The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 24 from this test.

The question is about a probability calculation concerning two coins, one of which is heads on both sides.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

To cut a long story short, my opinion is that the test is highly inappropriate. In this blog post, I will focus on the study of question 24.

This question is highly inappropriate as well, as are most (if not all) of the questions on this test. The reason this question is inappropriate is that it does not test cleverness but knowledge. If you have been taught probability theory and the concept of sample space, then you can find the solution. Otherwise, you are clueless. Probability theory and the concept of sample space is one of the domains that the following brilliant comic is right on the spot.

So, we have two coins. One coin is heads on both sides (um…. who comes up with these scenarios?). The other coin is heads on one side and tails on the other side. Ok, that’s the normal coin. The first coin is the abnormal one, the “rebel”.

If the question just stated that one coin is selected at random and we examine one of its faces at random, then we would have the following 4 cases (the following sample space):

Coin H-H is selected and its first face H is examined.

Coin H-H is selected and its second face H is examined.

Coin H-T is selected and its first face H is examined.

Coin H-T is selected and its second face T is examined.

But the question states something more: That the face that we examine is heads. So the sample space ends up having the following 3 cases:

Coin H-H is selected and its first face H is examined.

Coin H-H is selected and its second face H is examined.

Coin H-T is selected and its first face H is examined.

So, only the 3 cases above could have happened.

Now, from the 3 above cases, only the first 2 correspond to the other face being heads.

Thus, the probability we are looking for is 2/3.