Anonymous at Mon, 27 May 2024 11:40:42 UTC No. 16195669

>>16195649
Yes, it does change.

In the first run by symmetry you can expect half the cointhrows to be heads and the other half tails.
In the second run Bob is more likely to have thrown more heads than tails since he stops after a long heads sequence but not after a long tails sequence. Tyrone of course doesn't have a bias in either direction.

🗑️ Anonymous at Mon, 27 May 2024 11:42:27 UTC No. 16195670

>>16195669
>Bob is more likely to have thrown more heads than tails
*Bob is more likely to have thrown more tails than heads

Anonymous at Mon, 27 May 2024 11:45:00 UTC No. 16195672

>>16195669
>In the second run Bob is more likely to have thrown more heads than tails since he stops after a long heads sequence but not after a long tails sequence.
Proof?

Anonymous at Mon, 27 May 2024 11:58:43 UTC No. 16195682

>>16195672
Let [math]X_i[/math] be the amount of heads in the i-th throw, Let [math]N[/math] be the total number of rounds needed.


[eqn]E \left[ \frac{1}{2N} \sum_{i=0}^N X_i \right] = \sum_{n=3}^\infty E \left[ \frac{1}{2N} \sum_{i=0}^N X_i \middle | N=n \right] P(N=n) = \sum_{n=3}^\infty \frac{1}{2n} \sum_{i=0}^n E[X_i] P(N=n) = \sum_{n=3}^\infty \frac{1}{2} P(N=n) = \frac{1}{2}[/eqn]

Anonymous at Mon, 27 May 2024 12:06:17 UTC No. 16195689

Tyrone doesn't listen to instructions and tries to steal the money

Anonymous at Mon, 27 May 2024 12:19:20 UTC No. 16195698

>>16195682
probably the most complicated way of explaining something so simple.

Anonymous at Mon, 27 May 2024 12:20:34 UTC No. 16195700

>>16195682
In English, professor?

Anonymous at Mon, 27 May 2024 12:20:46 UTC No. 16195701

>>16195698
Can you explain it more simply?

Anonymous at Mon, 27 May 2024 12:30:27 UTC No. 16195708

>>16195669
>In the second run Bob is more likely to have thrown more heads than tails since he stops after a long heads sequence but not after a long tails sequence.
Doesn't preclude a long tails sequence tho
He can get ten tails in a row but never more than three heads

Anonymous at Mon, 27 May 2024 12:50:26 UTC No. 16195727

>>16195701
N=number of coin flips till condition is satisfied.
NJ=For Jane NB=For Bob NT=For Tyrone

Scenario one -
Bob number of head flips = ((NB-3)/2)+3
Jane number of head flips = ((NJ-3)/2)
Tyrone number of heads = NT/2
Thus, total number of heads = (NB+NJ+NT)/2

Scenario two -
Bob number of head flips = ((NB-3)/2)+3
Jane number of head flips = 0
Tyrone number of heads = NT/2
Thus, total number of heads = (((NB-3)/2)+3)+(NT/2)

Overall -
(((NB-3)/2)+3)+(NT/2)
Doesn't equal
(NB+NJ+NT)/2

THE END!

Anonymous at Mon, 27 May 2024 12:52:07 UTC No. 16195729

>>16195727
Pretty sure this is wrong in every aspect.

Anonymous at Mon, 27 May 2024 12:56:03 UTC No. 16195732

>>16195729
Nope.

Anonymous at Mon, 27 May 2024 12:57:49 UTC No. 16195733

>>16195701
Yeah. It's a fair coin. 50/50

Anonymous at Mon, 27 May 2024 13:01:14 UTC No. 16195736

>>16195727
>Bob number of head flips = ((NB-3)/2)+3
The fuck's this

Anonymous at Mon, 27 May 2024 13:02:15 UTC No. 16195738

>>16195733
p. sure this is right

Anonymous at Mon, 27 May 2024 13:06:33 UTC No. 16195744

>>16195682
X_i is dependent on n sorry buddy

Anonymous at Mon, 27 May 2024 13:08:51 UTC No. 16195746

>>16195727
>>16195732
On average, the number of flips until Bob's condition is satisfied would be 14. According to your explanation, shouldn't the expected proportion of heads for a sequence of length 14 be (((14-3)/2)+3)/14 ~= 0.607? If so, that's wrong because the actual value is something near 0.5.

Anonymous at Mon, 27 May 2024 13:21:04 UTC No. 16195754

>>16195746
I just simulated it 1000000 times and bob throws heads 60% of the time

Anonymous at Mon, 27 May 2024 13:21:51 UTC No. 16195755

>>16195746
>If so, that's wrong because the actual value is something near 0.5.
Can you show me how?

I mean I agree with you but in the Monty Hall thread where this was first posted, someone else wrote a script that returned 0.6~ and I can't identify the flaw.

Anonymous at Mon, 27 May 2024 13:25:12 UTC No. 16195760

>>16195754
>>16195755
Maybe I misunderstood. What is ((NB-3)/2)+3 supposed to calculate? Is it the expected number of heads for a sequence with length NB?

Anonymous at Mon, 27 May 2024 13:40:35 UTC No. 16195777

in an average round bob flips 60% heads

but when you add up the total number of heads bob throws across a lot of rounds and divide by the number of flips across those rounds you get 50%

it would appear bob throws high percentage of heads in short games and low percentage of heads in a long games, but I think the answer is 60%

vbs code:

randomize

rounds = 100000
percentage_sum = 0
heads_total = 0
flips_total = 0

for i = 1 to rounds

a = int(2*rnd)
b = int(2*rnd)
c = int(2*rnd)
flips = 3
heads = a + b + c

do while (a + b + c) < 3
c = b
b = a
a = int(2*rnd)
heads = heads + a
flips = flips + 1
loop

percentage_sum = percentage_sum + heads / flips

flips_total = flips_total + flips
heads_total = heads_total + heads

next

msgbox "average percentage of heads per round: " & percentage_sum / rounds
msgbox "total percentage of heads across all rounds: " & heads_total / flips_total

Anonymous at Mon, 27 May 2024 13:42:24 UTC No. 16195782

>>16195760
Looks like it, yes. The last three flips are necessarily heads so he calculates it as for N flips, N-3 are random, and then you add three heads. Which does not seem to account for their randomness.

Anonymous at Mon, 27 May 2024 13:42:48 UTC No. 16195784

>>16195777
>but when you add up the total number of heads bob throws across a lot of rounds and divide by the number of flips across those rounds you get 50%
Well, this is trivial because "across all rounds" is agnostic to the stopping condition of individual rounds, but the question concerns the expected value for a round, not across all rounds.

Anonymous at Mon, 27 May 2024 13:46:07 UTC No. 16195787

>>16195782
Well, I can say with certainty that it doesn't match the number of heads for sequences of length NB, nor does deriving the overall expected proportion of heads from it give an accurate result.

Anonymous at Mon, 27 May 2024 13:48:15 UTC No. 16195793

>>16195777
>in an average round bob flips 60% heads
>but when you add up the total number of heads bob throws across a lot of rounds and divide by the number of flips across those rounds you get 50%
These statements sound incompatible

Anonymous at Mon, 27 May 2024 13:49:30 UTC No. 16195795

>>16195793
They are perfectly compatible.

Anonymous at Mon, 27 May 2024 14:03:14 UTC No. 16195807

>>16195787
It accurately predicts the percentage of heads for Bob and makes logical sense.
Don't see the issue honestly.

Anonymous at Mon, 27 May 2024 14:04:34 UTC No. 16195809

>>16195807
>It accurately predicts the percentage of heads for Bob
It demonstrably doesn't.

>makes logical sense
Not really, because it incorrectly assumes the sequence preceding the 3 heads is unbiased.

Anonymous at Mon, 27 May 2024 14:06:00 UTC No. 16195815

>>16195809
>It demonstrably doesn't.
>>16195746
>shouldn't the expected proportion of heads for a sequence of length 14 be (((14-3)/2)+3)/14 ~= 0.607?
>>16195777
>in an average round bob flips 60% heads

?

Anonymous at Mon, 27 May 2024 14:07:21 UTC No. 16195816

>>16195815
Your point?

Anonymous at Mon, 27 May 2024 14:09:15 UTC No. 16195822

>>16195816
Are you serious?

Anonymous at Mon, 27 May 2024 14:11:56 UTC No. 16195824

>>16195815
it's close to 60% because 14 was a lucky choice, try it for other numbers

Anonymous at Mon, 27 May 2024 14:12:41 UTC No. 16195827

>>16195822
Yes. I'm dead serious. Make an actual point. I remember you from the other thread. You were a moron and you remain a moron. The guy arguing with you had godlike patience.

Anonymous at Mon, 27 May 2024 14:18:14 UTC No. 16195833

>>16195824
>try it for other numbers
Yep. It's easy to see that formula is wrong with a stopping condition of one head. It predicts a value of 3/4 whereas in reality it's 1*(1/2) + (1/2)*(1/4) + (1/3)*(1/8) + (1/4)*(1/16) + ...

Anonymous at Mon, 27 May 2024 14:21:21 UTC No. 16195840

>>16195827
No, the moron from the other thread who refused to make an actual point for a couple dozen posts settled on 60% in the end

Anonymous at Mon, 27 May 2024 14:21:23 UTC No. 16195841

>>16195824
>>16195827
Oh, you are serious.
How embarrassing.
14 isn't a choice.
It's the most common number of coin tosses needed to get three heads or tails in a row.
Simple as.
Checks out.

Anonymous at Mon, 27 May 2024 14:22:03 UTC No. 16195843

>>16195840
>>16195841
Retarded ape.

Anonymous at Mon, 27 May 2024 14:24:09 UTC No. 16195847

>>16195843
What's even the point of namecalling with nothing else? If anything it'll convince people that you have no idea what the right answer is.

Anonymous at Mon, 27 May 2024 14:24:11 UTC No. 16195848

no, it's 50% for both scenarios. no matter how you carve up the results of fair coin flips the expected ratio is always just 50%.

Anonymous at Mon, 27 May 2024 14:25:25 UTC No. 16195852

>>16195847
I've already provided definite proof that "((NB-3)/2)+3" is wrong. If you're not convinced by it, you're not "people".

Anonymous at Mon, 27 May 2024 14:25:34 UTC No. 16195853

>>16195833
>Yep. It's easy to see that formula is wrong with a stopping condition of one head
The formula wasn't made with that condition in mind. It's strictly for the condition of the 3 heads and 3 tails as described in the original meme problem given.

Anonymous at Mon, 27 May 2024 14:26:31 UTC No. 16195855

>>16195853
You were a mentally ill, dysgenic ape and you will forever remain one. I can tell you're the same cretin from the last thread because you're uniquely moronic.

Anonymous at Mon, 27 May 2024 14:27:04 UTC No. 16195856

>>16195852
I didn't even realise what your argument was. I agree with you, but when people start calling each other retarded and refuse to elaborate for several points in a row, it gets hard to keep track.

Anonymous at Mon, 27 May 2024 14:27:11 UTC No. 16195857

>>16195852
>I've already provided definite proof that "((NB-3)/2)+3" is wrong.
Why do the simulated results agree with the formula and disagree with you then?

Anonymous at Mon, 27 May 2024 14:28:12 UTC No. 16195861

>>16195855
>You were a mentally ill, dysgenic ape and you will forever remain one. I can tell you're the same cretin from the last thread because you're uniquely moronic.
That's not an experimental result or mathematical argument anon.
Just saying.

Anonymous at Mon, 27 May 2024 14:29:13 UTC No. 16195862

>>16195847
>What's even the point of namecalling with nothing else? If anything it'll convince people that you have no idea what the right answer is.
The truth is he doesn't know lol.
It's sad.

Anonymous at Mon, 27 May 2024 14:33:32 UTC No. 16195868

>>16195857
>Why do the simulated results agree with the formula
They don't, but this is besides the point. You don't need to simulate anything to see that your logic fails. Just try it with 1 instead of 3 heads.
>inb4 muh reasoning is correct but only for the magic number and only up to the 2nd digit

Anonymous at Mon, 27 May 2024 14:34:13 UTC No. 16195869

>>16195833
>Yep. It's easy to see that formula is wrong with a stopping condition of one head. It predicts a value of 3/4 whereas in reality it's 1*(1/2) + (1/2)*(1/4) + (1/3)*(1/8) + (1/4)*(1/16) + ...
We need another simulation with a one head stopping condition.
You were wrong with the three head stopping condition percentage of heads so you're probably also wrong with the one head stopping condition.

Anonymous at Mon, 27 May 2024 14:34:55 UTC No. 16195870

>>16195856
I refuse to elaborate because I immediately recognized this inbred. You could prove to it that black is black and white is white in 500 different ways over the course of the entire day and it won't budge from its trivial mistakes.

Anonymous at Mon, 27 May 2024 14:36:06 UTC No. 16195872

>>16195869
>We need another simulation with a one head stopping condition.
No, we don't. If you can't figure it out without a simulation you shouldn't be posting ITT.

Anonymous at Mon, 27 May 2024 14:36:07 UTC No. 16195873

>>16195868
>They don't, but this is besides the point. You don't need to simulate anything to see that your logic fails. Just try it with 1 instead of 3 heads.
You were completely wrong with 3 heads anon.
You are also completely wrong with 1.
You got your theory wrong with 3 and reality had to correct you.
Your theory is wrong with 1 and once again reality will correct you again.

Anonymous at Mon, 27 May 2024 14:36:21 UTC No. 16195874

>>16195855
I am not >>16195853 but I did spend a LOT of time arguing with someone who was completely unreasonable and arrogant and also thought that ((NB-3)/2)+3 was the correct formula, like he does. However, because you are not me, and because you yourself have the same attitude as the person I was arguing with, even though you apparently disagree with him, I am quite lost at this point.

Maybe we shouldn't drag drama from other threads into new threads on an anonymous board. Seems quite pointless.

Anonymous at Mon, 27 May 2024 14:37:07 UTC No. 16195876

>>16195873
>You were completely wrong with 3 heads anon.
Where? Quote it in your next post. You can't and won't. Mentally ill mongoloid.

Anonymous at Mon, 27 May 2024 14:37:51 UTC No. 16195877

>>16195872
>o, we don't. If you can't figure it out without a simulation
You failed and the simulation proved you wrong.
You don't get to dictate reality.
Reality dictates you.
Sorry anon.

Anonymous at Mon, 27 May 2024 14:38:36 UTC No. 16195879

>>16195870
You are a thoroughly unpleasant individual and you make threads a chore to read.

I just want to talk about the actual probability problem and if you people can't do that then maybe fuck off.

Anonymous at Mon, 27 May 2024 14:38:49 UTC No. 16195880

>>16195874
This board should just put up stats problems like this as captchas.

Anonymous at Mon, 27 May 2024 14:38:57 UTC No. 16195881

>>16195876
>Where? Quote it in your next post. You can't and won't. Mentally ill mongoloid.
>>16195746
>shouldn't the expected proportion of heads for a sequence of length 14 be (((14-3)/2)+3)/14 ~= 0.607? If so, that's wrong because the actual value is something near 0.5.
>>16195777
>in an average round bob flips 60% heads
?

Anonymous at Mon, 27 May 2024 14:39:56 UTC No. 16195883

>>16195879
>Comes to my thread
>Wants to talk about a probability problem I came up with
>Complains that I am le bad and don't belong in my own thread
Ok. :^(

Anonymous at Mon, 27 May 2024 14:43:12 UTC No. 16195887

>>16195883
>>Comes to my thread
Is actually possessive over a thread because they failed to predict the outcome of a simple probability simulation lol.

Anonymous at Mon, 27 May 2024 14:45:17 UTC No. 16195891

>>16195881
The expected portion of heads for sequences of length 14 is indeed close to 0.5. The formula looks like it's trying to calculate that and fails. By sheer luck, it manages to get close to the expected portion of heads in general (not specifically for sequences of length 14) but it's not exact and it only coincides for a 3-heads stopping condition.

Anonymous at Mon, 27 May 2024 14:45:50 UTC No. 16195893

>>16195874
>but I did spend a LOT of time arguing with someone who was completely unreasonable and arrogant and also thought that ((NB-3)/2)+3 was the correct formula, like he does
Literally didn't happen lol.

Anonymous at Mon, 27 May 2024 14:46:29 UTC No. 16195894

>>16195883
What was the point of this thread if it wasn't to DISCUSS the problem at hand? Just to angrily jerk yourself off whilst telling people they're ignorant cretins and refusing to elaborate?

Anonymous at Mon, 27 May 2024 14:47:51 UTC No. 16195899

>>16195891
>The expected portion of heads for sequences of length 14 is indeed close to 0.5.
It's 0.6 confirmed by simulation anon lol.
You're simply wrong.
>but it's not exact and it only coincides for a 3-heads stopping condition.
And that's wrong.
You were wrong with three and you are wrong with one. Simulate it or shut up.
Easy as.

Anonymous at Mon, 27 May 2024 14:48:52 UTC No. 16195901

>>16195894
He's a low functioning autistic psychopath.

Anonymous at Mon, 27 May 2024 14:49:08 UTC No. 16195902

>>16195893
Well it would be embarrassing to learn that he actually agreed with me and we argued for nothing, but it would be unsurprising given that he seemed to be allergic to making an actual point and loved to argue for the sake of it so most of the time it was impossible to make out what he was even getting at.

Anonymous at Mon, 27 May 2024 14:50:16 UTC No. 16195904

>>16195894
>What was the point of this thread if it wasn't to DISCUSS the problem at hand?
Well, we could have, but the obsessed moron started spamming it with his opinions.

Anonymous at Mon, 27 May 2024 14:51:25 UTC No. 16195906

>>16195904
Lol, you were wrong OP.

Anonymous at Mon, 27 May 2024 14:52:03 UTC No. 16195908

>>16195899
>It's 0.6 confirmed by simulation
Not for Bob's sequences in general. Not for Bob's sequences of length 14. Good thing you insist on shitting the bed like this because now everyone can see that you're utterly incompetent and there's no need to reply to anything you post.

Anonymous at Mon, 27 May 2024 14:53:04 UTC No. 16195910

>>16195908
Yes* for Bob's sequences in general, I mean.

Anonymous at Mon, 27 May 2024 14:53:49 UTC No. 16195911

The internet and its consequences have been a disaster for the human race

I just want to understand the problem man, I don't want to watch a bunch of spergs bicker about their bruised egos

Anonymous at Mon, 27 May 2024 14:54:09 UTC No. 16195913

>>16195908
Sucks that you were wrong OP.
Sucks that you're a sore loser OP.

Anonymous at Mon, 27 May 2024 14:55:04 UTC No. 16195914

>>16195911
Then come up with your own problems and solve them alone, fag.

Anonymous at Mon, 27 May 2024 14:56:14 UTC No. 16195916

>>16195908
>Not for Bob's sequences in general.
Lol, what does that even mean?
>Not for Bob's sequences of length 14.
Simulation proved you wrong OP lol.
>Good thing you insist on shitting the bed like this because now everyone can see that you're utterly incompetent and there's no need to reply to anything you post.
You literally predicted the wrong percentage idiot lol.
If you have an opinion on different stopping conditions you need to simulate it.

Anonymous at Mon, 27 May 2024 14:58:02 UTC No. 16195918

>>16195914
The entire problem is that we're not solving anything here and I might as well be on my own.

Fuck it, >>16195733 is correct, prove me wrong

Anonymous at Mon, 27 May 2024 15:00:01 UTC No. 16195922

>>16195891
>>16195916
he's right you know

Anonymous at Mon, 27 May 2024 15:00:51 UTC No. 16195924

>>16195918
>The entire problem is that we're not solving anything here and I might as well be on my own.
Well you're basically just talking to yourself anyway autismos.
You already have the solution.

Anonymous at Mon, 27 May 2024 15:02:32 UTC No. 16195926

>>16195922
>he's right you know
Aha... You're talking about a differing thing anon.
>>16195777
>in an average round bob flips 60% heads
I'm answering the original question.
You're not.
Because you were wrong lol.

Anonymous at Mon, 27 May 2024 15:03:00 UTC No. 16195928

Consider the situtation with only Bob in the case where you stop after the first head then you have

>50% chance for only heads
>25% chance for a 1/2 ratio of heads
>12.5% chance for a 1/3 ratio of heads
>6.25% chance for a 1/4 ratio of heads
...
>1/2^k chance for a 1/k ratio of heads
So the expected percentage of heads is
[eqn]\sum_{k=1}^\infty \frac{1}{k} \frac{1}{2^k} = \log(2) \approx 69.3 \%[/eqn]

It's not 50% here so it's likely also not 50% in OP's problem.

Anonymous at Mon, 27 May 2024 15:04:11 UTC No. 16195930

>>16195918
He's incorrect because for Bob's shorter sequences, there will be a significant bias in favor of heads, meanwhile for his long sequences, there will be a small bias in favor of tails. There is also a bias in favor of short sequences.

Anonymous at Mon, 27 May 2024 15:05:28 UTC No. 16195933

>>16195926
he said:
>expected portion of heads for sequences of length 14
that's what the simulation estimates

Anonymous at Mon, 27 May 2024 15:08:42 UTC No. 16195937

>>16195933
Yes, and...

Anonymous at Mon, 27 May 2024 15:09:43 UTC No. 16195940

>>16195928
>It's not 50% here so it's likely also not 50% in OP's problem.
It's already been simulated to be 60% lol.

Anonymous at Mon, 27 May 2024 15:10:22 UTC No. 16195941

>>16195940
A simulation is not a mathematical proof.

Anonymous at Mon, 27 May 2024 15:11:38 UTC No. 16195943

>>16195937
so he wasn't wrong about what he said, that formula doesn't actually calculate anything

Anonymous at Mon, 27 May 2024 15:12:06 UTC No. 16195945

>>16195924
>You already have the solution.
I have people bickering amongst each other insisting that the other is wrong and refusing to elaborate.

Anonymous at Mon, 27 May 2024 15:14:28 UTC No. 16195951

>>16195928
Nice. I've shown this earlier ITT. If nothing else, it demonstrates that the sloppy "reasoning" people use to argue it's 1/2 is invalid. Decent attempt but it doesn't quite settled the question.

Anonymous at Mon, 27 May 2024 15:18:27 UTC No. 16195955

>>16195943
>>16195941
It's 60% idiot anon lol.

Anonymous at Mon, 27 May 2024 15:20:02 UTC No. 16195959

>>16195955
it's not. i simulated it yesterday. it doesn't match the result of your formula

Anonymous at Mon, 27 May 2024 15:21:40 UTC No. 16195964

>>16195955
What makes you so sure it isn't 59.999999999% or 60.000000001% instead?

Anonymous at Mon, 27 May 2024 15:22:49 UTC No. 16195967

>>16195955
Wrong. It's neither 60% nor what your retarded formula computes.

Anonymous at Mon, 27 May 2024 15:27:51 UTC No. 16195972

>>16195967
>>16195964
>>16195959
It's already been confirmed to be 60% lol
>>16195777
>in an average round bob flips 60% heads

Anonymous at Mon, 27 May 2024 15:28:57 UTC No. 16195974

>>16195972
it's more like 60.2% which disagrees with your calculation

Anonymous at Mon, 27 May 2024 15:32:10 UTC No. 16195981

>>16195974
>it's more like 60.2% which disagrees with your calculation
Lol, that's closer to my calculation of 60.7%

Anonymous at Mon, 27 May 2024 15:36:01 UTC No. 16195986

>>16195981
your calculation is wrong, sorry. first digit after the decimal point should be 2

Anonymous at Mon, 27 May 2024 15:42:57 UTC No. 16195990

>>16195986
>your calculation is wrong, sorry. first digit after the decimal point should be 2
Or it's right and you need more tosses to get to 60.7?
Do you have a more correct answer?
nope...
sad

Anonymous at Mon, 27 May 2024 15:43:53 UTC No. 16195991

>>16195848
I believed I proved this. I'm a /g/tard so I dunno how to do whatever fancy formatting you use but here:

Calculate the expected value of number of heads for Bob:
Consider the T*, HT*, HHT* and HHH cases separately:
E(num_heads) = 1/2 * [E(num_heads)] + 1/4 * [1 + E(num_heads)] + 1/8 * [2 + E(num_heads)] + 1/8 * [3]
simplifying we get the following recurrence:
E(num_heads) = 7/8 * [E(num_heads) + 1]

Calculate the expected value of number of tails for Bob:
Like last time, consider the T*, HT*, HHT* and HHH cases separately:
E(num_tails) = 1/2 * [1 + E(num_tails)] + 1/4 * [1 + E(num_tails)] + 1/8 * [1 + E(num_tails)] + 1/8 * [0]
simplifying we get the following recurrence:
E(num_tails) = 7/8 * [E(num_tails) + 1]

These recurrences are identical so E(num_heads) = E(num_tails) meaning the expected ratio for Bob is 0.5. By symmetry, it's 0.5 for Jane. Tyrone's is trivially 0.5. The ratio doesn't change

Anonymous at Mon, 27 May 2024 15:47:09 UTC No. 16195997

>>16195986
You're arguing with a legit mental patient. Just stop.

Anonymous at Mon, 27 May 2024 15:48:55 UTC No. 16196000

>>16195991
Try your logic on a different stopping condition, like 1 head.

Anonymous at Mon, 27 May 2024 15:50:14 UTC No. 16196002

>>16196000
>>16195997
>>16195991
simulate it or shut up lol.

Anonymous at Mon, 27 May 2024 15:53:43 UTC No. 16196006

I still don't know what the correct solution is, but I do know what failed to convince me in the past:

copious namecalling
repeating that some answer is wrong
refusing to elaborate
refusing to acknowledge criticisms

Some of you try these for hours expecting different results and no matter how good you are at maths, that's pretty fucking dumb.

Anonymous at Mon, 27 May 2024 15:56:26 UTC No. 16196009

>>16196000
tried it and it also checks out. In both cases the recurrance simplifies to
E(x) = 1/2 * [E(x) + 1]

Anonymous at Mon, 27 May 2024 15:57:19 UTC No. 16196010

>>16196006
No one cares, homo.

Anonymous at Mon, 27 May 2024 16:00:14 UTC No. 16196016

>>16195991
Can you explain why >>16195777 is wrong?

Anonymous at Mon, 27 May 2024 16:01:51 UTC No. 16196020

>>16196016
It's not. His calculations are.

Anonymous at Mon, 27 May 2024 16:03:59 UTC No. 16196026

>>16196020
Why?

Anonymous at Mon, 27 May 2024 16:07:14 UTC No. 16196031

>>16195649
It doesn't, 50-50 change always. I believe this is called an "independent event."

Anonymous at Mon, 27 May 2024 16:11:33 UTC No. 16196035

>>16196026
Why what? Why the code he posted isn't wrong?

Anonymous at Mon, 27 May 2024 16:12:51 UTC No. 16196039

>>16196035
Why are his calculations wrong?

Anonymous at Mon, 27 May 2024 16:13:52 UTC No. 16196040

>>16196035
>>16196039
But sure, why isn't the code wrong works too. Then why is >>16195991 wrong?

Anonymous at Mon, 27 May 2024 16:16:27 UTC No. 16196044

>>16196039
>>16196040
I looked at his code and I see that 1. there's nothing wrong with it 2. it gives a result that I know to be correct. Why the other guy is wrong is for that guy to figure out.

Anonymous at Mon, 27 May 2024 16:18:32 UTC No. 16196049

>>16196044
Well, the other guy's code looks fine to me and gives a result that I know to be correct, so you figure out why you're wrong then.

Anonymous at Mon, 27 May 2024 16:18:50 UTC No. 16196050

>>16196049
Your mental illness is out of control.

Anonymous at Mon, 27 May 2024 16:19:33 UTC No. 16196052

>/sc/i - arguing with the most annoying people you know

Anonymous at Mon, 27 May 2024 16:23:09 UTC No. 16196058

>>16196050
There are several people in this thread who give the same answer I do, including one who has written code that produces that answer. So I once again refer you to >>16196006

I am extremely amenable to correction. I am open to admitting that I'm wrong. I am absolutely begging you to tell me why I am wrong. If only all out-of-control mental illness could be alleviated by simply saying "no, that's wrong, and here's why, see".

But you don't. You repeatedly, explicitly refuse to. I think if anyone here is mentally ill, it's you. Case in point: I'm going to once again explicitly point out to you that calling people mentally ill and refusing to elaborate is not going to convince anyone you're correct. You're probably going to insist on doing it anyway and grow increasingly agitated that people continue to be wrong.

Anonymous at Mon, 27 May 2024 16:29:18 UTC No. 16196068

>>16196058
>There are several people in this thread who give the same answer I do
What's your answer and whose code corroborates it?

Anonymous at Mon, 27 May 2024 16:30:25 UTC No. 16196070

Okay dumbest solution : 7 conditions where a heads can occur and plus 1 for no heads making it 87.5%

Next one person out so 3 conditions where a heads can occur plus 1 where no head occurred major it 75%

So yes it does change!

Anonymous at Mon, 27 May 2024 16:31:41 UTC No. 16196072

>>16196050
You know, you do us both a disservice. Being wrong is how we learn. Your insistence on only interacting with people you already agree with robs us all of the opportunity to improve, and furthermore does not inspire confidence that your opinion is actually well-considered at all.

Anonymous at Mon, 27 May 2024 16:32:42 UTC No. 16196074

>>16196068
see >>16195991
You know, the one I *just* asked about?

Anonymous at Mon, 27 May 2024 16:32:54 UTC No. 16196075

>>16196072
I don't care. So long as it triggers you to shit out paragraph after paragraph and samefag like this, I'm satisfied.

Anonymous at Mon, 27 May 2024 16:33:44 UTC No. 16196076

>>16196070
I don't know what you're trying to do here

Anonymous at Mon, 27 May 2024 16:33:57 UTC No. 16196078

>>16196074
Yeah, your mental illness is definitely out of control. Be sure to reply to my post pretending to be several different people.

Anonymous at Mon, 27 May 2024 16:34:45 UTC No. 16196080

>>16196075
>>16196078
What has made you like this, Anon? You pitiable excuse for a human being.

Anonymous at Mon, 27 May 2024 16:36:05 UTC No. 16196082

Anyway, back to sanity
>>16195991
Anon are you willing to explain what your code does? I want to know the case for both sides so I can understand them better and thus far neither is very cooperative.

Anonymous at Mon, 27 May 2024 16:37:32 UTC No. 16196085

>>16196080
Just take your meds or something.

Anonymous at Mon, 27 May 2024 16:37:41 UTC No. 16196087

>>16196076
Me neither!

Anonymous at Mon, 27 May 2024 16:39:16 UTC No. 16196091

>>16196085
Anon, you think being open to the possibility that you are wrong is a mental illness.

Anonymous at Mon, 27 May 2024 16:42:53 UTC No. 16196094

>>16196091
No, but replying to me over and over trying to win my approval, while spouting schizobabble, is.

Anonymous at Mon, 27 May 2024 16:45:41 UTC No. 16196100

>>16195991
nvm, E[heads/(heads + tails)] != E[heads]/(E[heads] + E[tails])

Anonymous at Mon, 27 May 2024 16:47:18 UTC No. 16196103

>>16196094
Your approval is worth less than nothing to me. I would think less of myself if I had somehow earned your approval. I am trying to see if you have one tiny trace of humanity left to appeal to. A tiny spark of cognizance that raises you above the level of an automaton. Thus far, I keep being disappointed.

I suppose it is useless after all to attempt to reason with you. Feel free to respond once more, hypocrite. It won't gain you my approval either, though you easily know how to get it.

Anonymous at Mon, 27 May 2024 16:48:19 UTC No. 16196105

>>16196100
What's the difference?

Anonymous at Mon, 27 May 2024 16:53:01 UTC No. 16196111

>>16196103
Didn't read. Keep begging for my approval. You'll never get it.

Anonymous at Mon, 27 May 2024 16:54:35 UTC No. 16196115

>>16196100
division is not linear so you can't just move the expectation inside like that. my computation for the expected number of heads and expected number of tails is right but the actual expected ratio is tricker

Anonymous at Mon, 27 May 2024 16:56:42 UTC No. 16196120

>>16196105
>>16196115

Anonymous at Mon, 27 May 2024 17:09:01 UTC No. 16196132

Just remembered someone in the other thread argued that it is impossible for Bob to get three tails in a row lol

Anonymous at Mon, 27 May 2024 17:48:32 UTC No. 16196177

Lol, you faggots are still going. Crazy.
Question was answered ages ago in a very simple manner and it is the only solution that agrees with simulations.
Stupidest thread ever.
Okay, here is a question for the Idiots here.
What's the simplest way of calculating the average number of coin flips needed to get N number of heads in a row.
The solution doesn't require a Markov chain either.
Are you smart enough!!!

Anonymous at Mon, 27 May 2024 17:53:04 UTC No. 16196179

>>16196177
>Question was answered ages ago in a very simple manner
Remind us?

Anonymous at Mon, 27 May 2024 17:56:33 UTC No. 16196181

>>16196179
What's the average number of coin flips needed to get 1000 heads in a row?
Are you smart enough to directly calculate that without markov chains anon?

Anonymous at Mon, 27 May 2024 18:00:00 UTC No. 16196185

>>16196181
21430172143725346418968500981200036211228096234110672148875007767407021022498722449863967576313917162551893458351062936503742905713846280871969155149397149607869135549648461970842149210124742283755908364306092949967163882534797535118331087892154125829142392955373084335320859663305248773674411336138750

Now what's the simple answer you referred to?

Anonymous at Mon, 27 May 2024 18:02:22 UTC No. 16196189

>>16196185
how did you calculate that anon?
that is the correct answer.

Anonymous at Mon, 27 May 2024 18:03:37 UTC No. 16196191

>>16196189
Lucky guess.

Anonymous at Mon, 27 May 2024 18:05:18 UTC No. 16196198

Reminder to ignore the mentally ill bot.

Anonymous at Mon, 27 May 2024 18:06:17 UTC No. 16196202

>>16195649
Since it's one person less flipping coins in the second run, that means 1/3 flips and therefore 1/3 less heads

Anonymous at Mon, 27 May 2024 18:08:04 UTC No. 16196204

>>16196191
>Lucky guess.
Come on anon.
>>16196198
>Reminder to ignore the mentally ill bot.
Your question was answered long ago anon.

Anonymous at Mon, 27 May 2024 18:10:51 UTC No. 16196209

>>16196198
The thing you do so well, huh

Anonymous at Mon, 27 May 2024 18:15:08 UTC No. 16196213

>>16196204
I just googled it desu

I'd google the answer to this bullshit too but it's harder to find.

Anonymous at Mon, 27 May 2024 18:16:06 UTC No. 16196216

>>16195727

Anonymous at Mon, 27 May 2024 18:24:11 UTC No. 16196230

>>16196213
>I just googled it desu
That doesn't explain anything.
Did you use a Markov chain in some code or a simple formula without code automation?
One is easy, the other isn't.

Anonymous at Mon, 27 May 2024 18:25:52 UTC No. 16196234

>>16196230
A simple formula.

I'm not sure why this is so important to you. I also asked you a question, remember?

Anonymous at Mon, 27 May 2024 18:27:44 UTC No. 16196239

>>16196234
>A simple formula.
Lol, ok so you just used a super long markov chain with some code.
Got ya.
Guess I was expecting a bit too much.

Anonymous at Mon, 27 May 2024 18:32:13 UTC No. 16196243

>>16196239
>Lol, ok so you just used a super long markov chain with some code.
I'm pretty sure I just told you the opposite.

I don't know what you're expecting here. This conversation seems entirely pointless. If you're not going to answer my queston I've no interest in your silly pastime.

Anonymous at Mon, 27 May 2024 18:33:13 UTC No. 16196245

>>16196243
>I'm pretty sure I just told you the opposite.
Where's the formula anon?
>>16196243
>I don't know what you're expecting here. This conversation seems entirely pointless. If you're not going to answer my queston I've no interest in your silly pastime.
Your question was answered long ago.

Anonymous at Mon, 27 May 2024 18:33:15 UTC No. 16196246

The expected number of coin flips until Bob is done is 14 as >>16195746 asserted (tons of proofs exists online). The expected number of heads for Bob is indeed 7 as >>16195991 proved. However nobody in this thread actually found the expected percentage of heads for Bob (or across the players in the second game), which isn't 50% and can't easily be deduced from Bob's expected number of heads

Anonymous at Mon, 27 May 2024 18:34:13 UTC No. 16196248

>>16196245
>Where's the formula anon?
Where's the answer, Anon?

Anonymous at Mon, 27 May 2024 18:35:14 UTC No. 16196252

>>16196246
The expected number of heads for Bob doesn't change. The expected percentage of total heads changes.

Anonymous at Mon, 27 May 2024 18:35:17 UTC No. 16196253

>>16196246
>>16195777
Almost forgot, this simulation does seem correct in brute forcing Bob's expected percentage of heads.

Anonymous at Mon, 27 May 2024 18:35:42 UTC No. 16196254

>>16196246
Wrong.
Lol, this has already been answered ages ago.
This is like a dude just mindlessly fighting against the fact that 1+1=2.
You're going to be here forever.
Literally forever.

Anonymous at Mon, 27 May 2024 18:37:26 UTC No. 16196258

>>16196254
So which answer do you hold to be correct, Anon? Because there are several in this thread.

Anonymous at Mon, 27 May 2024 18:58:10 UTC No. 16196282

>they are still replying to the mentally ill bot

Anonymous at Mon, 27 May 2024 18:59:10 UTC No. 16196283

>>16196254
>You're going to be here forever.
Only until someone gives a straightforward explanation of the right answer
so yes

Anonymous at Mon, 27 May 2024 19:46:57 UTC No. 16196331

>>16196283
>Only until someone gives a straightforward explanation of the right answer
I did so ages ago:

>>16195930
>for Bob's shorter sequences, there will be a significant bias in favor of heads, meanwhile for his long sequences, there will be a small bias in favor of tails. There is also a bias in favor of short sequences.

Anonymous at Mon, 27 May 2024 20:03:17 UTC No. 16196350

>>16196331
>There is also a bias in favor of short sequences.
We've established that Bob is expected to need 14 coin tosses, is that short?

Anonymous at Mon, 27 May 2024 20:07:33 UTC No. 16196360

>>16196331
>no proof
into the trash it goes

Anonymous at Mon, 27 May 2024 20:07:57 UTC No. 16196361

>>16196350
>We've established that Bob is expected to need 14 coin tosses, is that short?
We've established no such thing. I just stated it way at the beginning. And no, that's relatively long and pretty much the cutoff point.

Anonymous at Mon, 27 May 2024 20:15:58 UTC No. 16196380

>>16196361
No, it's the expected value, i.e. the mean

Anonymous at Mon, 27 May 2024 20:17:21 UTC No. 16196384

>>16196380
>it's the expected value, i.e. the mean
It is. So what?

Anonymous at Mon, 27 May 2024 20:18:41 UTC No. 16196391

>>16196384
So then it is not "pretty much the cut-off point" and if the mean is relatively long then there is no bias towards short sequences.

Anonymous at Mon, 27 May 2024 20:20:11 UTC No. 16196398

>>16196391
>So then it is not "pretty much the cut-off point"
Why not?

>if the mean is relatively long then there is no bias towards short sequences.
Why?

Anonymous at Mon, 27 May 2024 20:31:37 UTC No. 16196422

>>16196398
If most of the sequences are shorter than 14 and 14 is an upper bound then how is 14 ever going to be the mean?

Anonymous at Mon, 27 May 2024 20:35:41 UTC No. 16196429

>>16196422
>14 is an upper bound
Upper bound on what? That's just the point where the bias shifts.

>how is 14 ever going to be the mean?
Because the distribution is not at all symmetric.

Anonymous at Mon, 27 May 2024 20:39:25 UTC No. 16196437

>>16196429
>That's just the point where the bias shifts
You're not making any sense, man.

Anonymous at Mon, 27 May 2024 20:40:29 UTC No. 16196440

>>16196437
I'm not really talking to you. I know you're the same mouth-breathing moron. I'm just saying it for the record.

Anonymous at Mon, 27 May 2024 20:46:49 UTC No. 16196451

>>16196440
You say there is a bias towards shorter sequences. But the bias shifts?

You know if you would ever just straightforwardly make your point you would find me an eager learner. But you insist on being an obtuse narcissist instead and dragging this out for whatever perverse amusement you get out of playing the oracle.

Anonymous at Mon, 27 May 2024 20:49:33 UTC No. 16196456

>>16196451
There is a bias towards heads overall because there is a bias towards heads in short sequences and there is a bias towards short sequences. Sequences of length 14 are slightly biased towards tails. Anything longer is even more biased towards tails.

Anonymous at Mon, 27 May 2024 20:55:17 UTC No. 16196465

>>16196456
See, now how hard was it for you to express yourself in clear language?
>Sequences of length 14 are slightly biased towards tails.
Why is that the point where it shifts? Just because it's the mean? Those things are entirely different.

Why are you even here, man, if explaining things makes you so angry?
Why do you continue to answer questions but as unhelpfully as you can?

Anonymous at Mon, 27 May 2024 20:58:03 UTC No. 16196477

>>16196456
still no proof

Anonymous at Mon, 27 May 2024 21:24:24 UTC No. 16196532

>>16196465
>Why is that the point where it shifts?
There is no "straightforward" way to answer this. Here are all the possible sequences for Bob for sequence lengths from 5 to 8, where 1 represents a head:

>00111
>10111

>000111
>010111
>100111
>110111

>0000111
>0010111
>0100111
>0110111
>1000111
>1010111
>1100111

>00001110
>00010111
>00100111
>00110111
>01000111
>01010111
>01100111
>10000111
>10010111
>10100111
>10110111
>11000111
>11010111

For each of these matrices, sum each row:
>3, 4
>3, 4, 4, 5
>3, 4, 4, 5, 4, 5, 5
>3, 4, 4, 5, 4, 5, 5, 4, 5, 5, 6, 5, 6

For each of these sequences, count the occurrences of each number:
>3x1, 4x1
>3x1, 4x2, 5x1,
>3x1, 4x3, 5x3,
>3x1, 4x4, 5x6, 6x2

Leave only the counts:
>1 1
>1 2 1
>1 3 3
>1 4 6 2

Notice anything funny? Pic related shows more of these patterns. I'm not gonna spoil it completely but this is related to Pascal's triangle.

Anonymous at Mon, 27 May 2024 21:54:06 UTC No. 16196604

>>16196532
>Notice anything funny?
I see the first count is always 1; one sequence with only three heads.
The second count keeps going up by one
The third count is increased by as much as the second count of the previous line
Can't see the pattern yet for 4+

Anonymous at Mon, 27 May 2024 23:20:12 UTC No. 16196752

I came up with a much faster way to approximate the expected value of the heads percentage for Bob. Code and brief explanation shortly.
Results are consistent with the simulation anon posted above