/sci/ - Science & Math

[Anonymous No.12310976]

Well?

[Anonymous No.12311002]

>>12310976
50% either you do or don't

[Anonymous No.12312221]

>>12310976
1 in 3

[Anonymous No.12312227]

[Anonymous No.12313118]

There are two possibilities:
1. You took a ball from box 1, the next one will be golden.
2. You took a ball from box 2, the next one will be silver.
It boils down to answering what's the chance the ball was taken from box 1.
The chance of picking a golden ball from box 1 is 100%. The chance of picking a golden ball from box 2 is 50%. There are twice as many instances in which you'd pick a golden ball and it happened to be from box 1.
So the chance you picked it from box 1 is 2/3, which means the probability of picking a golden one next is 2/3.

[Anonymous No.12313121]

>>12311002
this

[Anonymous No.12313216]

>>12311002
XD

[Anonymous No.12313226]

>>12313118
OP has three gold balls in Box 1. So in this case it would be 75%.

[Anonymous No.12313239]

It's been a while since I've used conditional probability and I couldn't work out how to use any formulae to answer this so I just listed the possibilities.
Let Gi be the ith golden ball numbered from left to right and Si be the ith silver ball numbered from left to right.
Then the possible ways of getting first a gold ball and then another ball are
{(G1, G2), (G1, G3), (G2, G1), (G2, G3), (G3, G1), (G3, G2), (G4, S1)}
As 6 of the 7 possibilities involve a golden ball being picked second, the probability is 6/7.

[Anonymous No.12313281]

I think get the explanation for three gold balls, but I feel like that can't be the whole answer.

Let's say you have 9,999,999 gold balls in box #1 and the other boxes are all the same as before. It's an interdimensional bullshit box, so the balls in it are equally as large as the others and the process of pulling feels the exact same in all 3 boxes.

If you pull out one ball, and it's gold, does that mean you have a >99.9% chance of pulling another gold?

I feel like there has to be an upper limit to the probability, it can't get arbitrarily close to 100% if you keep adding things, right?

[Anonymous No.12313348]

>>12313281
Of course it approaches 100%. It's like I came back with a drop of water and ask you if I've been in a desert or the ocean. There is a very small chance I found some water in a desert, but it's far far more likely (approaching 100%) I got it from the ocean.

[Anonymous No.12313370]

>>12313348
It makes intuitive sense to me now, thanks!

[Anonymous No.12313399]

>>12313226
It's irrelevant how many gold balls are in box 1. The chance of picking a gold one from the box is still 100%.

>>12313239
The possibilities you listed don't have an equal chance of happening. G4 has a higher probability of being picked than either G1, G2 or G3. Box 1 and box 2 have an equal chance of being picked, but box 2 has fewer balls, so the probability of picking a desired ball from there is higher.

[Anonymous No.12313407]

>>12310976
Twice as likely you chose a gold ball from Box 1 than Box 2, so the answer is 2/3.

[Anonymous No.12313409]

>>12313399
Surely it is relevant. Given that you've picked a golden ball already, you're more likely to be in a box with many golden balls than one with only a few.

[Anonymous No.12313419]

>>12313409
But you picked a box at random first. Both boxes have the same probability of being chosen let's say box 2 had an equal number of balls to box 1, but half of them were gold and half were silver (assume the number of balls is even). It's exactly the same thing.

[Anonymous No.12313461]

>>12313399
>>12313419
I've just modelled it on Python, and you're right, the answer is 2/3. My apologies.
My code, if you're interested

import random
n = 1
second_gold_count = 0

while n <= 10000:
____BOXES = [["GOLD", "GOLD", "GOLD"], ["GOLD", "SILVER"], ["SILVER", "SILVER", "SILVER"]]
____picked_box = random.choice(BOXES)
____picked_ball = random.choice(picked_box)
____if picked_ball == "GOLD":
________picked_box.remove(picked_ball) # Remove picked ball from box.
________second_picked_box = random.choice(picked_box)
________if second_picked_box == "GOLD":
____________second_gold_count += 1
________n += 1


print(second_gold_count / n)

The dashes are just because I can't remember how to format code on here. I even tried an arbitrary number of gold balls in the first box and it didn't make a difference, it's always 2/3.

[Anonymous No.12313948]

>>12313461
>can't remember how to format code on here
texttt{}
[math]
\texttt{#include <stdio.h>} \\
\texttt{ } \\
\texttt{int main(int argc, char **argv)} \\
\texttt{\{} \\
\texttt{printf("Hello, World!\n");} \\

\texttt{return 0;} \\
\texttt{\}}
[/math]

[Anonymous No.12314093]

>>12310976

Choose box 1, first pick is gold, 1/3
Choose box 2, first pick is gold, 1/6
Choose box 2, first pick is silver, 1/6
Choose box 3, first pick is silver, 1/3

Only first two choices are valid upon the condition; renormalizing them, the probabilities that each occur based on the condition are:

Choose box 1, first pick is gold, 2/3
Choose box 2, first pick is gold, 1/3

If we picked box 1 at first, the second pick will invariably be gold.
If we picked box 2 at first, the second pick will invariably be silver

Thus there's a 2/3 chance that the second pick from the same box is gold.

[Anonymous No.12314283]

>>12313948
Your explanation was the most succinct, fair play.

>>12313948
texttt{print("Test")}

[Anonymous No.12314309]

>>12313461
Literally just apply Bayes' theorem or make a probability tree anon. Why code up a simulation?

[Anonymous No.12314369]

>>12314309
Practising my coding to be honest.

[Anonymous No.12314388]

>>12314283
[math]
\texttt{print("Test")}
[/math]

[Anonymous No.12314395]

>>12310976
This is the solution, with infinite iterations it will be 66%, shut up.

[Anonymous No.12314680]

>>12314283