/sci/ - Science & Math

[Anonymous No.4760450]

alright, me and a co-worker are having an argument and none of us know anything about math or odds so we were wondering if you could help us.

Imagine a lottery where you have to guess 5 numbers from a universe that ranges from 1 to 50.

If your lottery ticket has 5 numbers all above 37 (for example 37, 41, 42, 47, 48) are the odds of winning always the same? or does the fact that every number is above 37 reduce the odds?

I think it does but he thinks it doesn't because every number has the same odd. But isn't it less likely that all numbers would be in the "house" of the 30's and 40's because it has the same odds as beings in the 10's, 20's?

pic unrelated

[Anonymous No.4760455]

Doesn't matter. As long as the lottery is fair, all combinations of numbers have exactly the same chance of being selected.

[Anonymous No.4760456]

Your friend is right.

[Anonymous No.4760459]

>>4760455
>>4760456
fuck me.

I made such a huge deal about being right. He's going to give me so much shit...

[Anonymous No.4760460]

Think of cards. You have just as much landing a straight made from 10, J, Q, K, A as you do landing 4, 5, 6, 7, 8, 9

Pretty much the same thing.

[Anonymous No.4760461]

>>4760460
I understand his reasoning.

I just thought that if every number was above 37 that would reduce the "universe" and therefore reduce the odds.

Guess I'm an idiot

>>4760455
>>4760456
forgot to say thanks, so thanks

[Anonymous No.4760463]

>>4760461
>Guess I'm an idiot
It's okay. There are a lot of us out there. We are working on getting t-shirts.

[Anonymous No.4760464]

>>4760461
It really depends how you look at the problem. If you're simply asking what the odds are for a given combination of numbers, the answer is that all combinations are equally likely. If you however introduce additional rules like, say, how are the odds for combinations with only even numbers then it's clear that the odds are far lower because there are far less combinations of only even numbers than all other combinations.

[Anonymous No.4760468]

>>4760461
No your right theres a 13/50 of getting numbers higher than 37 and a 37/50 of getting numbers less than 37.

[Anonymous No.4760471]

>>4760463
send me one right away, I need it badly.

>>4760464
got it. we were actually talking about a real lottery, and I was mad that the machine randomly picked all numbers above 37 and he said the odds were the same.

>>4760468
what? can someone confirm this?

[Anonymous No.4760474]

>>4760468
Also if you pick a number over 37 the chances of getting another number over 37 will decrease even further to 16/49 15/48 14/47 etc.

[Anonymous No.4760481]

>>4760471
yeah, then the answer is all are equally likely to be drawn.

[Anonymous No.4760484]

>>4760455
>>4760456
>>4760460
These guys are all trolling.
>>4760464
This guy is also incorrect because there are much more combinations you can get from 37 numbers than 13

[Anonymous No.4760485]

>>4760481
but if every drawn number has the same odds of being in the 10's, 20's. 30's, 40's and number 50, doesn't the fact that the ticket's numbers are all only in the 30's and 40's reduce the odds?

this is making my head hurt, I should have payed more attention in math class.

[Anonymous No.4760489]

>>4760485
Okay here i go, Yes there is an equal chance of drawing 1 specific combination of numbers over 37 and 1 specific combination under 37. But there are many more combinations of numbers under 37 than over 37 so there is a greater likelyhood of getting 5 numbers under 37.

[Anonymous No.4760508]

>>4760489
This is exactly what I think.

After arguing with my co-worker again, I think I've figured out where we differ in thought process.
He says that every number is drawn individually and therefore, when the first number is drawn, it has a 1/50 chance of being x number. And when the second number is drawn, it has a 1/49 chance of being y number.

However, I still think that if the first number drawn is 49, the odds of the second one being 50 are much smaller than being any other number under 49. He disagrees.

[Anonymous No.4760526]

>>4760508
if the first one is 49 the odds of getting 50 is 1/49 the odds of getting anything else is 48/49.

[Anonymous No.4760545]

>>4760526
yes, this is exactly what I told my co-worker.

but he says that in a lottery you don't want ANY number, but a specific one, and that that specific number still has the same odds as any other of being drawn.

So I tried this: in a universe from 1 to 10, the number 7 is drawn first. Now, the odds of it being a number under 7 are obviously larger than it being over 7. Therefore, if what you need is a 9 then your are odds are slimmer than it being any number under 7 because the universe of 1-6 is larger than the one of 8-10.

But he says that the universe are not reduced and that they all still have the same choice. This has escalated to the point we're thinking of walking into a local university and approaching some professor for the answer.

[Anonymous No.4760567]

>>4760450

Disregard the primary problem for a second.

If I charged you the standard fee for a ticket in this lottery, would you feel cheated if you had all the numbers above 37?

[Anonymous No.4760568]

>>4760545
Man, you're just really missing this basic concept of probability. The probability of getting any number is always the same. If you're choosing from 1-10 and the first is 9, obviously the chance of getting 10 on the next is much smaller than the chance of getting 1 OR 2 OR 3 OR 4 OR 5 OR 6 OR 7 OR 8, but the chance of getting 1 is equal to the chance of getting 10, the chance of getting 2 is equal to the chance of getting 10, and so on.

No matter what constraints you apply, when you buy a lotto ticket you pick 5 numbers, and every single combination of numbers has exactly the same probability of being selected, 1 2 3 4 5 has the same probability as 37 39 48 43 50 which has the same probability as 12 17 29 32 8, and so picking all numbers above 37 has absolutely zero effect on whether the 5 specific numbers you pick are the same as the 5 specific numbers chosen.

[Anonymous No.4760580]

>>4760484
The question you're asking isn't what the OP asked.

The odds of all of the numbers selected by the lottery being above 37 are much lower than the odds of the numbers all being below 37. This is correct.

However, this does not change the fact that every combination of numbers is equally likely, regardless of if they're all above or below a certain number (or have no constraints).

[Anonymous No.4760603]

>>4760567
I did have all numbers over 37. I wouldn't say I feet cheated, I just feel I have a smaller chance of winning now, which prompted my co-worker to tell me the odds are all the same which led to this whole thing.

>>4760580
I understand your argument completely.

But that only seems correct at the start of the draw. Once a number is drawn, the odds of the numbers being drawn is affected. So if the first number is 37, the odds of all other 4 numbers drawn being within 37 and 50 are greatly reduced. I could be wrong though

[Anonymous No.4760607]

>>4760508
Okay, there's a higher chance of the winning combo to be all under 37 than over 37. HOWEVER, each individual combination has the same chance to win. It's just the ranges that have different probabilities, because they have multiple combinations. But you're only doing one combo anyway, so that doesn't matter. Equal probabilities

[Anonymous No.4760610]

Ok, just a little bit of math here.

Assuming that the lottery is 5 numbers from 1-50, no replacement (so the same number can't appear twice), and order doesn't matter, there are:

50! / ( 45! * 5! )

Possible outcomes. If I trust my poking it in to a calculator, that's 2,118,760 outcomes.

All are equally likely.

Now, the odds of all of them coming up greater than 37:

[ 13! / ( 8! * 5!) ] / [ 50! / ( 45! * 5! ) ]

Which comes out to... about 0.06%. Which is, as the OP noted, tiny.

Now!

The way those odds are calculated is, basically, by calculating the number of combinations with all numbers greater than 37 -- [ 13! / ( 8! * 5! ) ], which is a mere 1287.


The trick is, each of those 1287 combinations is equally likely to occur, and has exactly the same chance of coming up as any other specific combination.

The core thing here is that any specific combination of numbers has exactly the same odds of coming up; membership in an arbitrarily unlikely group has zero effect on those odds.

[Anonymous No.4760614]

>>4760603
I'm not sure what you're looking for -- that the odds change during the drawing is fairly meaningless, as you don't get to pick numbers during the draw, is it not?

[Anonymous No.4760617]

>>4760603
Okay, let's say there's a game where you pick a number from 1-10, and you win if you're correct. Each number has a 1/10 chance to win right?
So for numbers above 8, it's two numbers, so that's 2*1/10 = 2/10 chance of being over 8, and 8/10 of being below. But each number still only has 1/10 chance. 9 has a 1/10 chance. 4 has a 1/10 chance. It's only a group of numbers of different size that have different chances. But you're only picking one number, and all numbers have the same chance.

[Anonymous No.4760628]

Hey, OP, do a mental exercise. Switch around the numbers. Say 37 was called 1, and 1 was called 37. Now not all numbers are above 30. Does this change your chances of winning? Of course not, because every number has the same chance of coming up.

And what you say here: >>4760461

>I just thought that if every number was above 37 that would reduce the "universe" and therefore reduce the odds.

It would reduce the space of outcomes if you required that every number DRAWN was over 37. Now you require that every number drawn corresponds to one of YOUR numbers. And the chance is the same no matter which numbers they are.

[Anonymous No.4760637]

>>4760610
>>4760617
I get it now. The odds are the same for every number regardless of how small a category I group them in.

>>4760614
I see your point. And unfortunately no, I don't pick the numbers, otherwise I'd be a millionaire.

>>4760628
>It would reduce the space of outcomes if you required that every number DRAWN was over 37

This line confused me again for a second because I DO require every number to over 37, but I get it now.

Thanks everyone. Time to eat shit from my co-worker for all eternity.

[Anonymous No.4760793]

Whatabout 4-8-15-16-23-42?

[Anonymous No.4760814]

>>4760793
you will share you winnings with 8000 nerds