Random Thought Thread

[TRfromMT]

I have three boxes, one containing 2 gold coins, one containing 2 silver coins, and one containing 1 gold coin and 1 silver coin. I choose a random box, and a random coin from that box and pull out a gold coin. What is the chance that the other coin in the box is also gold?

Seems right... BUT WRONG

This is like when a 5 year old starts a joke out with the punch line to a different joke, has to start over, then forgets what joke they are telling, and ends up having to call an audible on a dumb knock knock joke. No one has any clue what just went down, the and the kid ain’t sure if you are laughing with them or at them.

 

[Synov]

This is like when a 5 year old starts a joke out with the punch line to a different joke, has to start over, then forgets what joke they are telling, and ends up having to call an audible on a dumb knock knock joke. No one has any clue what just went down, the and the kid ain’t sure if you are laughing with them or at them.

I just didn't want to give the answer away too early.

Here's how to rigorously solve conditional probability questions:

The chance of X given that Y occurred = (the general chance of X)*(the chance of Y given X occured)/(the general chance of Y)

So the chance that we chose from the all gold box given we chose a gold coin = (1/3)*(1)/(1/2) = 2/3

 

[bluemax_1]

Everything you said is correct, except for the conclusion. What you missed is that It's twice as likely you chose a gold coin from the box that was all gold coins, versus the box that only had a 1/2 chance of you choosing a gold coin.

Maybe changing the problem slightly would help your intuition. If we changed the GS box to having 999 silver coins and 1 gold coin, it would be highly unlikely you got the gold coin from that box right? You wouldn't say it's equally likely to be from either box.

Simplest way I can think of, is to break it down into the basics.

It’s not the box with 2 silver coins, so we can toss that one from the equation.

What remains are 2 boxes, that each had 2 coins, for a total of 4 coins. 1 Silver and 1 Gold in one box. 2 Gold in another box.

We drew 1 Gold coin.

While it seems logical that the chances are 50:50 (the other coin HAS to be either Gold, or Silver), the math isn’t just either, or.

There WERE 3 Gold coins and 1 Silver coin. We drew 1 of the Gold coins, leaving 3. 2 Gold. 1 Silver, so the possibility that the other coin is Gold, is 2 out of 3.

 

[Synov]

Simplest way I can think of, is to break it down into the basics.

It’s not the box with 2 silver coins, so we can toss that one from the equation.

What remains are 2 boxes, that each had 2 coins, for a total of 4 coins. 1 Silver and 1 Gold in one box. 2 Gold in another box.

We drew 1 Gold coin.

While it seems logical that the chances are 50:50 (the other coin HAS to be either Gold, or Silver), the math isn’t just either, or.

There WERE 3 Gold coins and 1 Silver coin. We drew 1 of the Gold coins, leaving 3. 2 Gold. 1 Silver, so the possibility that the other coin is Gold, is 2 out of 3.

I like your explanation but it wouldn't work very well if there were different amounts of coins in at least one box. In the variation of the problem I gave for example, would you say it's a choice between 2 remaining gold coins and 999 silver coins, therefore 2/1001? That would be incorrect since it's almost certain that you chose from the all gold box. Your method fails there because it would give undue weight to the box with 1000 coins.

The actual answer is more like a weighted average:

(1/3)(1)/( (1/3)(1)+(1/3)(1/1000)+(1/3)(0) ) = 1000/1001

 

[yoko]

I like your explanation but it wouldn't work very well if there were different amounts of coins in at least one box. In the variation of the problem I gave for example, would you say it's a choice between 2 remaining gold coins and 999 silver coins, therefore 2/1001? That would be incorrect since it's almost certain that you chose from the all gold box. Your method fails there because it would give undue weight to the box with 1000 coins.

The actual answer is more like a weighted average:

(1/3)(1)/( (1/3)(1)+(1/3)(1/1000)+(1/3)(0) ) = 1000/1001

My brain hurts lol

 

[TRfromMT]

The way the question was set up eliminates the SS box from consideration. Then, the second coin is coming from the same box as the first coin, which eliminates the other box from consideration. You are down to the odds of the second coin in one box.

 

[Synov]

The way the question was set up eliminates the SS box from consideration. Then, the second coin is coming from the same box as the first coin, which eliminates the other box from consideration. You are down to the odds of the second coin in one box.

Exactly, and the odds of the second coin being gold is twice as likely as silver, because it's twice as likely you found a gold coin in the all gold box than the one that is only half gold.

The information that we got a gold coin increases our suspicion that we chose from the first box rather than the second. That's the part most people miss.

 

[ferider]

It's called "Bertrand's Paradox" for a reason.



From the web somewhere:

• The Bertrand Paradox, which concerns probabilities, isn’t actually a paradox at all.
• The problem is with wording, not with actual situations.

 

[Synov]

It's called "Bertrand's Paradox" for a reason.

View attachment 2680398

From the web somewhere:

• The Bertrand Paradox, which concerns probabilities, isn’t actually a paradox at all.
• The problem is with wording, not with actual situations.

I think it's called a paradox because the answer is unintuitive. It's not a wording issue, just an issue of people not being used to conditional probability. If you eliminate the all silver box from contention because of the information you chose a gold coin, you should also downgrade the probability that you chose from the half silver box for the same reason. But most don't.

 

[ferider]

I have three boxes, one containing 2 gold coins, one containing 2 silver coins, and one containing 1 gold coin and 1 silver coin. I choose a random box, and a random coin from that box and pull out a gold coin. What is the chance that the other coin in the box is also gold?

The information that we got a gold coin increases our suspicion that we chose from the first box rather than the second. That's the part most people miss.

Note your wording in the original question.

The reason why I showed some code is because it defines/simulates my interpretation of your question.

 

[Synov]

Note your wording in the original question.

It's the same question. If you chose from the first box, the remaining coin is gold. If you chose from the second, it's silver.

For a paradox that's based on ambiguous wording, look at the boy girl paradox:

Mr. Jones has two children. At least one of them is a boy. What is the probability that the other is a boy?

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

 

[tinfoil hat timmy]

Everything you said is correct, except for the conclusion. What you missed is that It's twice as likely you chose a gold coin from the box that was all gold coins, versus the box that only had a 1/2 chance of you choosing a gold coin.

Maybe changing the problem slightly would help your intuition. If we changed the GS box to having 999 silver coins and 1 gold coin, it would be highly unlikely you got the gold coin from that box right? You wouldn't say it's equally likely to be from either box.

no its 50 percent that you get a gold coin and 50 percent that you get silver.

because you already grabbed a gold one, that means box 3 that had only silver is out of the equation. so theres only the boxes with gold left.

once you take out the 1 gold that you have from the box the odds of getting gold again are 50/50

 

[tinfoil hat timmy]

It's the same question. If you chose from the first box, the remaining coin is gold. If you chose from the second, it's silver.

For a paradox that's based on ambiguous wording, look at the boy girl paradox:

Mr. Jones has two children. At least one of them is a boy. What is the probability that the other is a boy?

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

STAY AWAY

 

[Synov]

no its 50 percent that you get a gold coin and 50 percent that you get silver.

because you already grabbed a gold one, that means box 3 that had only silver is out of the equation. so theres only the boxes with gold left.

once you take out the 1 gold that you have from the box the odds of getting gold again are 50/50

Again, everything you said is correct except for the conclusion that it's 50/50. You didn't take into account that, just like finding a gold coin reduces the probability you chose from the all silver box from 1/3 to 0, finding a gold coin increases the probability you chose from the all gold box from 1/3 to 2/3.

 

[tinfoil hat timmy]

Again, everything you said is correct except for the conclusion that it's 50/50. You didn't take into account that, just like finding a gold coin reduces the probability you chose from the all silver box from 1/3 to 0, finding a gold coin increases the probability you chose from the all gold box from 1/3 to 2/3.

it doesnt matter if you get the gold coin from the gg or gs box.

the odds AFTER CHOOSING a box at random and finding a gold coin, means that there is one coin left that can either be gold or silver.

50 50

 

[tinfoil hat timmy]

I'll be here fighting to the death on this tiny hill all day boys

 

[Richard338]

I once tried to explain the Monty Hall problem to an organic chemist...

 

[Synov]

it doesnt matter if you get the gold coin from the gg or gs box.

Of course it matters, that's what the question is asking the probability of!

the odds AFTER CHOOSING a box at random and finding a gold coin, means that there is one coin left that can either be gold or silver.

It's the odds after choosing a box at random AND finding a gold coin. The boxes remain equally likely until you are given information about your choice that changes their likelihood. You understand this for the all silver box, but not for the all gold box for some reason.

The only other way I can explain it is that there are three equally likely possibilities:

You chose the first coin in the GG box
You chose the second coin in the GG box
You chose the first coin in the GS box

2 out of 3 times, the remaining coin is gold.

 

[Nathan the Machinist]

It's a demonstrable fact that can be (and is) shown experimentally.

The Monty Hall paradox

Some things are counterintuitive

For example, if you're riding on a perfectly smooth train at a thousand miles an hour, the track is perfect and there's no wind, a perfectly smooth train. There is no way experimentally that you could know that you were moving. Even though you might think, moving at a thousand miles an hour, it would be obvious. It isn't.

But that's a different theoretical situation isn't it. My bad...

 

[TRfromMT]

Let’s say there are 100 boxes, One with GG, one with GS and 98 with only SS. Take it to infinity if you want to. The way the question was phrased, once the first event happened, all silver/silver boxes are out of contention. I think we agree here. Then that event is over, and our second choice is constrained to the same box. You know the gold boxes had one of two options for the second coin. The second gold-coin-containing box is not part of the population.

However, if you flip things and say there are 100 boxes, one has SS, a second box has GS and 98 have GG, then your odds are looking pretty good. In this case your odds on Gold for the second coin are 196 out of 197.

I think I get it now.