Maths boffins Odds and Probability

Bald-Eagle22 · 11 Mar 2023 at 12:10

Guys need some help to work the last part of this out, probability was not my strong point at school.

Scenario. Literally a bag with some numbers in it and the same number gets pulled out three times in a row three weeks apart.

1. 17 different numbers go in the bag and i Pull out number 6
2. 18 different numbers go in the bag and I pull out number 6
3. 17 different numbers go in the bag and I pull out Number 6

I was 8th in turn to pull a Number out and each number that was pulled out was not put back in the bag .

Odds of this happening are ?

Dirk Diggler · 11 Mar 2023 at 12:32

Wouldn't you just multiply the odds?

17*18*17= 5202/1

Mr C · 11 Mar 2023 at 12:45

Bald-Eagle22 said:
Guys need some help to work the last part of this out, probability was not my strong point at school.

Scenario. Literally a bag with some numbers in it and the same number gets pulled out three times in a row three weeks apart.

1. 17 different numbers go in the bag and i Pull out number 6
2. 18 different numbers go in the bag and I pull out number 6
3. 17 different numbers go in the bag and I pull out Number 6

I was 8th in turn to pull a Number out and each number that was pulled out was not put back in the bag .

Odds of this happening are ?

With the current amount of information, completely unknown, you mention 17/18 numbers go into a bag, are they all numbered 6 or are they numerically numbered 1-17/18, you mention the number pulled out isn't put back but take out 3 6's in the 3 scenarios are there multiple drawings from the same bag in one weekly lottery or are these three seperate lotteries with a singular 6 in each time.

Assuming that we are talking 3 seperate lotteries, the chances that a 6 comes out is the same as any other number (assuming 1 of every number is in the bag, i.e. 1-18). This also assumes that the 6 is drawn first, is this in the middle of a draw, do we draw how many times per lottery, if we pull 7 or 8 numbers per lottery then the chances of seeing a 6 every time is relatively high considering we are withdrawing half the bag almost.

One final note is that you mention you was 8th in turn on a bag with 17/18 numbers, that would imply the chances of you drawing a 6 on each time would have been 1/9, 1/10 and 1/9 for each weekly draw, they cannot (reasonably) be multiplied as they are completely independent events and therefore the chance of you drawing a 6 in next weeks draw in the same position (e.g. 8th out of a bag of 18) is going to be once again 1/10. Unless you are placing a bet on drawing a 6 for 3 weeks in a row, only then would it be reasonable to infer the odds of such a thing.

The human mind is a wonderful thing and sees many great patterns but when it comes into statistical modelling it's dull as dishwater and attempts to see statistical correlations (especially in random distributions) that simply do not exist.

Bald-Eagle22 · 11 Mar 2023 at 12:54

Mr C said:
With the current amount of information, completely unknown, you mention 17/18 numbers go into a bag, are they all numbered 6 or are they numerically numbered 1-17/18, you mention the number pulled out isn't put back but take out 3 6's in the 3 scenarios are there multiple drawings from the same bag in one weekly lottery or are these three seperate lotteries with a singular 6 in each time.

17 or 18 different numbers ( 1 to 17/18 )each drawn out by different people and they don’t go back in the bag ,I’ve been 8th in turn each week

Mr C said:
Assuming that we are talking 3 seperate lotteries, the chances that a 6 comes out is the same as any other number (assuming 1 of every number is in the bag, i.e. 1-18). This also assumes that the 6 is drawn first, is this in the middle of a draw, do we draw how many times per lottery, if we pull 7 or 8 numbers per lottery then the chances of seeing a 6 every time is relatively high considering we are withdrawing half the bag almost.

Its 1 draw per person , ie 17 numbers 17 people each get 1 draw . Same happened three weeks on the trot !

Mr C said:
One final note is that you mention you was 8th in turn on a bag with 17/18 numbers, that would imply the chances of you drawing a 6 on each time would have been 1/9, 1/10 and 1/9 for each weekly draw, they cannot (reasonably) be multiplied as they are completely independent events and therefore the chance of you drawing a 6 in next weeks draw in the same position (e.g. 8th out of a bag of 18) is going to be once again 1/10. Unless you are placing a bet on drawing a 6 for 3 weeks in a row, only then would it be reasonable to infer the odds of such a thing.

The human mind is a wonderful thing and sees many great patterns but when it comes into statistical modelling it's dull as dishwater and attempts to see statistical correlations (especially in random distributions) that simply do not exist.

Was just wondering if it had any bearing .

Bald-Eagle22 · 11 Mar 2023 at 12:54

Dirk Diggler said:
Wouldn't you just multiply the odds?

17*18*17= 5202/1

What I thought but can’t be that simple .

Mysterae_ · 11 Mar 2023 at 13:18

Bald-Eagle22 said:
Guys need some help to work the last part of this out, probability was not my strong point at school.

Scenario. Literally a bag with some numbers in it and the same number gets pulled out three times in a row three weeks apart.

1. 17 different numbers go in the bag and i Pull out number 6
2. 18 different numbers go in the bag and I pull out number 6
3. 17 different numbers go in the bag and I pull out Number 6

I was 8th in turn to pull a Number out and each number that was pulled out was not put back in the bag .

Odds of this happening are ?

I sense the Devil may be at work here

.

dowie · 11 Mar 2023 at 13:23

Bald-Eagle22 said:
I was 8th in turn to pull a Number out and each number that was pulled out was not put back in the bag .

Odds of this happening are ?

It's unclear what you're asking, I think from the first line you're asking about the probability of an event occurring but you've then thrown in some extraneous information and later mention odds, odds are not the same probability: https://towardsdatascience.com/probability-vs-odds-f47fbc6789f4

If you're just asking what the probability of drawing a 6 in three consecutive draws out of 17, 18 and 17 would be then Dirk's answer is along the right lines but wrong:

Dirk Diggler said:
Wouldn't you just multiply the odds?

17*18*17= 5202/1

The probbaility is (1/17) * (1/18) * (1/17) = 1/5202

that's equivalent to odds of 5201:1

You've also thrown in extraneous info in about your position in each of the the draws but not how that was selected and it doens't seem like you're really asking anything relevant to that.

Mr C said:
With the current amount of information, completely unknown, you mention 17/18 numbers go into a bag, are they all numbered 6 or are they numerically numbered 1-17/18

It doesn't matter if they're numbered numerically from 1 to 17, he stated in the OP that they're different numbers, they could have 100, 428, 699 etc.. on them, it's irrelevant, you only need to know that they're different.

Vanquish-Storm · 11 Mar 2023 at 13:34

Is it not 1/9, 1/10 & 1/9 if OP is 8th pick

Derek W · 11 Mar 2023 at 13:36

OP you need mrwong to help you out. He had a sterling display of odds and probability in the last few pages of the premium bonds topic

jpaul · 11 Mar 2023 at 13:37

for first draw probability you get 6 on the 8th withdrawal is product of not getting it on first 7 withdrawals times probability on 8th ie
first draw 16/17*15/16*14/15*13/14*12/13*11/12*10/11*1/10

and then multiply by similar products for other 2 draws

dowie · 11 Mar 2023 at 16:46

Vanquish-Storm said:
Is it not 1/9, 1/10 & 1/9 if OP is 8th pick

No

jpaul said:
for first draw probability you get 6 on the 8th withdrawal is product of not getting it on first 7 withdrawals times probability on 8th ie
first draw 16/17*15/16*14/15*13/14*12/13*11/12*10/11*1/10

and then multiply by similar products for other 2 draws

You're over complicating things; if it's a fair process then the person in position 1 has the same chance of being allocated the 6 as the OP or indeed the person in position 17.

You don't need to add extra steps here to calculate that that is 1/17 for a draw where there are 17 numbers and he's allocated 1.

Think about it

positon 1: 1/17
positon 2: 16/17*1/16 = 1/17
position 3: 16/17*15/16*1/15 = 1/17

and so on...

You just need to know he's getting 1 number allocated at random from 17 as he hasn't asked anything about his positon or told us how that's allocated.

Minusorange · 11 Mar 2023 at 17:01

Dirk Diggler said:
Wouldn't you just multiply the odds?

17*18*17= 5202/1

You have to square the 18 and hexagon the last 17

Bald-Eagle22 · 11 Mar 2023 at 17:07

dowie said:
No

You're over complicating things; if it's a fair process then the person in position 1 has the same chance of being allocated the 6 as the OP or indeed the person in position 17.

You don't need to add extra steps here to calculate that that is 1/17 for a draw where there are 17 numbers and he's allocated 1.

Think about it

positon 1: 1/17
positon 2: 16/17*1/16 = 1/17
position 3: 16/17*15/16*1/15 = 1/17

and so on...

You just need to know he's getting 1 number allocated at random from 17 as he hasn't asked anything about his positon or told us how that's allocated.

Position is called by the organiser who takes names on a list from the prior week , your name then gets called out on that list so it’s probably going to stay the same each time until I don’t attend ( it’s a sporting event of sorts ).

Grrrrr · 11 Mar 2023 at 17:12

dowie said:
The probbaility is (1/17) * (1/18) * (1/17) = 1/5202

This is correct for the odds of drawing any specific number (e.g. 6).

But if the question is "What are the odds of getting the same number three weeks in a row?" It's just:

(1/18) * (1/17) = 1/306

dowie · 11 Mar 2023 at 18:39

Grrrrr said:
This is correct for the odds of drawing any specific number (e.g. 6).

But if the question is "What are the odds of getting the same number three weeks in a row?" It's just:

(1/18) * (1/17) = 1/306

Yeah, that's true.

I guess it depends what the OP is asking as he's not clarified... "I got a 6 this week, what were the chances of that?": 1/17 "A 6 was also what I got the week before and the week before that too, what were the chances of all three draws giving me a 6" etc...

Bald-Eagle22 said:
Position is called by the organiser who takes names on a list from the prior week , your name then gets called out on that list so it’s probably going to stay the same each time until I don’t attend ( it’s a sporting event of sorts ).

OK so I guess there isn't really anything being asked about the positon ergo it's just one of the two answers above depending on whether you're asking what were the chances of you specifically getting a 6 in the three draws or, as per @Grrrrr 's answer re: what are the chances you simply had 3 numbers in a row. I guess perhaps it's actually that which you're asking, you've noticed the same number three times ergo you're surprised.

Bald-Eagle22 · 11 Mar 2023 at 18:46

dowie said:
Yeah, that's true.

I guess it depends what the OP is asking as he's not clarified... "I got a 6 this week, what were the chances of that?": 1/17 "A 6 was also what I got the week before and the week before that too, what were the chances of all three draws giving me a 6" etc...

Yes exactly this thanks .

dowie said:
OK so I guess there isn't really anything being asked about the positon ergo it's just one of the two answers above depending on whether you're asking what were the chances of you specifically getting a 6 in the three draws or, as per @Grrrrr 's answer re: what are the chances you simply had 3 numbers in a row. I guess perhaps it's actually that which you're asking, you've noticed the same number three times ergo you're surprised.

Vanquish-Storm · 11 Mar 2023 at 18:48

dowie said:
No

You're over complicating things; if it's a fair process then the person in position 1 has the same chance of being allocated the 6 as the OP or indeed the person in position 17.

You don't need to add extra steps here to calculate that that is 1/17 for a draw where there are 17 numbers and he's allocated 1.

Think about it

positon 1: 1/17
positon 2: 16/17*1/16 = 1/17
position 3: 16/17*15/16*1/15 = 1/17

and so on...

You just need to know he's getting 1 number allocated at random from 17 as he hasn't asked anything about his positon or told us how that's allocated.

Why 1/17 when OP gets to his pick there’s 9 balls in the pot. He’s said he was 8th.

AndrewNR · 11 Mar 2023 at 18:52

I guess because he was asking the odds prior to any balls being picked.
Yes, if he was asking for the odds when he picked it after the first 8 were not a 6 it would be 1/9.

dowie · 11 Mar 2023 at 22:38

Vanquish-Storm said:
Why 1/17 when OP gets to his pick there’s 9 balls in the pot. He’s said he was 8th.

No one said it was 1/17 when OP gets to his pick given the number 6 hasn't been picked by then but also no one asked that in the first place.

And regardless of that, it's not clear what scenario you were trying to represent when combining probabilities.

Vanquish-Storm · 11 Mar 2023 at 23:26

dowie said:
No one said it was 1/17 when OP gets to his pick given the number 6 hasn't been picked by then but also no one asked that in the first place.

And regardless of that, it's not clear what scenario you were trying to represent when combining probabilities.

Each time the OP went to pick the 6 hadn’t been picked. So at the time the OP went to pick those were the odds that he would pick the 6 no?