Pagename: mathematics - Mathematics
Post Count: 2Category:Mathematics
Pagename: mathematics - Mathematics

Missing convention?

Recently, I came across this youtube video by ‘MindYourDecisions’…

In this clip from years ago, (SPOILER ALERT…) Presh explains how sticking to the ‘operations order’ determines the correct answer to be 9. I was quite shocked at this result. My mental calculation arrived at ‘1’ without any uncertainty. Going through the comments made me feel a little less concerned about my mathematical understanding. Many people were battling for a ‘1’ as the answer.

A quick inspection of the equation shows the source of the ambiguity. It would be common, in my travels, to interpret 6/2(1+2) as 6/(2(1+2)) with the lack of a multiplication sign signalling that the ‘2/(1+2)’ should be treated as a denominator. Adding the multiplication sign… 6/2*(1+2) makes it more clearer that the parentheses should be treated as a separate term. Even with the added ‘*’ I find the equation to be a touch ambiguous.

There is ambiguity in the equation to the point that I would suggest that the real answer should not be defined. Order of operations is a clear convention but it is clear from the comments generated by the clip that other conventions are in use even if they are not strictly taught.

I expect, because equations are rarely put on a single line, as this one is, it is probably not necessary to state that ‘2(1+2)’ would be treated as a whole in a modern math class. I find it amusing to consider that a mathematics lecturer would probably be more likely to get this wrong than someone who learnt their math at primary school and then left it at that. The lecturer could be expected to apply the usual conventions and that would be their undoing.

What are those usual conventions I am referring to? As far as I know, they are not explicity stated. They are followed due to the need to keep one’s mathetmatical workings free of ambiguity. I like to think it is quite unlikely that a mathematician would ever produce such a confusing equation.

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Pagename: mathematics - Mathematics

The Lost Boarding Pass puzzle

The puzzle:

One hundred passengers are about to board a plane. The first passenger lost his boarding pass and decides to choose a seat randomly. Each subsequent passenger sits in their assigned seat if it is free or chooses a seat randomly from those remaining.

What is the probability that the last passenger will get to sit in his assigned seat?


Despite the fact that this puzzle lead me on a wild goose chase, I do have a soft spot for it. The reason is that after reading the solution given by Peter Winkler in his book:Mathematical Puzzles (A Connoisseur’s Collection)

I was so convinced he was wrong I decided to test it using a simple computer model. Thus started my adventure with Excel VBA in earnest, which continues to this day. The principle was simple, follow the rules of the puzzle in an excel sheet and see how many times the last passenger had their own seat.

If you have Excel on your computer, feel free to give it a whirl: lost-boarding-pass-puzzle However, keep in mind that it uses VBA so security warnings are likely.


Unfortunately I only succeeded in proving Winkler right and myself wrong. The answer is: as Winkler states, 50%.

In hindsight it is hard to see how I could of thought otherwise. It seems the puzzle uses a bit of misdirection and leads the puzzler to focus on an area of zero relevance.

Considering how the seating pattern affects the outcome is offtrack.

The result is determined when someone looking for a new seat either sits in their own seat or the seat of the last passenger to board. Right from the first passenger, the probability of sitting in either of these two seats is equal and the puzzle only ends when one of the two seats is taken. All other passengers who select a different seat have no impact on the outcome. They simple kick the can further down the line.

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