State target p8 sol

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EaZ_Shadow

1197 posts

EaZ_Shadow

#1 h Apr 6, 2025, 3:53 PM • 3 Y

Y by PikaPika999, Soupboy0, jkim0656

This solution was so educational! It helped me understand how to solve this problem in many ways I couldn't, and I feel so enlightened!

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greenplanet2050

1313 posts

greenplanet2050

#2 h Apr 6, 2025, 3:55 PM • 1 Y

Y by PikaPika999

ikr i literally kinda guessed this one correct

This post has been edited 1 time. Last edited by greenplanet2050, Apr 6, 2025, 3:55 PM

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EaZ_Shadow

1197 posts

EaZ_Shadow

#3 h Apr 6, 2025, 3:56 PM • 1 Y

Y by PikaPika999

greenplanet2050 wrote:

ikr i literally kinda guessed this one correct

Lowk mathcounts was def cooking with that sol!

This post has been edited 1 time. Last edited by EaZ_Shadow, Apr 6, 2025, 3:56 PM

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evt917

2291 posts

evt917

#4 h Apr 6, 2025, 3:56 PM • 1 Y

Y by PikaPika999

i didn't take mathcounts cuz i'm only in 5th grade

but i also checked the problems

the solution was way too good

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DreamineYT

280 posts

DreamineYT

#5 h Apr 6, 2025, 4:01 PM • 1 Y

Y by PikaPika999

EaZ_Shadow wrote:

This solution was so educational! It helped me understand how to solve this problem in many ways I couldn't, and I feel so enlightened!

our team got p1 team wrong :noo:

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sadas123

1172 posts

sadas123

#7 h Apr 6, 2025, 7:39 PM • 1 Y

Y by PikaPika999

It is actually a good solution and I am not being sarcastic skull

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maxamc

530 posts

maxamc

#8 h Apr 6, 2025, 7:47 PM • 1 Y

Y by PikaPika999

i did all 10 team round in 19 min but failed target p8 and i am not being sarcastic skull

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sadas123

1172 posts

sadas123

#9 h Apr 6, 2025, 8:19 PM • 1 Y

Y by PikaPika999

maxamc wrote:

i did all 10 team round in 19 min but failed target p8 and i am not being sarcastic skull

skull i did 11 team rounds and got a perfect on target and sprint to get 46 overall ¿como estas?

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DottedCaculator

7331 posts

DottedCaculator

#10 h Apr 6, 2025, 8:41 PM • 3 Y

Y by PikaPika999, jkim0656, scannose

https://www.eecs.qmul.ac.uk/~norman/papers/probability_puzzles/equal_sequences.shtml

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maxamc

530 posts

maxamc

#11 h Apr 6, 2025, 9:03 PM • 1 Y

Y by PikaPika999

DottedCaculator wrote:

https://www.eecs.qmul.ac.uk/~norman/papers/probability_puzzles/equal_sequences.shtml

https://www.youtube.com/watch?v=t8xqMxlZz9Y

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huajun78

60 posts

huajun78

#12 h Apr 6, 2025, 9:44 PM • 1 Y

Y by PikaPika999

Quote:

The probability of getting consecutive coin tosses with the pattern HTT is (2/3)(1/3)(1/3)=2/27 . The expectation value of the number of tosses needed to first achieve that pattern is the reciprocal of that probability, thus 27/2.

I don't get the solution (why can you just reciprocate it?)

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Craftybutterfly

335 posts

Craftybutterfly

#13 h Apr 6, 2025, 9:51 PM • 1 Y

Y by PikaPika999

Is MATHCOUNTS only for middle school students or also high school students or elementary students as well?

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DottedCaculator

7331 posts

DottedCaculator

#14 h Apr 6, 2025, 9:56 PM • 4 Y

Y by PikaPika999, huajun78, iwastedmyusername, vincentwant

huajun78 wrote:

Quote:

The probability of getting consecutive coin tosses with the pattern HTT is (2/3)(1/3)(1/3)=2/27 . The expectation value of the number of tosses needed to first achieve that pattern is the reciprocal of that probability, thus 27/2.

I don't get the solution (why can you just reciprocate it?)

Consider doing many coin tosses. Each appearance of HTT is disjoint, and the total number of HTTs is 2/27 times the number of tosses, so the expected distance to the next one is 27/2.

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mdk2013

596 posts

mdk2013

#15 h Apr 7, 2025, 12:19 AM

Y by

i did it like dottedcaculator did, everyone was like did you p8 and they were like god that problem was actually hard, and i was like bruh its just that the probability is 2/3*1/3*1/3= 2/27 so the expected number is 27/2 lol you guys shouldve seen my friends expression when i told him that

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scrabbler94

7551 posts

scrabbler94

#16 h Apr 7, 2025, 12:27 AM

Y by

DottedCaculator wrote:

Consider doing many coin tosses. Each appearance of HTT is disjoint, and the total number of HTTs is 2/27 times the number of tosses, so the expected distance to the next one is 27/2.

I like how this solution fails if HTT is replaced with, say, TT. The expected number of flips is not 4 in this case.

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pingpongmerrily

3558 posts

pingpongmerrily

#69 h Friday at 7:13 PM

Y by

bro its only easy if you guess 27/2
i'm pretty sure most people at states aren't able to do states (events with states i mean)

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Soupboy0

317 posts

Soupboy0

#70 h Friday at 7:21 PM

Y by

matthcounts ahh moment

what if they want to trick people in the future so they always reciprocate probabilities

This post has been edited 1 time. Last edited by Soupboy0, Friday at 7:22 PM

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Inaaya

252 posts

Inaaya

#71 h Friday at 8:52 PM

Y by

Soupboy0 wrote:

matthcounts ahh moment

what if they want to trick people in the future so they always reciprocate probabilities

nobody would fall for that because probabilities are all under 1 (self-explanatory)

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Aaronjudgeisgoat

863 posts

Aaronjudgeisgoat

#72 h Yesterday at 2:21 AM

Y by

i think they mean for expected value stuff

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Charizard_637

98 posts

Charizard_637

#73 h Yesterday at 2:32 AM

Y by

I put down 1/(2/27+4/729+8/19683) and got 25/2 (took this and learned ev the next day smh)

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SI113

1 post

SI113

#74 h Yesterday at 10:44 AM

Y by

Inaaya wrote:

Soupboy0 wrote:

matthcounts ahh moment

what if they want to trick people in the future so they always reciprocate probabilities

nobody would fall for that because probabilities are all under 1 (self-explanatory)

Yea but that means you get a value over 1 for the expectation value (the expected number of trials needed on average to attain success). The only case which reciprocating doesn't really work is where p = 0 in 1/p, which gives undefined. This idea comes from geometric distribution.

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MC_ADe

175 posts

MC_ADe

#75 h Yesterday at 2:08 PM • 1 Y

Y by giratina3

I am sure it looks like a coincidence but in this case the reciprocal actually works. To the people that are confused this a repost and a solution.

It's a theorem. Let $X$ be the random variable representing the number of trials needed to get the first success. If $X$ follows the geometric distribution with parameter $p$ the expected value of that geometric distribution is $\frac{1}{p}$ .

A standard derivation:
We know that the expected value $E[X]$ of a discrete random variable $X$ is $\sum$ $x \cdot P(X=x)$
Given that $k$ is the number of trials until the first success we have $P(X=k) = (1-p)^{k-1} \cdot p$
Also the sum of an infinite geometric series is $\left[\sum_{k=0}^{\infty} kr^{k-1}\right] = \frac{1}{(1-r)^2}$

Now, we know that $E[X]$ is given by $\left[\sum_{k=0}^{\infty} kP(X=k) \right]$
Since, $E[X]$ $=$ $\sum_{k=0}^{\infty} k(1-p)^{k-1}p$
Factoring out the $p$ results in $E[X]$ $=$ $p \cdot \sum_{k=0}^{\infty} k(1-p)^{k-1}$

After this, we substitute $r = 1 - p$
Then simplifying gives $\sum_{k=0}^{\infty} k(1-p)^{k-1}$ $=$ $\frac{1}{p^2}$

Now we combine our past two results to get the following:
$E[X] = P \cdot \frac{1}{p^2} = \frac{1}{p}$

We are done.

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giratina3

486 posts

giratina3

#76 h Yesterday at 2:20 PM

Y by

MC_ADe wrote:

I am sure it looks like a coincidence but in this case the reciprocal actually works. To the people that are confused this a repost and a solution.

It's a theorem. Let $X$ be the random variable representing the number of trials needed to get the first success. If $X$ follows the geometric distribution with parameter $p$ the expected value of that geometric distribution is $\frac{1}{p}$ .

A standard derivation:
We know that the expected value $E[X]$ of a discrete random variable $X$ is $\sum$ $x \cdot P(X=x)$
Given that $k$ is the number of trials until the first success we have $P(X=k) = (1-p)^{k-1} \cdot p$
Also the sum of an infinite geometric series is $\left[\sum_{k=0}^{\infty} kr^{k-1}\right] = \frac{1}{(1-r)^2}$

Now, we know that $E[X]$ is given by $\left[\sum_{k=0}^{\infty} kP(X=k) \right]$
Since, $E[X]$ $=$ $\sum_{k=0}^{\infty} k(1-p)^{k-1}p$
Factoring out the $p$ results in $E[X]$ $=$ $p \cdot \sum_{k=0}^{\infty} k(1-p)^{k-1}$

After this, we substitute $r = 1 - p$
Then simplifying gives $\sum_{k=0}^{\infty} k(1-p)^{k-1}$ $=$ $\frac{1}{p^2}$

Now we combine our past two results to get the following:
$E[X] = P \cdot \frac{1}{p^2} = \frac{1}{p}$

We are done.

Wait, this actually works. Oh wow. They should have made the solution this, not just assume it's the reciprocal.

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DottedCaculator

7331 posts

DottedCaculator

#77 h Yesterday at 2:46 PM

Y by

The problem here is that reciprocal requires that the events are independent, but in this case they're not.

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stjwyl

1264 posts

stjwyl

#78 h Yesterday at 2:50 PM

Y by

pingpongmerrily wrote:

i'm lowkey so mad bc all these random people who didn't know how to do anything guessed 27/2 and i tried to like model it with states and stuff and couldn't solve it

Exactlyyy
:sob:
it made the difference between
20th place and like 9th T-T
MA is that tuff

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Craftybutterfly

335 posts

Craftybutterfly

#79 h Yesterday at 3:06 PM

Y by

EaZ_Shadow wrote:

Quote:

As for the people saying the reciprocation of the probability is unjustified, no ... That's how it works, and that is the whole point of probability. If you have 1/x, that means that on average, you will have one result in x tries, and thus 1/x = 2/27 => x = 27/2.

"Thats how it works" I did the expected value and i did the reciprocation for HH, HHH, HHHH, TTTT, HTH. They all contradict. So your logic is incorrect. Additionally probability is not like that. If probability is 1/x, then over a long time you should expect 1 success out of x trials, its not always oh 1 in x trials, so it takes x trials to get 1 success.

Why do you have to reciprocate it?

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EaZ_Shadow

1197 posts

EaZ_Shadow

#80 h Yesterday at 4:46 PM

Y by

Craftybutterfly wrote:

EaZ_Shadow wrote:

Quote:

As for the people saying the reciprocation of the probability is unjustified, no ... That's how it works, and that is the whole point of probability. If you have 1/x, that means that on average, you will have one result in x tries, and thus 1/x = 2/27 => x = 27/2.

"Thats how it works" I did the expected value and i did the reciprocation for HH, HHH, HHHH, TTTT, HTH. They all contradict. So your logic is incorrect. Additionally probability is not like that. If probability is 1/x, then over a long time you should expect 1 success out of x trials, its not always oh 1 in x trials, so it takes x trials to get 1 success.

Why do you have to reciprocate it?

Works sometimes not always

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Shan3t

230 posts

Shan3t

#81 h Yesterday at 4:49 PM

Y by

EaZ_Shadow wrote:

This solution was so educational! It helped me understand how to solve this problem in many ways I couldn't, and I feel so enlightened!

when target p8 sol is as fast sprint p1 sol xD

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Inaaya

252 posts

Inaaya

#82 h Today at 2:16 AM

Y by

I sillied this problem. Don't even ask.

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RandomMathGuy500

57 posts

RandomMathGuy500

#83 h Today at 3:01 AM

Y by

I tried to model it with a geometric series and got 9/2 by some rounding mistake, and there's this kid who wouldn't have even qualified for states if it wasn't for me, got it right by guessing (im 70% sure he copied it off the person next to him)

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