# Foo - Statistics question

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View Full Version : Statistics question

timmyquest
11-19-07, 10:49 PM
So i went out with my cousin and some friends her friends...to the church, for some bingo (yeah baby).

Anyways, after a game where we didn't win (again), one of her friends said "It was probably one of those people with a ton of cards." This lead to a discussion on statistics to which i claimed that your odds were the same regardless of how many cards you have. She wouldn't have any of it.

I don't think she's entirely wrong, and i don't think I'm entirely wrong. It's been a few years since I've taken a stats class and none of the statistics i deal with for my schoolwork is related to odds/probability etc.

So, if you have one dice you have a 1/6 chance of rolling a six. If you have two dice, each dice still has a 1/6 chance of rolling a dice. So, your chance is still 1/6 of rolling a 6 right? Or is it 2/12...which is the same thing. Right, i just figured it out didn't I (why didn't i bring that fact up earlier).

Tom Stormcrowe
11-19-07, 10:53 PM
Each dice individually has a 1:6, but to roll simultaneous 6's it'd be 1²:6² or 1:36

So i went out with my cousin and some friends her friends...to the church, for some bingo (yeah baby).

Anyways, after a game where we didn't win (again), one of her friends said "It was probably one of those people with a ton of cards." This lead to a discussion on statistics to which i claimed that your odds were the same regardless of how many cards you have. She wouldn't have any of it.

I don't think she's entirely wrong, and i don't think I'm entirely wrong. It's been a few years since I've taken a stats class and none of the statistics i deal with for my schoolwork is related to odds/probability etc.

So, if you have one dice you have a 1/6 chance of rolling a six. If you have two dice, each dice still has a 1/6 chance of rolling a dice. So, your chance is still 1/6 of rolling a 6 right? Or is it 2/12...which is the same thing. Right, i just figured it out didn't I (why didn't i bring that fact up earlier).

KingTermite
11-19-07, 10:55 PM
So, if you have one dice you have a 1/6 chance of rolling a six. If you have two dice, each dice still has a 1/6 chance of rolling a dice. So, your chance is still 1/6 of rolling a 6 right? Or is it 2/12...which is the same thing. Right, i just figured it out didn't I (why didn't i bring that fact up earlier).
I believe it would be 1/6 + 1/6 or 2/6 = 1/3.

More cards equals N number of more chances to get the number called. Now there are more factors involved with a bingo card...there are only so many possible numbers in each column, and there is a possibility of repeat numbers, but still your chances are definitely going to be increased with more bingo cards.

KingTermite
11-19-07, 10:56 PM
Each dice individually has a 1:6, but to roll simultaneous 6's it'd be 1²:6² or 1:36I don't think he referred to simultaneous...I think he meant the chance of rolling one 6 with 2 rolls (e.g. 2 chances).

timmyquest
11-19-07, 10:57 PM
Each dice individually has a 1:6, but to roll simultaneous 6's it'd be 1²:6² or 1:36

If we are to relate it to BINGO, i don't think you'd apply this rule because you don't need a bingo on each card, just one card.

jschen
11-19-07, 10:57 PM
Argh... timmyquest, of course the number of attempts matters. Let's take a simple example to convince ourselves of this. If you flip a fair coin, you have a 50% chance of getting heads. If you flip 100 coins, you are virtually assured of getting at least one coin coming up heads. Right?

My next post will analyze your two dice issue. You can generalize from there.

jschen
11-19-07, 11:00 PM
Your chances with one die are as following:

1/6 you roll a 6
5/6 you roll not a 6

Since the results on the two dice are independent events, you can simply multiply probabilities:

5/6 * 5/6 = 25/36 you roll not a 6 on both dice
1/6 * 5/6 = 5/36 you roll a 6 on the first die but not the second
5/6 * 1/6 = 5/36 you roll a 6 on the second die but not the first
1/6 * 1/6 = 1/36 you roll a 6 on both dice

So 11/36 chance you get at least one 6. Which should be clear without analyzing the entire situation by simply asking how many ways you don't get any 6's.

Next post: How this applies to a classic math problem involving birthdays...

jschen
11-19-07, 11:10 PM
Suppose you ask what the probability is that in a group of 30 people picked at random, at least two share a birthday (not counting the year). And for ease of math, assume no one born on Feb 29, though the results don't change significantly (and it's not hard to calculate).

You can build a complex equation, but let's just analyze the approach using common sense. With one person, obviously there can't be a shared birthday. Probability 1 that there is no shared birthday. With two people, there is a 364/365 chance the second person doesn't share the first person's birthday. Suppose that's the case. (Because if it isn't the case, you can stop checking the others' birthdays.) Now two dates out of 365 are already claimed, and there is a 363/365 chance the third person doesn't share a birthday with either the first person or the second person. So on and so forth...

So for the Nth person you add to the group of people, there is a (366-N)/365 chance that if you started with no shared birthdays, after adding that person, you still don't have a shared birthday.

So for N people, you have (365/365) * (364/365) * ... * (366-N)/365 chance, simplifying to 365!/(365-N)!/365^N chance, of having no shared birthdays. So you have 1 - [365!/(365-N)!/365^N] chance you have a shared birthday.

For N = 30, there's a ~70% chance of a shared birthday.

jschen
11-19-07, 11:11 PM
Hope that helps. Let me know if you have any questions.

jschen
11-19-07, 11:20 PM
I believe it would be 1/6 + 1/6 or 2/6 = 1/3.

This is a good approximation if the odds of a single event are very low and the number of trials is very low. (How low depends on how good an approximation you want.) For the case of two dice, 1/3 = 12/36 is a pretty good back of the envelope estimate for the real answer of 11/36.

When those criteria don't hold, you'll overestimate your chances. If you flip a fair coin twice, your chances of getting at least one heads is 3/4. But this estimate predicts probability 1. The reason you overestimate is easily analyzed in this model case. You're counting the times you flip two heads as two successful cases.

Without getting into a lot of math (I have to get back to writing for work, and besides, I don't want to bore Foosters more than I already have) this is why it is a good approximation for low probability events. You aren't likely to get more than one hit (if you get one at all), so you can basically just multiply the probability for a single attempt by the number of attempts. But if there is a significant possibility of multiple hits, then this approximation becomes screwy.

Okay, as much fun as this is compared to writing for work, I should get back to work. :o

mezza
11-19-07, 11:40 PM
When I studied marketing I had a statistics exam.

I rushed through the paper answering all the questions I knew instantly.
I rushed through a second time answering all the questions I'd have to do a maximum of 20 seconds working out on.
I then worked out that I had pretty much got the pass mark already and all I had to do was guess the rest, so I did.

I aced a 3 hour exam in 20 minutes :)

deraltekluge
11-20-07, 12:06 PM
If you flip a fair coin, you have a 50% chance of getting heads. If you flip 100 coins, you are virtually assured of getting at least one coin coming up heads. Right? It depends on who's doing the flipping. There are people who can flip a coin and get what they want every time.

And of course having more cards in bingo improves your chances of winning, just as playing more sets of numbers in a lottery improves your chances there.

Think of it this way: Suppose that there are 100 cards being played. Divide them into two sets of 50. There is an equal chance of the winner being in each set. Now, suppose one player has 50 cards, and the other cards are distributed among 50 other players. The player with 50 cards will win half the time, and each of the 1-card players will win 1/100 of the time.

deraltekluge
11-20-07, 12:09 PM
Some terminology: If you have two or more, they are dice. If you have just one, it's a die...remember the phrase, "The die is cast."

timmyquest
11-20-07, 12:18 PM
And of course having more cards in bingo improves your chances of winning, just as playing more sets of numbers in a lottery improves your chances there.

This is actually where my confusion came from. The odds in most lotto games are so astronomical that buying a few more tickets does not significantly improve your odds.

1 in 30,000,000 is about the same as 10 in 30,000,000

CliftonGK1
11-20-07, 12:22 PM
Some terminology: If you have two or more, they are dice. If you have just one, it's a die...remember the phrase, "The die is cast."

What if they were press-form molded, rather than cast?

squegeeboo
11-20-07, 12:25 PM
Some terminology: If you have two or more, they are dice. If you have just one, it's a die...remember the phrase, "The die is cast."

I always thought the plural of thingen was always en. so it would be dien, or perhaps dicen. Just like micen or goosen.

zbicyclist
11-20-07, 12:35 PM
[QUOTE=deraltekluge;5666299]

And of course having more cards in bingo improves your chances of winning, just as playing more sets of numbers in a lottery improves your chances there.
[QUOTE]

But playing more cards, or playing more sets of lottery numbers, also increases the amount you lose. Since both are negative sum games, your expected winnings are higher in both cases if you don't play at all.

deraltekluge
11-20-07, 12:36 PM
This is actually where my confusion came from. The odds in most lotto games are so astronomical that buying a few more tickets does not significantly improve your odds.

1 in 30,000,000 is about the same as 10 in 30,000,000Roughly speaking, you have twice the chances of winning with two tickets as you do with one, and ten times the chances with ten tickets.

Take it to an extreme. Suppose the odds are 14 million to one (that's about what it is for picking 6 out of 49). Suppose you buy 14 million lottery tickets, all different...you're certain to win.

Some people actually tried to do that in a lottery where the pay-off had grown so much that it was much greater than the cost of the tickets. The didn't succeed in buying all those tickets...they ran out of time...but they won anyway. The state tried to refuse to pay, but the last I heard was that a court had awarded them the money.

ModoVincere
11-20-07, 12:37 PM
what's it matter anyway....if its a gambling game, the odds are gonna be stacked against you.
You know the old saying..."What ever you bring to vegas, stays in vegas" :D

deraltekluge
11-20-07, 12:39 PM
But playing more cards, or playing more sets of lottery numbers, also increases the amount you lose. Since both are negative sum games, your expected winnings are higher in both cases if you don't play at all. But that's no fun.

squegeeboo
11-20-07, 12:41 PM
Roughly speaking, you have twice the chances of winning with two tickets as you do with one, and ten times the chances with ten tickets.

Take it to an extreme. Suppose the odds are 14 million to one (that's about what it is for picking 6 out of 49). Suppose you buy 14 million lottery tickets, all different...you're certain to win.

Some people actually tried to do that in a lottery where the pay-off had grown so much that it was much greater than the cost of the tickets. The didn't succeed in buying all those tickets...they ran out of time...but they won anyway. The state tried to refuse to pay, but the last I heard was that a court had awarded them the money.

Last time I heard of someone doing something like that, there ended up being 3-4 winners, so the one guy who bought several mill in tickets ended up in the red a few hundred thousand.

timmyquest
11-20-07, 01:05 PM
Roughly speaking, you have twice the chances of winning with two tickets as you do with one, and ten times the chances with ten tickets.

Take it to an extreme. Suppose the odds are 14 million to one (that's about what it is for picking 6 out of 49). Suppose you buy 14 million lottery tickets, all different...you're certain to win.

Some people actually tried to do that in a lottery where the pay-off had grown so much that it was much greater than the cost of the tickets. The didn't succeed in buying all those tickets...they ran out of time...but they won anyway. The state tried to refuse to pay, but the last I heard was that a court had awarded them the money.

twice of nothing is still nothing ;)

Statistically insignificant increases in your odds are essentially nothing. So if your odds are essentially nothing, and you increase them by essentially nothing, then again you are still left with nothing. I'm not saying that applies with BINGO, i don't know the odds there, but lets just think of this:

I was at a table with 8 people and 10 cards all together.

That would cost \$90 to buy those cards.

No one at my table won.

...seems as though the increase in odds wasn't very important was it.

jschen
11-20-07, 01:25 PM
twice of nothing is still nothing ;)

True, but it's still twice as much. Also, for the lottery players in aggregate (or put another way, for the person running the lottery), it does matter since the sample size is so huge. Whether having more chances increases odds is something mathematically quantifiable. It most definitely does. Whether or not that amount matters is a matter of perspective.

...seems as though the increase in odds wasn't very important was it.

Statistics doesn't predict with accuracy what will happen any one time in a small sample size. It predicts averaged behavior over very large numbers. Your "experiment" didn't involve very large numbers.

eubi
11-20-07, 01:44 PM
. If you flip a fair coin, you have a 50% chance of getting heads.

Generally true, but twice in my life, with God as my witness, I have flipped a coin that landed on its edge.

So it's slightly less than 50%.

:D

How many sides does a coin have? Three!

jschen
11-20-07, 01:55 PM
Fair enough. :)

CdCf
11-20-07, 02:03 PM
This is actually where my confusion came from. The odds in most lotto games are so astronomical that buying a few more tickets does not significantly improve your odds.

1 in 30,000,000 is about the same as 10 in 30,000,000

I'm a member of the council for the municipal water production and distribution organisation in Gothenburg. The tech guys briefed us on waterborne diseases a while ago.

With the level of water processing currently in use, the calculated risk of infection is 1 in 10 000 per year.

That means that on an individual level, I pretty much need to drink the water for 10 000 years to be certain of getting infected. Doesn't sound so bad, right?

Now think about it from another perspective. The city of Gothenburg has around 500 000 people. 1 in 10 000 per year means on average 50 cases of infection per year. If you were that likely to win the lottery, I bet you'd buy a ticket right away... :D

KingTermite
11-20-07, 02:05 PM
Generally true, but twice in my life, with God as my witness, I have flipped a coin that landed on its edge.

So it's slightly less than 50%.

:D

How many sides does a coin have? Three!

You been talking to Darrin from Bewitched during his spin on the Twilight Zone?

erraticrider
11-20-07, 02:33 PM
On the bingo game, you must not forget that it is not just a game of chance. There is the potential for human error -- that is to say, your skill cannot increase your probability of winning, but your lack of skill can definitely decrease it.

So, if you are playing one card sitting at a table with five beautiful women (or hot guys if that is your preference), you very well may have a lower probability of winning than the person two tables over does on each of the 10 cards she is playing with intense interest.