Kittens - fin!

I'd bet no statistical conundrum has been so widely debated in both professional and armchair circles, than the famous Monte Hall problem. Circulated in 1990, again in 1996 (when it was first presented to me), it appears to go back at least to 1975, according to Car Talk! The “Let's make a deal” official website actually has a page devoted to it as well!

Here’s a brief summary of the scenario: you’re on the “Let's make a deal” show, and you’re the lucky contestant to choose one of the three doors – and behind one of those doors is the big deal of the day. After you choose, Monte Hall offers to show you what’s behind one of the other doors – and it’s a goat! Your tension is mounting, as he asks you, “do you want to change your selection?”

When I first heard of this, my thought as an ex-aspiring statistician, was, “how absurd: of course not! I chose the first door completely at random; it’s still got the same chance as the other one.” I was 100% wrong! So were so many others – some of whom have used all kinds of vitriol to describe those who think there’s a difference. But the contrary view (that is: that it’s advantageous to switch) is right – proven by many analyses, simulators, and you name it.

I won’t belabor what is analyzed (to near death) all over the internet nowadays – you can research it if you’d like. But the bottom line involves remembering that Monte Hall knows which door has the grand prize, and realizing that he would never prematurely rob the thunder of the show by displaying the big deal or your door! Therefore, his action is far from random – he’s adding information to the system. In the end, if you stick with your original pick, you retain your original odds, 1/3, of having the big prize; switch, knowing that the “non-prize” door has been eliminated, and your odds spike to 2/3 – which is the probability that you initially chose a non-prize door!

This scenario is the answer to question #4 I posed in “Kittens.” Sort of! I claimed it was isomorphic with one of my scenarios, yet when I reviewed it to write this, I realized I’d set up the scenario wrong – sorry! There’s not an exact correspondence, but rather, the fact that information is added to the system – the essential underpinning of the Monte Hall scenario – that is in common.

Back to kittens:

You might think, having chosen one of three hidden kittens at random, that the probability of finding no others of the same sex would be ¼ (½ a chance that the next one you pull is not of the same sex, and another ½ chance that the final kitten is not the same sex). But my purposeful choice in these scenarios adds information and tinkers with the numbers. Let’s see how.

Using either full stochastic analysis, or an information theory approach as was suggested by my friend, Heavy D'luxe, you can arrive at the answers to questions #2 and #3. Whew – that’s a relief: the statistical and information theory analyses lead to the same conclusion! I’m going to borrow D’luxe’s mode of argument now – it’s a refreshing change from fraction multiplication in the Bayesian game!

Answer #2; my daughter has chosen one of the three hidden kittens at random and announced its sex. I, wanting to at least temporarily give the impression that they’re all the same gender, reach in, and purposely select, if possible, another kitten of the same sex. Given that this has just occurred, what’s the probability that the final kitten is of the same gender?

Let's say, with no loss of generality, that my daughter chose the first kitten in the possible sequences below – my choice is the highlighted one (if two are highlighted – it means I chould have chosen either):

GGG
GGB
GBG
GBB - I couldn’t do it here
BGG - … or here
BGB
BBG
BBB

Of the eight original scenarios, the middle two are “knocked out” – I wouldn’t have been able to select a kitten of the same sex and the scenario would have ended prematurely. Of the remaining 6, there are 2 cases where the final kitten’s sex is the same as the first one – A: 1/3 of the time.

Answer #3: in which I have purposefully pulled a kitten of opposite sex – what is the probability that the final kitten matches my daughter’s kitten’s sex? I’ll let you do the analysis yourself (you can use the 8 scenarios above, knock out the two that don’t apply). You should reach the bottom line A: 2/3 of the time.