Probability problem for statisticians/mathematicians

[TimberWolf]

These 2 are not at all the same thing. That is where your scenario differs from the goats and the doors.

If the number that will fail is fixed, then it makes sense to change your choice after 1 failure other than yours is announced; but if the probablility is an independednt probability on each action, it doesn't matter whether you change or not.

This confusion in the other direction is why most people find it hard to analyse the goats/doors problem.

It makes no difference, this is not like rolling a dice for each door. On average there are two failures in 10 missions, so you had a 2 in 10 chance of picking a bad mission, on average. Gaining the knowledge that mission 10 is a bad one means that if you re-pick you only have a 1 in 9 chance of picking the remaining bad mission.

[DimPrawn]

A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues
were:

1. The fifth number plus the third number equals fourteen.
2. The fourth number is one more than the second number.
3. The first number is one less than twice the second number.
4. The second number plus the third number equals ten.
5. The sum of all five numbers is 30.

What were the five numbers and in what order?

Not a probability problem is it?

Just some Masters Degree level algebra*

*2008 Masters Degree level

[Diver]

Not a probability problem is it?

Just some Masters Degree level algebra*

*2008 Masters Degree level

Wasn't supposed to be!

Just something to do if you're bored

[Bagpuss]

U2 are on an Island with one boat, how many Africans die needlessly at the click of Bono's fingers, until the boat reaches shore?

[BrilloPad]

A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues
were:

1. The fifth number plus the third number equals fourteen.
2. The fourth number is one more than the second number.
3. The first number is one less than twice the second number.
4. The second number plus the third number equals ten.
5. The sum of all five numbers is 30.

What were the five numbers and in what order?

C+E=14
D=B+1
A=2B-1
B+C=10
A+B+C+D+E=30

(2B-1)+B+C+(B+1)+(14-C)=30
4B=16

A=7
B=4
C=6
D=5
E=8

?

[expat]

It makes no difference, this is not like rolling a dice for each door. On average there are two failures in 10 missions, so you had a 2 in 10 chance of picking a bad mission, on average. Gaining the knowledge that mission 10 is a bad one means that if you re-pick you only have a 1 in 9 chance of picking the remaining bad mission.

Only if exactly 2 missions in this group of 10 fail.

If OTOH every mission has a 0.2 probability of failing (i.e. on average there are two failures in 10 missions), that is not the same situation. Then it doesn't matter what you know about mission 10, that doesn't affect the probability for any other mission.

That is precisely my point: there is a critical difference between
(1) an independent probability for each trial
(2) a fixed number of failures.

I suggest that the space launches are likely to be type (1), whereas the game show is type (2). That is, for the launches, if you know that launch 10 fails, you can not talk of the other failure: there might be no more, or there might be 2 or more other failures, though on average there will be 1 more. Each launch is independent (or so I am presuming: if you set the case that, e.g. there are 2 known faulty components at 1 per vehicle, then I allow that if you knwo 1 faulty vehicle it tells you something: but if it is a random independent chance, then knowing 1 case tells you nothing new about the others).

[NotAllThere]

Take a sphere. Drill a hole along a diameter of the sphere that has walls exactly 1m long.

How much of the sphere is left?

[Diver]

C+E=14
D=B+1
A=2B-1
B+C=10
A+B+C+D+E=30

(2B-1)+B+C+(B+1)+(14-C)=30
4B=16

A=7
B=4
C=6
D=5
E=8

?

[TimberWolf]

Only if exactly 2 missions in this group of 10 fail.

If OTOH every mission has a 0.2 probability of failing (i.e. on average there are two failures in 10 missions), that is not the same situation. Then it doesn't matter what you know about mission 10, that doesn't affect the probability for any other mission.

That is precisely my point: there is a critical difference between
(1) an independent probability for each trial
(2) a fixed number of failures.

I suggest that the space launches are likely to be type (1), whereas the game show is type (2). That is, for the launches, if you know that launch 10 fails, you can not talk of the other failure: there might be no more, or there might be 2 or more other failures, though on average there will be 1 more. Each launch is independent (or so I am presuming: if you set the case that, e.g. there are 2 known faulty components at 1 per vehicle, then I allow that if you knwo 1 faulty vehicle it tells you something: but if it is a random independent chance, then knowing 1 case tells you nothing new about the others).

If OTOH every mission has a 0.2 probability of failing

Nope, 0.2 chance of failure overall was specified (2 in 10). What I'm not certain about is whether your emphasis on average is even relevant, since though in a non-average scenario you may get one bad mission in 10, you may also have the misfortune of getting a 3 bad missions in 10 scenario, and all the more reason to change the original choice.