Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.

“Now,” he says, turning toward you, “do you want to keep door #1, or do you want to switch to door #2?”

More here – Pricenomics

In the long winded explanation nowhere seems to show that one door is eliminated and as only two are left behind either of which could be the prize I still see chance as 50/50.

Then you would be wrong.

Why ?

Because the host “knows” Which is behind the three doors and WILL open one with a smelly goat no matter what.

You’re always better to change your choice. See another explanation below.

Turn the problem around a bit.

You have 3 doors behind one of which is $100. You can buy one of two options, option A is $33 dollars and allows you to open door 1. Option B is $66 dollars and allows you to open doors 2 & 3. You buy option B for $66. You have a 2 in 3 chance to win $100 or 66% chance. Now they open the first door, door number 2 and it is empty. You still have one more door to go that has your chance to win. The money was never behind both your doors, just likely one of them. The fact we have opened door 2 and it is empty does not all of a sudden change the odds of door 1 to a $50 door. You still paid fair price of $66 for your option to open two doors.

Hope this helps.

Two thoughts as I read the article and the response by readers to the solution.One question was,

I wonder how many people attacking the female author’s answer were male?

The other was, I wonder how many people later apologizing and acknowledging they were wrong were academics?

Not focussing on the maths problem but behavioural response by readers. Given this example, are men more likely to be vocal about their difference of opinion to a woman than to a man, ie Eddie McGuire difference of opinion to Carolyn Wilson as to her journalistic talent.

And are people in areas such as mathematics or academia more skilled and open to the idea of being wrong and publicly acknowledging it.

In essence these are questions about what behaviour norms do each of these groups have. It is easier to accept a vocal critical group who are also vocally apologetic.

Whilst seemingly an odd train of thought these questions have implications for the public contributions by women to forums of discussion. A google scholar search on this area shows a 2012 study of wikepedia contributions as only being 15% by women. Given the strength and personal nature of the criticism faced by the female author at the centre of this article, despite the accuracy of her answer, it does question what inherent biases we have. Why do so few women feel like contributing to public discussion and what are we missing out on, when intelligence/wisdom is distributed equally among men and women.

My guess from looking into the problem is that all of the critics were men and that very few apologies came from academics.

Yeah I had guessed that a dickhead had a higher probability of being a male but I was hoping that education level might correlate positively to behaviours like saying “I am wrong”.

Reflectively I know there must be times when I am that dickhead and will not say I am wrong. So I am looking for my own “cure”.

A simple way is to use 100 boxes. After you chose one, then I open 98 of the ones that I know don’t contain anything. Then ask if you’d like to change your choice. Surely you won’t say your chances have gone from 100:1 to 50:50? Will you?

Try the twelve coins problem.

You have 12 otherwise indistinguishable coins, but one of the coins is either heavier or lighter than each of the rest. The only way you have to find the odd coin out is the simple beam balance. Condition #1: You are only allowed to use the balance beam no more than THREE times. Condition #2: You cannot rely on luck. You must use a repeatable method which will be able to give the correct answer 100% of the time and tell if the odd coin is heavier or lighter than each of the rest.

.

Now before you evaluate that this is impossible, I’ll make it a little easier.

1) Label the coins A, B, C, D, E, F, G, H, I, J, K, L. (This will make it easier to keep track of the coins).

2) Make your first weighing; A, B, C, D AGAINST E, F, G, H.

One of three things will happen : 1) Both sides balance. 2) the left hand side goes down and the right side goes up. 3) The right side goes down and the left side goes up.

3) No matter what happens on the first weighing, make the second weighing: A, B, C, E AGAINST I’ J’ K’ D.

After each weighing you will gather some information about each of the coins.

I’ll leave it with you if you feel like nutting it out, including the third weighing.

Vic.