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/sci/ - Science & Math

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5117255 No.5117255 [Reply] [Original]

Is the answer to this 0?
>Complete the following statement:
>In a city of 2,384 people, at least _______ have the same pair of first and last initials.

This is all the information given

>> No.5117268 [DELETED] 

>>5117255
2384/26^2 = ~3.5

26 letters in the alphabet, yo
so that's 676 combinations

so any more than that, and they gotta duplicate initials, rite?

spread out evenly, thats about 3 and a half people sharing.

gotta round up

at least 4 have the same first and last initials

>> No.5117277

>>5117268
>gotta round up

Not this troll again.

>> No.5117285 [DELETED] 
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5117285

>>5117277
come on, bitch
after the 2028th person even if all names are distributed evenly, thats everyone sharing with 2 other people
anyone after that has gotta be in a foursome, and OP's asking for 'at least'

and you cant have half a person, brah

and btw, 3.52 would round up to 4, as nearest whole number anyway

but FYI, even as 3.01, gotta round up in this case

>> No.5117291

>>5117285
>3.52 would round up to 4
No, 3.52 rounds down to 3.5 rounds down to 3

>> No.5117292 [DELETED] 
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5117292

>>5117291
hah!

>> No.5117293

I'm an idiot it just dawned on me. Would the answer be 3 or 4?
26^2 676
2384 / 676

>> No.5117297 [DELETED] 

>>5117293
gotta be 4
he asks for at least, and you know evenly spread, you'll have loads of foursomes and threesomes (about half and half)

course, you could have more, you could have 2000 people all being AA by coincidence, and like only 1 ZZ, but you don't know

but you DO know, that its AT LEAST 4,
you dig?

>> No.5117302

The answer is 1. There could be one person with unique initials.

>> No.5117306

>>5117302
ANY initials that repeat.

>> No.5117307 [DELETED] 

>>5117302
yeh, but what about every other fucker??

far too many to not have any matchups, yo
4 at LEAST

>> No.5117309

>>5117291
epic torll

>> No.5117319 [DELETED] 
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5117319

>>5117309
just a joke, he's making fun of the other thread

>> No.5117326

>>5117319
>>5117307
>>5117292
>>5117297
>>5117285
quit shitting up the board. go back to /soc/

>> No.5117338 [DELETED] 

>>5117326
hey! i gave the right fucking asnwer, bitch!
and i aint even ON /soc/!

>> No.5117366

The question cannot be answered without knowing the relative frequencies of different initials. For sure names starting stupid letters like x or q are less frequent than names starting with cool letters like c or s.

>> No.5117375

>>5117366
Doesn't matter at all for setting a lower bound. Kill yourself.

>> No.5117376

>>5117293
There are no 3.5 people. There are 3 people though.

>> No.5117384

>>5117375
It does matter alot. It can raise the lower bound significantly.

>> No.5117395

>>5117338
why are you so vulgar? Is this how you guys talk in scotland?

>> No.5117397

>>5117384
You're stupid. the LEAST possible amount of people sharing initials would occur with perfectly distributed initials. That's what's being asked

>> No.5117398

>>5117384
No, it can't. Frequencies change the expected number of repeats, but NOT the lower bound on them. There's no reason an imaginary town can't be full of Quentins and Xaviers. Kill yourself.

>> No.5117400

>>5117397
Please take a course on statistics. You're wrong.

>> No.5117404

>>5117398
>There's no reason an imaginary town can't be full of Quentins and Xaviers.

Then it's more likely that 5 or 6 people have these names and this would become the new lower bound.

>> No.5117408

>>5117400
Not sure if troll or reddit-style stupid

>> No.5117409

>>5117404
We're not dealing with what's LIKELY. We're talking about the fewest POSSIBLE repeats.
This is a combinatorics problem, not a statistical problem. KILL YOURSELF. KILL YOURSELF. KILL YOURSELF.

>> No.5117413

What if, in this society, people don't use the Roman alphabet? Or don't have last names?

>> No.5117418

>>5117408
Math says I'm right.

>>5117409
It's inherently a statistics problem. Without knowing the frequency distribution it cannot be solved.

>> No.5117426

>>5117418
>Without knowing the frequency distribution it cannot be solved.
notice it's asking for the THEORETICAL MINIMUM that would be possible under circumstances where people would have lowest possible shared initials

>> No.5117432

>>5117426
Of course it asks the theoretical minimum. For a non-theoretica minimum we would have data and count them. But to calculate the theoretical minimum we need to know the frequencies.

>> No.5117443

Think about it this way, asking what the LEAST number of people DO share initials is the same as asking what is the MOST number of people who do NOT share initials....the MAXIMUM POSSIBLE number of people who DON'T share initials is obviously 675, and for that to occur, the rest ALL have the same initials. 1,709 people are left, so that's your answer (2,384 - 675).

>> No.5117448

>>5117432
that's not the fucking purpose of the exercise

>> No.5117453

>>5117443
What statistical distribution are you using?

>>5117448
How do you know? You weren't able to solve it.

>> No.5117460

>>5117453
Yes I was, the question ask what the lowest amount of people is that could share initials for ANY distribution

>> No.5117462

>>5117460
Then the answer has to show the dependency on the choice of distribution.

>> No.5117474

This question reminds me of the census riddle. Where data is hidden in the wording or phrasing of the question.

A census taker approaches a house and asks the woman who answers the door "How many children do you have, and what are their ages?"

Woman: "I have three children, the product of their ages are 36, the sum of their ages are equal to the address of the house next door."

The census taker walks next door, comes back and says "I need more information."

The woman replies "I have to go, my oldest child is sleeping upstairs."

Census taker: "Thank you, I now have everything I need."

Granted, the one I read, she only says that she'd give him the sum of her children's ages, but that wouldn't help. Couldn't find it online and I don't have the book handy.

>> No.5117520

>>5117474
I don't think I get it... does the fact that the oldest child is sleeping mean that he most be young, limiting the number of solutions?

>> No.5117529

>>5117255
What language/country are we talking about?

>> No.5117541

>>5117520
I always thought this was a bit silly, two children could be the same age in years, but one could be 11 months younger than the other.

>> No.5117546

Worst case scenario: The initials are evenly distributed so they fit 2384/26²~3.5 times into the population. This means there must be groups of 4.

(If I give out the set of initals 3 times - now there are groups of 3 - I will have provided 2028 people with initials, which means there are 320 left for which I have to give out another set. That means there must be groups of 4)

So 4 is our upper boundry, however I would say 0,1,2 and 3 are also valid answers, because of "at least".

>> No.5117577

The minimum is definitely four.

26*26=676

|--------- 2384 ------------------|
|--676--||--676--||--676--||-356-|

Just apply pigeonhole.

>> No.5117642
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5117642

>>5117285
>you cant have half a person
Can too

>> No.5117688

Assuming a mathematically perfect random distribution of letters?

26 * 26 = 676

So every 676 people should have the same initials.

2384 / 676 = 3.52, and you can't have .52 of a person, so that would be at least 3 people.

>> No.5117692

>>5117642
2/3 of a person

>> No.5117708

1708

>> No.5117727

>>5117285
>>5117319
tits or it didnt happen

>>5117255
1809..... thats right /sci/borgs it doesnt matter what the pairs its how many pairs there are. it is just everything after 24^2 -1

it is "the same pair of first and last initials"
if you do the whole pairs of 576 thing you will have at least 4 people in a group with the same first and last initial, all those groups can classify themselves as having "the same pair of first and last initials"...... which is the minimum? ....... well when you redefine each group as "same pair of first and last initials" - in order to answer the question of which is larger - you would need to combine all the like terms after you redefine the groups.
the lowest of the possibilities would thus be 1808+1=1809

checkmate bitches

>> No.5117730

>>5117727
sorry using the English language it would be 1709 peoples

>> No.5117736

just put two

it's still correct