The two kittens problem...

[TheFallen]

Okay here's my latest brain fritz. It's superficially at least not as bad as the two envelope problem, but it's still a little freaky. Here goes...

You want to buy two kittens, but for reasons best known to yourself, you want two male kittens. So you call your friendly neighbourhood petshop and the owner tells you that, although he's got no kittens in stock today, he's getting a black one and a white one delivered in at 9 am tomorrow.

Tomorrow dawns. You're not too excited, since – and we'll assume that there's an exact 50/50 chance of a kitten being born male or female – your chances of the two kittens both being male are only 25%. We can list the possible and equally likely permutations thus:

Bm + Wm, or Bm + Wf, or Bf + Wm, or Bf + Wf.

But then you notice you've got a missed call on your mobile...

Scenario A.
The pet shop owner has left a message on your voicemail. He tells you that he's had time to look at the black kitten and it's male, but he hasn't got time to look at the white kitten. Your excitement rises. You've now got a 50/50 chance of getting the two male kittens you so badly want, right? From our eager cat-buying POV, we can now list the possible and equally likely permutations thus:

Bm + Wm, or Bm + Wf,

...since we've eliminated the possibilities of Bf + Wm and Bf + Wf.

Scenario B.
The pet shop owner has left a message on your voicemail. He tells you that he's had time to look at one of the kittens and it's male, but he hasn't got time to look at the other kitten. That's all he says... his message offers no further information. Your excitement rises, but not so much. Again from our eager cat-buying POV, we can now list the possible and equally likely permutations thus:

Bm + Wm, or Bm + Wf, or Bf + Wm,

...can't we? All we've done is eliminate the single Bf + Wf possibility, so doesn't this now merely leave a 33.3% chance of there being two male kittens?

Am I right so far?

[I'm Murrin]

It means you have a 1 in 3 chance of correctly guessing the gender and colour combination of both cats, but still a 50% chance both cats are male.

[Hashi Lebwohl]

When you first call and learn of tomorrow's delivery you are correct in that he could be getting Bm/Wm, Bm/Wf, Bf/Wm, or Bf/Wf so you have a 1/4 chance of correctly guessing and receiving what you want.

Scenario A: black kitten male, so now you are down to the 50/50 white male or white female. Odds of correctly guessing color/gender that you want: 50%. Odds that both kittens are male: 50%.

Scenario B: one kitten is male, so you can rule out Bf/Wf. Odds of correctly guessing color/gender that you want: 33.3%. Odds of both kittens being male: 50%.

Regardless of which phone call is left you still have a 50% chance of receiving two male kittens given that one kitten is clearly identified as male.

[Vraith]

Be prepared:
There is a chance between 20% and 80% [roughly] that the white cat will be wholly or partially deaf. [lower end if it has no blue eyes, high end if it has both blue eyes, in them middle if it has one blue eye.]

[wayfriend]

I think the second scenario with the 33% number is wrong. My hunch is that it should be 50%.

To do things correctly, you have to consider two non-orthogonal cases.
You problem doesn't state it, so we will assume the pet owner chose randomly, with a 50% chance to choose each one for examination.

1. The pet man chose the black kitten to examine - 50%
2. The pet man chose the white kitten to examine - 50%

Within each of those subcases, we can break it out by result.
Because the odds of the sex are completely orthogonal to the odds of the chosen kitty, the odds of each scenario are 50% x 50% = 25%.

1a The pet man examined the black kitten, and it was male - 25%
1b The pet man examined the black kitten, and it was female - 25%
2a The pet man examined the white kitten, and it was male - 25%
2b The pet man examined the white kitten, and it was female - 25%

The phone call comes in, and the pet man rules out scenarios 1b and 2b. Only scenarios 1a and 2a are now possible. Since each is equally probable, their probability is now 50% each.

1a The pet man examined the black kitten, and it was male - 50%
2a The pet man examined the white kitten, and it was male - 50%

Now, in each of the remaining scenarios, the sex of the unexamined kitten is 50/50. Since this is orthogonal to all other odds discussed so far, we now have the following scenarios, each with 25% odds.

1a' Examined the black kitten, it was male, white kitten is male - 25%
1a'' Examined the black kitten, it was male, white kitten is female - 25%
2a' Examined the white kitten, it was male, black kitten is male - 25%
2a'' Examined the white kitten, it was male, black kitten is female - 25%

As you can see, of these scenarios, two of them are an all-male scenario, with a total odds of 50%.

The reason it runs counter to your first guess is that you did not factor in selection bias. The fact that the manager ruled out one cat being female not only eliminated the female-female possibility, but it also elevated the male-male possibility to be more possible than the male-female possibilities. They were not equally possible at 33%, they were 50%, 25%, and 25%.

[TheFallen]

Thanks to all and especially you, WF, for laying it out like I would have. I had given this some thought overnight, despite statistics and probabilities really not being my strong point (could you have guessed?) and had come to the same conclusion via a more convoluted route. (for the record, although I was sure that my 33% postulate was wrong, since it flew in the face of common sense, in order to understand fully why and where it was wrong, I needed to conceptualise the problem linguistically, rather than purely mathematically – not really a surprise, given my own educational leanings).

Anyhow I finally reasoned that, although there were only three possible combinations once the petshop owner had left the vaguer message in Scenario B, there were actually four possible outcomes, those being exactly as you delineated, WF. It's just that two of the four possible outcomes resulted in the exact same combination. Hence yes, the probability of getting the two male kittens that you've set your heart on is indeed 50/50.

This sort of comprehension may be instinctual to you guys, but as I said, I kind of need to have it verbalised to myself in order to get my brain around it entirely. Now I just need to do a little research to understand what "non-orthogonal" means in terms of probabilities/scenarios...

[Hashi Lebwohl]

I think what wayfriend calls "non-orthogonal" I would call "independent"--the outcome of probability A has no bearing on the probable outcomes of instance B.

At first I thought the pet shop owner had made both phone calls but then I read it again and realized that they were two completely different scenarios.

[wayfriend]

Sorry for the term, but it's habituated in me. Hashi has it. I should say "independent".

BTW, the best plain language description of the phenomenon I can devise is this: the second call did not just eliminate the f-f outcome, but it also halved the probabilities of the m-f and f-m outcomes, because either the f-m or m-f outcome was eliminated as well, but we do not know which one.

[Hashi Lebwohl]

I figured you had picked it up from a textbook or instructor who used it. It isn't incorrect, only different than the term I use.

I would wager that wayfriend's skill at statistics is more honed than mine.

[wayfriend]

BTW, this is non-authoritative, but helpful:

Q: In other contexts, orthogonal means "at right angles" or "perpendicular". What does orthogonal mean in a statistical context?

A: It means they are independent to each other. Independent variables are often considered to be at right angles to each other. For example on the X-Y plane the X and Y axis are said to be orthogonal because if a given point's x value changes, its y value remains the same, and vice versa (i.e. they are independent).

Another term that might be used is "uncorrelated", meaning that there is no correlation. Statistics are correlated when they are somewhat predictive of each other; when the value of one doesn't predict the range of the other, they are uncorrelated.

The opposite is when two events are dependent. An example of that is a deck of cards. If you randomly draw a card, put it back, and then randomly draw a second card, they are independent events. But if you randomly draw a card, keep it, and then randomly draw a second card, they are dependent events. Because removing one card from the deck changes the odds of what you can draw the second time. Particularly, the odds of drawing the same card again went down to 0%.

[sgt.null]

math is hard.