Stoney
• Posts: 30
Joined: Sep 22, 2012
November 16th, 2012 at 9:36:18 AM permalink
i was going to come just to ask a simple question, but my mind started going crazy. i'll try to keep things as clear and directed as i can... so i guess it's best to tell it as it happened.

for those who don't know, football squares is a game where a player chooses any number of squares from a 10x10 grid. then, after all squares are picked, numbers 0 through 9 are assigned to each column and each row. the column represents the home team, and the row the visitor (or vise-versa). if the last digit of each teams score matches the square you picked, you win. for expamle, you picked the 0-7 square. you would win on any combination of 0-7, 10-17, 10-37, etc. most games pay out at the end of each quarter and final score. you can win multiple times with the same square.

my questions are...

what are the chances per square? at first i want to say 1/100, but since they pay out 4 times, would it be 1/25?

how many would i have to buy to give a reasonable chance to win, while maintaining profitability? and would this number be doubled if i wanted the same expectation of winning twice? triple for thrice?

then, i got to thinking, what are the chances of not breaking even if i bought 75? not that i'd do it, but spend \$150 to have a high probablility of making \$50 and a small chance of losing my ass.

that reminded me of the birthday paradox that says you only need a group of 23 to have a better than 50% chance to have 2 people with the same birthday. they calculate it by figuring the chance of NOT having the same and the inverse must be the solution. the wikipedia page says you only need 57 people to have a 99% chace to 2 people sharing a birthday!

then that made me think that the gamblers phallacy (mainly the martingale system)isn't all that wrong. the chance of red/black is 50% (not including green), so the chance of NOT winning is exceeding low with very small sample sizes.

both of the last 2 can't be right, can they?

FinsRule
• Posts: 3917
Joined: Dec 23, 2009
November 16th, 2012 at 12:00:40 PM permalink
Quote: Stoney

i was going to come just to ask a simple question, but my mind started going crazy. i'll try to keep things as clear and directed as i can... so i guess it's best to tell it as it happened.

for those who don't know, football squares is a game where a player chooses any number of squares from a 10x10 grid. then, after all squares are picked, numbers 0 through 9 are assigned to each column and each row. the column represents the home team, and the row the visitor (or vise-versa). if the last digit of each teams score matches the square you picked, you win. for expamle, you picked the 0-7 square. you would win on any combination of 0-7, 10-17, 10-37, etc. most games pay out at the end of each quarter and final score. you can win multiple times with the same square.

my questions are...

what are the chances per square? at first i want to say 1/100, but since they pay out 4 times, would it be 1/25?

how many would i have to buy to give a reasonable chance to win, while maintaining profitability? and would this number be doubled if i wanted the same expectation of winning twice? triple for thrice?

then, i got to thinking, what are the chances of not breaking even if i bought 75? not that i'd do it, but spend \$150 to have a high probablility of making \$50 and a small chance of losing my ass.

that reminded me of the birthday paradox that says you only need a group of 23 to have a better than 50% chance to have 2 people with the same birthday. they calculate it by figuring the chance of NOT having the same and the inverse must be the solution. the wikipedia page says you only need 57 people to have a 99% chace to 2 people sharing a birthday!

then that made me think that the gamblers phallacy (mainly the martingale system)isn't all that wrong. the chance of red/black is 50% (not including green), so the chance of NOT winning is exceeding low with very small sample sizes.

both of the last 2 can't be right, can they?

If there is no takeout, and there is no choice in picking squares, then it's a 0% house edge / even ev game. There is no optimal number of squares to buy, because there is no advantage/disadvantage of buying numbers.
FinsRule
• Posts: 3917
Joined: Dec 23, 2009
November 16th, 2012 at 12:02:21 PM permalink
Also, I don't think Gambler's Fallacy or the "Birthday Paradox" comes into play at all.
AxiomOfChoice
• Posts: 5761
Joined: Sep 12, 2012
November 16th, 2012 at 12:43:51 PM permalink
Quote: Stoney

what are the chances per square?

1/100, each quarter

Quote:

at first i want to say 1/100, but since they pay out 4 times, would it be 1/25?

Not quite.

If they re-shuffle the numbers after each quarter, then the chances of winning each quarter are independent. You can't just add them to figure out the probability of winning at least once, though, because you are double-counting the probability that you win twice.

If they don't re-shuffle the numbers each quarter, then the quarters are correlated (because some numbers are better than others). In other words, if you win once, you have a better chance of winning again. This makes it difficult to calculate your chances of winning multiple times (you need statistical information about game scores)

Quote:

how many would i have to buy to give a reasonable chance to win,

It depends what your definition of reasonable is. Do you want a 1/100 chance of winning? 1/25? 1/10? 1/3?

Quote:

while maintaining profitability?

There is no profitability here. Your EV is 0 no matter how many squares you buy (assuming that all the entry fees are paid out as prizes).

Quote:

and would this number be doubled if i wanted the same expectation of winning twice? triple for thrice?

No. You will get significantly different numbers depending on whether or not they reshuffle the numbers after each quarter, but it's not going to be a simple linear thing.

Quote:

then, i got to thinking, what are the chances of not breaking even if i bought 75? not that i'd do it, but spend \$150 to have a high probablility of making \$50 and a small chance of losing my ass.

Your chances of winning all 4 are not that high. If they reshuffle the numbers each quarter, you have a 3/4 chance of winning each quarter (and these are independent) so your chances of winning all 4 are (3/4)^4 = 3^4 / 4^4 = 81 / 256, which is worse than 1/3.

Quote:

that reminded me of the birthday paradox that says you only need a group of 23 to have a better than 50% chance to have 2 people with the same birthday. they calculate it by figuring the chance of NOT having the same and the inverse must be the solution. the wikipedia page says you only need 57 people to have a 99% chace to 2 people sharing a birthday!

Yup.

Quote:

then that made me think that the gamblers phallacy (mainly the martingale system)isn't all that wrong. the chance of red/black is 50% (not including green), so the chance of NOT winning is exceeding low with very small sample sizes.

Nope.

First, these two are more or less unrelated. If you actually have a fair game (which pays 1:1 on a result that has a 50% chance of succeeding) then no matter what you do, your EV is 0. You can certainly structure things so that you win most of the time, but you do that by losing more when you do lose.

Say you have \$127 in your pocket. You decide to play this fair game using a martingale system, trying to win \$1. So, you bet \$1, and if you lose you bet \$2, etc, etc, until you either win and quit or lose your final \$64 bet and leave broke.

In order for you to lose, you need to lose 7 bets in a row. The probability of this is 1 / 2^7, which is 1 / 128.

So, your chances of winning \$1 are 127 / 128. Your chances of losing \$127 are 1 / 128.

So, on average, every 128 times, you can expect to win 127 times (for a net win of \$127) and lose once (for a loss of \$127) leaving you with \$127 - \$127 = \$0.

"Short term" does not help you. If you only do it once, it is true that most likely you will win \$1, but it is also true that there is a small chance that you will lose \$127. The losing outcome is just as likely to happen the first time you try as it is the 10th, 50th, 100th, or 128th time.

And, of course, in a game with a house edge (double-zero roulette is over 5%, BTW -- "ignoring green" is not a good idea) then you will lose in the long term instead of breaking even. This is why the house doesn't care what betting system you use. If everyone used Martingale, then most people would walk away small winners, and the occasional chump would lose enough to cover all the winners, plus some profit for the house.

Quote:

both of the last 2 can't be right, can they?

They sure can, and they are. If they weren't the martingale bettors would have bankrupted the casinos long ago (this betting system has been well-known for, literally, hundreds of years)

Quote:

I'm sure you will have no trouble finding a place that does football squares for the super bowl. Some online sportsbooks do them (taking a cut, of course)

FWIW, if there is a place that does it without taking a cut, you may as well do it. It's a fair gamble, which is not easy to find. If you enjoy gambling it's free entertainment -- and a lot better than buying a scratch-off lottery ticket. Just don't bet more than you want to lose.
Stoney
• Posts: 30
Joined: Sep 22, 2012
November 16th, 2012 at 2:10:15 PM permalink
thanks a lot axiom. lots of info to digest.... and some new questions....

first off, it's a workplace pool (for the ohio state/michigan game), so all the money is payed out and the numbers are the same for the whole game. i know that having some combination of 0,3,7 are optimal, and research on the internet says 2-2 and 8-8 have never come up at the end of any quarter in the last 6 years of nfl play. so these numbers (and a few others) are extremely bad. so my assumtion is that more squares means better chance to get those coveted combinations, but at the cost of reduced return on investment. but that's where the birthday paradox came to mind....

23 people seems oftly low but it made me believe there might be some optimal number of squares that the odds of not winning vs rate of return were favorable... but knowing it's a 0 edge game, i should have realized the profit vs losses cancel each other out.

now on the birthday/phallacy issue. if we bet on the birthdays of a random 57 people, a smart person would say they would share a birthday and come out ahead +98% (99 wins - 1 loss). but lets say we have 56 people with all different dates and we bet on #57 matching with one of them. phallacy states that each trial is independant and #57 only has 56/365 of matching. how can this be? 1 says 99% the other 15%. the odds of red coming up 7 times in a row is just under 1%, and the chance of it coming up 15 times in a row is .003%. i would bet against red coming up 15 times. so martingaling would work if you waited for those instances when a streak has already exceded the 99% win percentage... right?
thecesspit
• Posts: 5936
Joined: Apr 19, 2010
November 16th, 2012 at 2:22:21 PM permalink
Wrong.

Each spin of the wheel is independent.

In the birthday party game, 23 is the number as each person compares against -each other person-, which is a lot of comparisons, each one have a 1 in 365 chance.

If all the birthdays are different, adding a new person will be 23/365 of matching. But that's using prior related information. The roulette wheel, the previous spins don't matter. Saying it's hit a 1 in 100 series of results previously is immaterial, and using it to predict the future IS the gambler's fallacy.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
AxiomOfChoice
• Posts: 5761
Joined: Sep 12, 2012
November 16th, 2012 at 3:36:24 PM permalink
Quote: Stoney

now on the birthday/phallacy issue. if we bet on the birthdays of a random 57 people, a smart person would say they would share a birthday and come out ahead +98% (99 wins - 1 loss). but lets say we have 56 people with all different dates and we bet on #57 matching with one of them. phallacy states that each trial is independant and #57 only has 56/365 of matching. how can this be? 1 says 99% the other 15%.

The issue here is that in one case you are picking 57 people with randomly chosen birthdays, and in your example you are picking 56 people with different birthdays and then adding one to them. If you pick 56 people with different birthdays, then the probability of the 57th matching is, indeed 56/365.

But, if you pick 56 random people, then there is a good chance that at least two of them already match (so the result of the 57th is irrelevant). In your example, you don't have that chance.
kubikulann
• Posts: 905
Joined: Jun 28, 2011
November 20th, 2012 at 6:19:51 AM permalink
Quote: Stoney

now on the birthday/Fallacy issue. if we bet on the birthdays of a random 57 people, a smart person would say they would share a birthday and come out ahead +98% (99 wins - 1 loss). but lets say we have 56 people with all different dates and we bet on #57 matching with one of them. Fallacy states that each trial is independant and #57 only has 56/365 of matching. how can this be?

First case : ANY two match. 57*56/2= 1596 possible pairs (not counting the triples, etc.)
Second case : only the match of #57 with another. 56 possibilities only.

This is the difference between a marginal and a conditional probability.

Reminds me of the joke of the fearful guy who takes a bomb on his airplane trip, because the probability of two bombs on the same plane is virtually nil. He confuses marginal and conditional. He is a victim of the gambler's fallacy, putting correlation where there is none.
Reperiet qui quaesiverit
vendman1
• Posts: 1034
Joined: Mar 12, 2012
November 20th, 2012 at 6:46:56 AM permalink
Quote: Stoney

thanks a lot axiom. lots of info to digest.... and some new questions....

first off, it's a workplace pool (for the ohio state/michigan game), so all the money is payed out and the numbers are the same for the whole game. i know that having some combination of 0,3,7 are optimal, and research on the internet says 2-2 and 8-8 have never come up at the end of any quarter in the last 6 years of nfl play. so these numbers (and a few others) are extremely bad. so my assumtion is that more squares means better chance to get those coveted combinations, but at the cost of reduced return on investment. but that's where the birthday paradox came to mind....

23 people seems oftly low but it made me believe there might be some optimal number of squares that the odds of not winning vs rate of return were favorable... but knowing it's a 0 edge game, i should have realized the profit vs losses cancel each other out.

now on the birthday/phallacy issue. if we bet on the birthdays of a random 57 people, a smart person would say they would share a birthday and come out ahead +98% (99 wins - 1 loss). but lets say we have 56 people with all different dates and we bet on #57 matching with one of them. phallacy states that each trial is independant and #57 only has 56/365 of matching. how can this be? 1 says 99% the other 15%. the odds of red coming up 7 times in a row is just under 1%, and the chance of it coming up 15 times in a row is .003%. i would bet against red coming up 15 times. so martingaling would work if you waited for those instances when a streak has already exceded the 99% win percentage... right?

If the numbers of the block pool are randomly assigned, and I assume they are. Axiom is correct, there is no strategy and no EV other than zero. Just pick some squares and hope you get lucky. Buying more or less squares doesn't matter. Yes you increase your chances of winning by buying more, but the amount you are spending exactly equals the expected amount you would win.