- X has two children. One of them is a girl. What is the probability that the other is also a girl ? If you have think something is fishy, compare it with this one:
- X has two children. The younger is a girl. What is the probability that the elder one is also a girl ?
- X wants to shoot Y. He puts two bullets in (
*edited*adjacent slots in) a six-barrell revolver. Then he rolls the barrell. He pulls the trigger at Y, but luckily, no shot is fired. He gives two options to Y, either X will fire again or X will roll the barrell and fire once more. Which is better for Y ?

It looks like in the pursuit of freedom and exposing the creative self in myself, I am losing my precious little grey-cells. Eh bien! Hope it's not too late.

## 4 comments:

it's never tooo late :=)

the 1st should be probability at 0, the two-shots-in-a-barrel at 0.5? I am not so sure..

and please, while you're probably reading this take a look at kde#142370?

Children puzzle-2: Elder and younger being girls are independent. So, the probability is 1/2.Now,

children puzzle-1: Consider all cases - GG, GB, BG, BB.Pr[GG | at least one girl]=1/3This is different from the other case because the event that one being girl is not independent of both being girl. A classic example of conditional probability.

The

shots puzzle: If rolled again,P[shot fires bullet]=1/3(2 out of 6 possibilities give a bullet at the next shot). So, chances of death=1/3.If X fires again, we need to compute

P[next shot is bullet | first shot is not bullet]There are 4 cases when first shot is not a bullet. And out of the 4, only 1 has bullet in second . So, the probability is 1/4. So the probability is less if X fires again.On a different note, I fixed the kde bug :)

Regarding the shots puzzle:

Ah, but your solution makes an assumption not in the original puzzle, that is, that the bullets were loaded adjacently. If they were not adjacent, then the probability is actually 1/2, so rolling again is safest.

Or (since the original problem said to assume equal probabilities wherever it makes sense), assume the bullets were loaded uniformly at random. Then there are 2 bullets uniformly distributed among 5 remaining chambers, so the chance of death is now 2/5 > 1/3, so rolling again is still safest...

On an unrelated topic:

if (dbera.blog.references("Poirot")) {

dbera.bonus_points += 1000;

}

## Post a Comment