Need help with a probability problem.

[ABanks]

Hey everyone,

Happy Thanksgiving. I am dealing with a probability/math problem that I can't even begin to wrap my head around. I figured if anyone could come up with the answer it would probably be someone from The Tank.

I'm wondering what the odds are of something occurring randomly by coincidence. Let's say there have been 572 basketball games played. Person X has has made a wager on a side involving 22 of those 572 games. Person Y has made a wager on a side involving 28 of those 572 games. In total they chose 7 of the same sides. What would the odds be that this occurred randomly and was just a coincidence?

I figure you have to start by realizing that there are 1144 possible choices (572 x 2) but after that I get stuck.

Can anyone help me solve this problem? If you could show how you figured it out it would be much appreciated as I am really curious what the math involved would be.

Thank you so much to anyone who attempts to solve this and once again Happy Thanksgiving.

[matthew919]

You want to use a hypergeometric test. A slight twist is involved if you require that both people cannot select both sides of the same game. However, I'll just make the simplifying assumption that both person X and person Y are allowed to play both sides in a single game. Otherwise the math becomes a little trickier, as the usual combinatorial functions need to be modified. Though given the relatively large number of games, the result will not be too far off...

This can be formulated as an urn problem, as follows.

We start by figuring out the probability that both person X and person Y choose exactly the same teams. The "urn" in this case contains 572*2 total marbles (total #games * 2), of which 28 are "owned" by person Y (i.e. the teams that have been picked by person Y). Person X, will sample 22 of those marbles (i.e. choose 22 teams) without replacement (since he cannot select a team more than once). The probability for these types of problems can always be formulated as:

(#successesful outcomes)/(#total possible outcomes)

The total number of possible outcomes is like asking how many ways we select 22 marbles out of 1144. In mathematical notation this is: C(1144,22), or 1144!/((1144-22)!(22)!). This is our denominator.

The number of successes is like saying: given that we've selected exactly 7 marbles that were owned by person Y and 15 that were not, how many ways can we arrange them all? For this, we multiply the two combinatorial functions: C(28,7)*C(1144-28, 15). The first part is the ways to arrange the selected marbles that are "owned" by person Y, the second part is the number of ways to arrange the marbles "not owned" by Y. This term goes in the numerator.

Putting them together we get:

C(28,7)*C(1116,15) / C(1144,22)

which gives: 3.050937e-07

Note that this is the probability that exactly 7 of the same sides were selected. Determining the probability that 7 or more of the same sides were selected would involve similar steps- though I'd suggest calculating the probability for 6, 5, ... 1, 0 overlaps, summing them, and subtracting the result from 1. It will be very close to 0, in any case, meaning that this is extremely unlikely to happen due to chance.

[NunyaBidness]

All of this assumes that each game and each side have an equal chance of being chosen by a 'typical' bettor. Certain games receive a higher amount of attention than others.

[matthew919]

All of this assumes that each game and each side have an equal chance of being chosen by a 'typical' bettor. Certain games receive a higher amount of attention than others.

That's really not how the problem was posed. But if you want to account for that, then just weight them with a prior.

[ABanks]

Thank you so much Matthew919. You're right, I guess I should have said 7 or more, not exactly 7. Just curious, what does the e mean in your final answer: 3.050937e-07?

So basically you are saying it would be a real long-shot that the two people would pick 7 or more of the same sides?

I appreciate you taking the time to solve this. The math is fascinating and even more complex than I imagined, lol.

[matthew919]

Scientific notation.

0.000000305

[ABanks]

I wish I paid more attention during math class, hah.

So how would you fill in this blank?:

The odds of the 2 people randomly choosing 7 or more of the same sides is *blank* to 1?

[Kaabee]

about 327,000 to 1

[ABanks]

Thanks!

[matthew919]

about 327,000 to 1

I think you missed a zero.

[Kaabee]

yep

[ABanks]

What about the fact that there are limited choices each day in terms of available sides? In other words all 1,144 choices aren't available at once. Would that change the odds?

[ABanks]

OK so the 572 games occurred over a 21 day period where only a certain amount of games were available to select from each day. How would this change the odds of the 7 similar games happening by chance?

Here is the breakdown of the total amount of picks, the picks that matched, and the total number of games per day:

Day 1:
Person X: 2 picks
Person Y: 3 picks
Number of matching picks: 1
Total number of available games: 52 (104 different sides)

Day 2:
Person X: 1 pick
Person Y: 1 pick
Number of matching picks: 1
Total number of available games: 22 (44 different sides)

Day 3:
Person X: 1 pick
Person Y: 0 picks
Number of matching picks: 0
Total number of available games: 8 (16 different sides)

Day 4:
Person X: 2 picks
Person Y: 2 picks
Number of matching picks: 0
Total number of available games: 17 (34 different sides)

Day 5:
Person X: 2 picks
Person Y: 3 picks
Number of matching picks: 2
Total number of available games: 40 (80 different sides)

Day 6:
Person X: 1 pick
Person Y: 2 picks
Number of matching picks: 0
Total number of available games: 15 (30 different sides)

Day 7:
Person X: 1 pick
Person Y: 1 pick
Number of matching picks: 0
Total number of available games: 17 (34 different sides)

Day 8:
Person X: 1 pick
Person Y: 3 picks
Number of matching picks: 0
Total number of available games: 24 (48 different sides)

Day 9:
Person X: 0 picks
Person Y: 0 picks
Number of matching picks: 0
Total number of available games: 36 (72 different sides)

Day 10:
Person X: 0 picks
Person Y: 0 picks
Number of matching picks: 0
Total number of available games: 30 (60 different sides)

Day 11:
Person X: 0 picks
Person Y: 0 picks
Number of matching picks: 0
Total number of available games: 22 (44 different sides)

Day 12:
Person X: 2 picks
Person Y: 1 pick
Number of matching picks: 0
Total number of available games: 22 (44 different sides)

Day 13:
Person X: 1 pick
Person Y: 1 pick
Number of matching picks: 0
Total number of available games: 15 (30 different sides)

Day 14:
Person X: 1 pick
Person Y: 1 pick
Number of matching picks: 0
Total number of available games: 27 (54 different sides)

Day 15:
Person X: 1 pick
Person Y: 1 pick
Number of matching picks: 1
Total number of available games: 45 (90 different sides)

Day 16:
Person X: 0 picks
Person Y: 0 picks
Number of matching picks: 0
Total number of available games: 45 (90 different sides)

Day 17:
Person X: 0 picks
Person Y: 0 picks
Number of matching picks: 0
Total number of available games: 31 (62 different sides)

Day 18:
Person X: 3 picks
Person Y: 2 picks
Number of matching picks: 1
Total number of available games: 26 (52 different sides)

Day 19:
Person X: 1 pick
Person Y: 3 picks
Number of matching picks: 1
Total number of available games: 37 (74 different sides)

Day 20:
Person X: 0 picks
Person Y: 4 picks
Number of matching picks: 0
Total number of available games: 23 (46 different sides)

Day 21:
Person X: 2 picks
Person Y: 0 picks
Number of matching picks: 0
Total number of available games: 18 (36 different sides)

[SteveRyan]

Okay....what exactly happened that would drive you to figure this out?

Were you involved in some type of gambling contest with multiple people and you suspect people cheating or something? Like what are the odds that these two people made the same picks?

[thehoorse]

Okay....what exactly happened that would drive you to figure this out?

Were you involved in some type of gambling contest with multiple people and you suspect people cheating or something? Like what are the odds that these two people made the same picks?

That's exactly what I was thinking. Or he's the one about to cheat

[shackfu99]

That's exactly what I was thinking. Or he's the one about to cheat

At Pregame a poster DubV accused Greg Shaker of selling his plays. That is what prompted this post.

[shackfu99]

At Pregame a poster DubV accused Greg Shaker of selling his plays. That is what prompted this post.

For the record I have heard this before. Shaker selling other peoples plays. I have no idea if it is true or not.

http://pregame.com/pregame-forums/f/6/t/779344.aspx?pi10417=1

[matthew919]

For the record I have heard this before. Shaker selling other peoples plays. I have no idea if it is true or not.
http://pregame.com/pregame-forums/f/...aspx?pi10417=1

I just read this thread. Now that I know the context it made me chuckle.

Johnny Detroit's response was mathematically incoherent- the number of games per day's card has absolutely no bearing, since they are all statistically independent events. You can take all the games, toss them in a bag, shake them up, order them however you want- you can play all 572 on one day, or one per week for the next ten years- it doesn't change the math, because they are all statistically independent events.

But he's correct on one point: there was an error in your "327,000-1" quote- it should have read 3.27 million to one.

The bottom line is that in the real world, there is no possible way to justify that this is a coincidence. I wouldn't blame someone for feeling slighted when you discover someone is selling your publicly posted plays for profit. But there will always be lazy people feeding off the scraps/steam of the originators. That's just how it is. Try to take it as a compliment, and stop posting publicly if it bothers you.

[SteveRyan]

Putting them together we get:

C(28,7)*C(1116,15) / C(1144,22)

which gives: 3.050937e-07

How do you calculate those factorials? Do you have a calculator for it?

I spent some time trying to figure this out with some help at another forum. One of the people who helped me came up with the same answer you did, but he moved the decimal one place to the left and said -6 instead of -7.

Here is what he came up with:

[Jayvegas420]

I think Bobby Shaker is BennyBigNutz!

[Donkey]

How do you calculate those factorials? Do you have a calculator for it?

I spent some time trying to figure this out with some help at another forum. One of the people who helped me came up with the same answer you did, but he moved the decimal one place to the left and said -6 instead of -7.

Here is what he came up with:

thats because the decimal is before the 3

[matthew919]

How do you calculate those factorials? Do you have a calculator for it?

I spent some time trying to figure this out with some help at another forum. One of the people who helped me came up with the same answer you did, but he moved the decimal one place to the left and said -6 instead of -7.

Here is what he came up with:

I used R's generic choose function. If you write your own, you need to use term by term ratios which will "cancel on the fly", and keep the integers from getting too large.