | | 3 | 1 | 0.9 | | 7 | 7 | 0.9 | | 7 | 5 | 0.9 | | 9 | 7 | 0.9 | | 9 | 4 | 0.9 | | 3 | 2 | 0.9 | | 4 | 2 | 0.8 | | 7 | 1 | 0.8 | | 9 | 1 | 0.8 | | 8 | 0 | 0.8 | | 5 | 0 | 0.8 | | 6 | 0 | 0.8 | | 6 | 6 | 0.7 | | 1 | 0 | 0.7 | | 9 | 5 | 0.7 | | 3 | 3 | 0.7 | | 7 | 2 | 0.7 | | 9 | 2 | 0.7 | | 9 | 0 | 0.6 | | 5 | 5 | 0.6 | | 2 | 1 | 0.6 | | 4 | 4 | 0.6 | | 1 | 1 | 0.6 | | 6 | 2 | 0.6 | | 2 | 0 | 0.5 | | 8 | 8 | 0.5 | | 0 | 0 | 0.2 | | 9 | 9 | 0.2 | | 2 | 2 | 0.0 | +------+------+------+
Look where '0/0' is now!
I don't think fairer is the right word, but this seems to me to be a clearly more interesting pool. It is less determined by the random assignment of squares and more determined by the random actions that happen as the game unfolds. And that's how it ought to be.
For the extra geeky, one way to improve this (in my opinion) would be to, at the end of the game, flip a coin to determine whether the criterion to be used is "last digit" or "last digit of sum of digits." In other words, say the game ends up at 23-14. If the coin comes up heads, it's a '3/4'. If it comes up tails, it's a '5/5'. The point of the pool is to keep people interested. With the coin flip rule in place, I'd guess that, even into the fourth quarter, just about anyone (except the poor suckers with '2/2' and '9/9'; they're beyond help) could invent a reasonable scenario whereby he or she collects the prize.
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February 4th, 2010 at 3:58 pm
I'm sure there are lots of ways to do this, but a bit of googling indicates that a standard payout structure is something like 10% of the pot after each of the first three quarters, and 70% for the final. This doesn't alter things too drastically, but it does have a couple of effects.
Really? Every single pool I've ever been in weights each of the first three quarters at 25% and the final score at 25%. No one has ever even suggested weighting the final so heavily.
---
Also, here's an easy way to do a squares-ish pool at a smaller party (for those of you interested):
1. Take a deck of cards and pull out one of each of the following cards: A, 2, 3, 4, 5, 6, 7, 8, 9, 10.2. Up to ten people can pay one unit of cost for one of the above ten cards.
3. At the end of each quarter, take the combined total points scored in the game. The final digit of that score is the winning card (with A=1 and 10=0).
ex) If the score is 14-10, the sum is 24. Whoever has the 4 card wins. If the score is 10-10, the sum is 20, and whoever has the 10 card wins.
This is a nice game for a smaller party, as you don't need to fill in 100 slots like you would for a true squares pool. Additionally, due to the easy nature of drawing one of ten cards, you can re-choose a card every quarter, so someone stuck with a bad card in the first quarter can get a better card in the second, and so forth.
(And, as with a regular squares pool, you don't have to limit to one entry per person. If you have five people at your party, each person could buy two cards, or one could buy six and everyone else buy one, etc.)
February 4th, 2010 at 3:59 pm
There seem to be a few rounding/truncating hiccups in Doug's 4 quarters and overall EV table, but using those numbers:
Assume 10 "owners" chosing squares in a typical fantasy-style "snake draft", with each owner taking the highest available EV square, the cumulative EVs of their 10 squares would be:
owner Total
B 11.07
A 10.92
C 10.77
D 10.19
E 10.06
F 9.46
J 9.25
I 9.18
H 9.1
G 9.07
Where owner A picked 1st (and 20th), owner B picked 2nd (and 19th), etc. That would be a reasonably fair distribution of squares.
And of course there are the possibilities of auctioning the squares, with the amounts bid going into the pot - hold the auction at halftime, which would be much better than watching the TV (commercials excepted).
February 4th, 2010 at 11:07 pm
I calculated this out just last night, using our superbowl squares scheme... 20% for each of the first 3 quarter scores, 40% for the final. 0-0 is the best square by a fair margin.
But the 0-9 numbers were filled in randomly after everybody picked their square, so it was fair regardless... Unfortunately I ended up with 4-2 and 8-9. Ech!
February 5th, 2010 at 1:17 pm
It's not clear to me that you are aware that the whole point of the squares game is to part the statistically disinclined from their money.
I'd be interested to know how to go the other way: seemingly innocuous adjustments to the rules that would allow my minority stake to do better than 4:1 average return it currently pulls.
February 5th, 2010 at 5:16 pm
Nice post!
Do you know where I could get the raw data (scores after each quarter for all games since '94) for this calculation in a single file?
February 7th, 2010 at 2:29 pm
This may be a dumb question, but why doesn't the 'ev' probability add up to 100%? It adds up to 55.7%
February 7th, 2010 at 11:20 pm
Kevin (#6): You need to count the "double" squares (0-0, 1-1, etc) twice, since the non-double squares (7-0, 3-6, etc) count twice (once as 7-3 and once as 3-7, for example).
(Disclaimer: I have not done this and checked the numbers, so the sum could still be off.)