• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Probability Question about Lottery

Turkish

Turkish

Lifer
Lets say, for every colon, you are supposed to pick 6 numbers from 1 to 49 (including both). If you play 100 colons with randomly choosen numbers, what's the probability that you will get the whole 6 numbers correct? Probability for 5 numbers correct? And probability for 4 numbers correct?

Anyone wanna help me? Been a long time since I took statistics 😱
 
hehe, well, i am gonna play some tomorrow, probably for around $60. Just wanted to calculate my chances.

Oh well, anyone wanna help me formulate this?
 
ok i still don't know if i really understand your question but:

assuming you need to pick six numbers from 1 to 49, that gives 49 possibilities per "slot."

that means there are 49^6 possible combinations. that's 13841287201 ways of picking six numbers, 1-49.

if i'm not mistaken (been a long time since i took statistics too) the probability you pick the right numbers in the right order is 1/13,841,287,201
 
Originally posted by: konichiwa
ok i still don't know if i really understand your question but:

assuming you need to pick six numbers from 1 to 49, that gives 49 possibilities per "slot."

that means there are 49^6 possible combinations. that's 13841287201 ways of picking six numbers, 1-49.

if i'm not mistaken (been a long time since i took statistics too) the probability you pick the right numbers in the right order is 1/13,841,287,201
Click to expand...

assuming you are correct, maybe i should save my money :Q

maybe if my chances were in millions, i would try, but not if in billions 😱
 
your odds do not increase if you pick 60 different numbers on all 60 lotto tickets.
i would play the same numbers for 60 drawings.
 
Originally posted by: konichiwa
ok i still don't know if i really understand your question but:

assuming you need to pick six numbers from 1 to 49, that gives 49 possibilities per "slot."

that means there are 49^6 possible combinations. that's 13841287201 ways of picking six numbers, 1-49.

if i'm not mistaken (been a long time since i took statistics too) the probability you pick the right numbers in the right order is 1/13,841,287,201
Click to expand...


<----Pulls math degree out of drawer and blows dust off of it.....let's see here......

You are a little off in your logic here. There are only 49^6 possible numbers if you could pick the same number all 6 times (ex. a draw of 12 12 12 12 12 12), which can't happen. You can only pick each number once, without replacement. This means that you could have 49 * 48 * 47 * 46 *45 *44 = 10,068,347,520 different draws.

Ahhh, but we are not finished yet. We have to take in account of the fact that we could get the same numbers in different orders. (Are we having fun yet!!!??) For example, (1 2 3 4 5 6) is the same as (6 5 4 3 2 1). So we must find out how many ways are possible. This is 6 *5 *4 *3 *2, or 6! = 720.

Now take the 10,068,347,520 and divide it by 720. You get 13,983,816. So your chances of winning are 1 in 13,983,816, or roughly .0000000715112.

I should get back to work, but I'll try to post the other odds (getting 4 or 5) later. Or you could just follow the logic and do it yourself!

 
Originally posted by: SpazzyChicken
Originally posted by: konichiwa
ok i still don't know if i really understand your question but:

assuming you need to pick six numbers from 1 to 49, that gives 49 possibilities per "slot."

that means there are 49^6 possible combinations. that's 13841287201 ways of picking six numbers, 1-49.

if i'm not mistaken (been a long time since i took statistics too) the probability you pick the right numbers in the right order is 1/13,841,287,201
Click to expand...


<----Pulls math degree out of drawer and blows dust off of it.....let's see here......

You are a little off in your logic here. There are only 49^6 possible numbers if you could pick the same number all 6 times (ex. a draw of 12 12 12 12 12 12), which can't happen. You can only pick each number once, without replacement. This means that you could have 49 * 48 * 47 * 46 *45 *44 = 10,068,347,520 different draws.

Ahhh, but we are not finished yet. We have to take in account of the fact that we could get the same numbers in different orders. (Are we having fun yet!!!??) For example, (1 2 3 4 5 6) is the same as (6 5 4 3 2 1). So we must find out how many ways are possible. This is 6 *5 *4 *3 *2, or 6! = 720.

Now take the 10,068,347,520 and divide it by 720. You get 13,983,816. So your chances of winning are 1 in 13,983,816, or roughly .0000000715112.

I should get back to work, but I'll try to post the other odds (getting 4 or 5) later. Or you could just follow the logic and do it yourself!
Click to expand...

I think I just developed a cerberal hemorrhage reading your post 😀

Sysadmin
 
Originally posted by: SpazzyChicken
Originally posted by: konichiwa
ok i still don't know if i really understand your question but:

assuming you need to pick six numbers from 1 to 49, that gives 49 possibilities per "slot."

that means there are 49^6 possible combinations. that's 13841287201 ways of picking six numbers, 1-49.

if i'm not mistaken (been a long time since i took statistics too) the probability you pick the right numbers in the right order is 1/13,841,287,201
Click to expand...


<----Pulls math degree out of drawer and blows dust off of it.....let's see here......

You are a little off in your logic here. There are only 49^6 possible numbers if you could pick the same number all 6 times (ex. a draw of 12 12 12 12 12 12), which can't happen. You can only pick each number once, without replacement. This means that you could have 49 * 48 * 47 * 46 *45 *44 = 10,068,347,520 different draws.

Ahhh, but we are not finished yet. We have to take in account of the fact that we could get the same numbers in different orders. (Are we having fun yet!!!??) For example, (1 2 3 4 5 6) is the same as (6 5 4 3 2 1). So we must find out how many ways are possible. This is 6 *5 *4 *3 *2, or 6! = 720.

Now take the 10,068,347,520 and divide it by 720. You get 13,983,816. So your chances of winning are 1 in 13,983,816, or roughly .0000000715112.

I should get back to work, but I'll try to post the other odds (getting 4 or 5) later. Or you could just follow the logic and do it yourself!
Click to expand...

I see what you're saying but I don't think you're right. Generally in a lottery you can repeat numbers, so you could pick 12 12 12 12 12 12. And as far as I know (I don't play the lottery but...) in terms of getting the winning ticket, if the balls come up the chute as 1 2 3 4 5 6 and your card says 6 5 4 3 2 1, you don't win. You have to get it in order I think...
 
Originally posted by: konichiwa
Originally posted by: SpazzyChicken
Originally posted by: konichiwa
ok i still don't know if i really understand your question but:

assuming you need to pick six numbers from 1 to 49, that gives 49 possibilities per "slot."

that means there are 49^6 possible combinations. that's 13841287201 ways of picking six numbers, 1-49.

if i'm not mistaken (been a long time since i took statistics too) the probability you pick the right numbers in the right order is 1/13,841,287,201
Click to expand...


<----Pulls math degree out of drawer and blows dust off of it.....let's see here......

You are a little off in your logic here. There are only 49^6 possible numbers if you could pick the same number all 6 times (ex. a draw of 12 12 12 12 12 12), which can't happen. You can only pick each number once, without replacement. This means that you could have 49 * 48 * 47 * 46 *45 *44 = 10,068,347,520 different draws.

Ahhh, but we are not finished yet. We have to take in account of the fact that we could get the same numbers in different orders. (Are we having fun yet!!!??) For example, (1 2 3 4 5 6) is the same as (6 5 4 3 2 1). So we must find out how many ways are possible. This is 6 *5 *4 *3 *2, or 6! = 720.

Now take the 10,068,347,520 and divide it by 720. You get 13,983,816. So your chances of winning are 1 in 13,983,816, or roughly .0000000715112.

I should get back to work, but I'll try to post the other odds (getting 4 or 5) later. Or you could just follow the logic and do it yourself!
Click to expand...

I see what you're saying but I don't think you're right. Generally in a lottery you can repeat numbers, so you could pick 12 12 12 12 12 12. And as far as I know (I don't play the lottery but...) in terms of getting the winning ticket, if the balls come up the chute as 1 2 3 4 5 6 and your card says 6 5 4 3 2 1, you don't win. You have to get it in order I think...
Click to expand...

Hey both, thanks for your answers again. In this specific situation, you can't pick the same numbers twice and they don't have to be in any specific order. Thanks again for your answers, after seeing that my chance is not in billions, I still may play tomorrow 😀
 
Originally posted by: konichiwa
Originally posted by: SpazzyChicken
Originally posted by: konichiwa
ok i still don't know if i really understand your question but:

assuming you need to pick six numbers from 1 to 49, that gives 49 possibilities per "slot."

that means there are 49^6 possible combinations. that's 13841287201 ways of picking six numbers, 1-49.

if i'm not mistaken (been a long time since i took statistics too) the probability you pick the right numbers in the right order is 1/13,841,287,201
Click to expand...


<----Pulls math degree out of drawer and blows dust off of it.....let's see here......

You are a little off in your logic here. There are only 49^6 possible numbers if you could pick the same number all 6 times (ex. a draw of 12 12 12 12 12 12), which can't happen. You can only pick each number once, without replacement. This means that you could have 49 * 48 * 47 * 46 *45 *44 = 10,068,347,520 different draws.

Ahhh, but we are not finished yet. We have to take in account of the fact that we could get the same numbers in different orders. (Are we having fun yet!!!??) For example, (1 2 3 4 5 6) is the same as (6 5 4 3 2 1). So we must find out how many ways are possible. This is 6 *5 *4 *3 *2, or 6! = 720.

Now take the 10,068,347,520 and divide it by 720. You get 13,983,816. So your chances of winning are 1 in 13,983,816, or roughly .0000000715112.

I should get back to work, but I'll try to post the other odds (getting 4 or 5) later. Or you could just follow the logic and do it yourself!
Click to expand...

I see what you're saying but I don't think you're right. Generally in a lottery you can repeat numbers, so you could pick 12 12 12 12 12 12. And as far as I know (I don't play the lottery but...) in terms of getting the winning ticket, if the balls come up the chute as 1 2 3 4 5 6 and your card says 6 5 4 3 2 1, you don't win. You have to get it in order I think...
Click to expand...

in each game you can only pick a number once.... so if he puts in 60 dollars, for 60 games for each combination numbers it's still 1 number per each game
 
haha well I think my math was right in the situation I had assumed. And i think for big lotteries (powerball and the like) both of those cases are true...

but who knows 😉
 


EDIT: i had it all wrongn

Redeemed myself.

Figure out the odds

5 out of 6 balls = 54201
4 out of 6 balls = 1032

For fun
3 out of 6 balls = 56.7

so with your $60 statisically you should win at least $1 if you get that for 3 right.
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top