Animals & Nature3 mins ago
So i got these three balloons
4 Answers
ABC. One of them got Special Gas in it, inert. Others have Flammable sensitive gas in, detected by placing candle under balloons.
I test them out in binary fashion: AB, etc.
I test AB to find which it is. Equal chance of it being A or B. I put the candle under A. It explodes.
I then go to test B or C.
I was told that chances of it being B/C are not 50/50. More likely to be B, since it already underwent a 'binary test', even though it did not feel the flame.
Is this right?
I test them out in binary fashion: AB, etc.
I test AB to find which it is. Equal chance of it being A or B. I put the candle under A. It explodes.
I then go to test B or C.
I was told that chances of it being B/C are not 50/50. More likely to be B, since it already underwent a 'binary test', even though it did not feel the flame.
Is this right?
Answers
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No it is not correct.
It doesn't matter if you group the balloons: AB and C
or A and BC
or A and B and C
or ABC
They are the same balloons and are not linked in any way.
The chance of each balloon being the special one is the same 1/3
When you explode the first one the chances of each one of the others being special are still the same but the chances have improved from 1/3 each to 1/2 each.
Chance has no consciousness and B doesn't know if it was beside A or B when A exploded.
Now, what would make a difference would be if a chain reaction was possible. If A exploding in close proximity to another explosive balloon would cause that to explode as well, then the fact that B didn't explode would indicate 100% that it was the special inert one.
It doesn't matter if you group the balloons: AB and C
or A and BC
or A and B and C
or ABC
They are the same balloons and are not linked in any way.
The chance of each balloon being the special one is the same 1/3
When you explode the first one the chances of each one of the others being special are still the same but the chances have improved from 1/3 each to 1/2 each.
Chance has no consciousness and B doesn't know if it was beside A or B when A exploded.
Now, what would make a difference would be if a chain reaction was possible. If A exploding in close proximity to another explosive balloon would cause that to explode as well, then the fact that B didn't explode would indicate 100% that it was the special inert one.
Actually, I think it is correct. Think about it. Of the three possible paired combinations (AB, AC, BC), the is a 2/3 chance of one of those balloons being the one with inert gas. Therefore, there is a higher probability of picking one flammable balloon and the inert balloon than there is of picking a two flammable balloons. So, for the first two balloons that you picked, if the one you heat up first explodes, then the remaining balloon (in you example, that would be balloon B) is more likely to be the inert balloon than the balloon that was not picked in the first pair.
This seems to be the '3 doors' probability question:
http://www.stat.sc.edu/~west/javahtml/LetsMake aDeal.html
http://www.stat.sc.edu/~west/javahtml/LetsMake aDeal.html
Be careful with the English or the maths will not resolve.
Chance and probability are not always the same. When considering the overall probability in a scenario such as this there is a difference in the odds, because you have already carried out one test. Chances imply a future tense to the question so the chances are based on how you set the question.
If you do test 1 first. A explodes. The CHANCES, in the future, of B or C exploding in the next test are equal 50:50 as a future test. So, like in "Deal or no deal" it makes no difference whether you swap the box (or balloon!).
The OVERALL PROBABILITY of B or C exploding are not the same because of the first test on A.
To resolve this properly consider selecting 2 balloons from 3 at random and working the probability as a function of both selecting (or not) and then exploding in two rounds of action and the full probability can be seen. You can build this as an event tree of probability.
Great question - good fun to argue about. I have seen it used to good effect to provoke discussion about the differences between chance and probability.
Chance and probability are not always the same. When considering the overall probability in a scenario such as this there is a difference in the odds, because you have already carried out one test. Chances imply a future tense to the question so the chances are based on how you set the question.
If you do test 1 first. A explodes. The CHANCES, in the future, of B or C exploding in the next test are equal 50:50 as a future test. So, like in "Deal or no deal" it makes no difference whether you swap the box (or balloon!).
The OVERALL PROBABILITY of B or C exploding are not the same because of the first test on A.
To resolve this properly consider selecting 2 balloons from 3 at random and working the probability as a function of both selecting (or not) and then exploding in two rounds of action and the full probability can be seen. You can build this as an event tree of probability.
Great question - good fun to argue about. I have seen it used to good effect to provoke discussion about the differences between chance and probability.
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