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Calculate the probability of death

Anonymous Tue 23 Nov 13:39:44 2021 No.13882619 View Reply Original Report

Quoted By: >>13882645 >>13882753 >>13882887 >>13882955 >>13882960 >>13882974 >>13883038 >>13883168

There are 10 bottles. One of them contains poison. A bottle is drank after a person rolls dubs that specify its number (00 is 1, 11 is 2, ..., 99 is 10). A random remaining bottle is drank after a person rolls trips. The bottle with poison is drank after a person rolls quads. Anything above quads has the same effect as them
Calculate the probability of death after x posts

Anonymous

Anonymous Tue 23 Nov 2021 13:41:35 No.13882625 Report

Quoted By:

roll

Anonymous

Anonymous Tue 23 Nov 2021 13:50:40 No.13882645 Report

Quoted By: >>13882648 >>13882724

>>13882619
After x posts meaning your post is post no. x+1, or meaning you make x number of posts that are assigned random post numbers?

Anonymous

Anonymous Tue 23 Nov 2021 13:51:59 No.13882648 Report

Quoted By: >>13882724

>>13882645
as on 4chan. 1st post in a thread is 1, 2nd is 2, ... yth is y.

Anonymous

Anonymous Tue 23 Nov 2021 14:26:35 No.13882724 Report

Quoted By: >>13882743

>>13882648
You didn't understand my question at all. What do you mean by "after x posts?"

You haven't given enough info to answer your question. It depends on what the last known post is, the rate of posting on the board, how fast it takes you to post, etc.>>13882645

Anonymous

Anonymous Tue 23 Nov 2021 14:35:48 No.13882743 Report

Quoted By: >>13882755

>>13882724
>Anything above quads has the same effect as them
Suit the data to the question. If you want to, consider it to start at 1, this being the only thread, posting being instaneous, etc.

Anonymous

Anonymous Tue 23 Nov 2021 14:41:02 No.13882753 Report

Quoted By:

>>13882619
Either you drink poison or you don't. 50/50

Anonymous

Anonymous Tue 23 Nov 2021 14:41:58 No.13882755 Report

Quoted By: >>13882859

>>13882743
>still not answering the question
I'm out.

Anonymous

Anonymous Tue 23 Nov 2021 15:41:01 No.13882859 Report

Quoted By:

>>13882755
Alright bitch drink bottle 6

Anonymous

Anonymous Tue 23 Nov 2021 15:49:25 No.13882887 Report

Quoted By: >>13882983 >>13882992

>>13882619
the problem does not specify the death of whom, nor who is supposed to drink the bottles after dubs/trips/quads are rolled.
Nor does the problem specify exactly when the bottle should be drank, it only mentiones that it should occur after posting, could well be a matter of years or generations.
It is not a well posed problem as is.

Anonymous

Anonymous Tue 23 Nov 2021 16:21:00 No.13882955 Report

Quoted By: >>13882956

>>13882619
1/1000 for the quads or higher
(1/100-1/1000)/5 for the trips
(1/10-1/100)/10 for the dubs

Anonymous

Anonymous Tue 23 Nov 2021 16:22:01 No.13882956 Report

Quoted By:

>>13882955
1.18%

Anonymous

Anonymous Tue 23 Nov 2021 16:24:38 No.13882960 Report

Quoted By:

>>13882619
I don't care, give me the quads so I can escape this garbage.

Anonymous

Anonymous Tue 23 Nov 2021 16:29:19 No.13882974 Report

Quoted By: >>13882978

>>13882619
This would require a lot of conditional probabilities

Anonymous

Anonymous Tue 23 Nov 2021 16:29:52 No.13882978 Report

Quoted By: >>13883017

>>13882974
add more if you want to

Anonymous

Anonymous Tue 23 Nov 2021 16:32:30 No.13882983 Report

Quoted By: >>13882992 >>13882994

>>13882887
Doesn't really matter for the solution. A possible ambiguity that I see that the bottles are not restored, so a drunk bootle remains empty, that would make it far more complicated.

Anonymous

Anonymous Tue 23 Nov 2021 16:35:25 No.13882992 Report

Quoted By:

>>13882887
>>13882983
Dubs would remain the same I guess, but trips would change value depending on how many bottles were emptied by the dubs, assuming it would mean any of the remaining bottle. (instead of drink a numbered bottle + another random bottle)

Anonymous

Anonymous Tue 23 Nov 2021 16:37:01 No.13882994 Report

Quoted By:

>>13882983
In the OP it is specified that
>A random remaining bottle is drank after a person rolls trips

Anonymous

Anonymous Tue 23 Nov 2021 16:46:00 No.13883017 Report

Quoted By: >>13883031

>>13882978
$\sum_{i=1}^{10}{\left( p_{iD}\left(X=x-\sum_{k=1}^{i-1}{x_k}\right)* \prod_{j=1}^{i-1}{\left( p_{jN}\left(X=x-\sum_{k=1}^{j-1}{x_k}\right) \right)} \right)}$

Here's a rough estimate of the formula.
Where p_{AB} has A representing the number of bottles left and B representing the kind of bottle. The probabilities would be some kind of hybrid geometric to account for the difference between dubs, trips, and quads. The product probabilities would require conditioning on the point between drinking safe bottles, the memoryless makes this easier to calculate.

Anonymous

Anonymous Tue 23 Nov 2021 16:49:44 No.13883031 Report

Quoted By:

>>13883017
Wait I just realized this system is easily modeled by discrete time markov chains. Give me a minute and I will come back with an answer.

Anonymous

Anonymous Tue 23 Nov 2021 16:53:12 No.13883038 Report

Quoted By: >>13883371 >>13883700

>>13882619
Roughly 1 in every 10 posts are dubs, so at 100 posts the chance of death is really close to 100%

Anonymous

View Same Google iqdb SauceNAO poison.png, 252KiB, 1268x938

Anonymous Tue 23 Nov 2021 17:47:16 No.13883168 Report

Quoted By: >>13883700

>>13882619
You can model this system as a discrete time Markov chain.
p(quads) = 1/1000
p(trips) = 1/100 - 1/1000 = 9/1000
p(dubs) = 9/100
p(Death | N bottles) = p(quads) + (1/N)*p(trips) + (1/10)*p(dubs)
p(Safe bottle | N bottles) = (N-1 / N)*p(trips) + (N-1 / 10)*p(dubs)
p(Stay | N bottles) = 1 - p(Safe) - p(Death)

Above is the transitional probability matrix and plot of the probabilities by post count.

Anonymous

Anonymous Tue 23 Nov 2021 18:38:45 No.13883371 Report

Quoted By:

>>13883038
The posts are incremental so all of the bottles are used before quads in the typical scenario.

Anonymous

Anonymous Tue 23 Nov 2021 20:05:37 No.13883700 Report

Quoted By:

>>13883038
>>13883168
At 100 posts there is a 68.8% chance of death

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