Photos courtesy of Tekko 

Last Saturday was my first Olympic Biathlon attempt. Organised by NIE and RunningLab, it was considered to be a very small-scale event as there’re not even a hundred participants in the Men’s Category. The weather didn’t look good at all, and it rained heavily just before my wave started. Fortunately, it was just a passing shower and my wave start was delayed only by about 15 minutes.

It was 2 laps of 750m in the open water, followed by a 10km run along Ecp. I can’t explain why, but I enjoy swimming in the sea so much more than in the pool. I also felt that I could swim more effortlessly in the sea. Perhaps it is the salt level that causes more buoyancy?

Anyway, since I was using breast-stroke, I got over-taken by most participants, including majority of the female swimmers who started 5 minutes after my wave. But that didn’t matter, as I really enjoyed the swimming.

Out of the water after my 2nd lap, I put on my socks and shoes for my 10km run. It was tiring as I’ve not done any brick training prior to this race. However, my objective was just to complete, so I went easy at a steady pace. I was to run 2 loops before reaching the finishing point. Personally, I felt it would’ve been better if they planned just a loop. The run turned out to be shorter than 10km, but I’m glad that was the case. I’ve got a valid reason to run less than 10km!

All in all, a fairly low publicity event, but it did help me gain confidence in joining next year’s Singapore Biathlon next year. And with this race in hand, I can skip the 30-lap time trial next year! Enjoyable race.

Towards the finishing point! 

Position: 54/65 (Men’s Cat)

Swim Time: 0:36:44

Transition Time: 0:02:18

Run Time: 0:55:32

Total Time: 1: 34: 33 

(See Question) 

So, have you decided whether to stick to the door that you had initially chosen, or make a switch? Or perhaps you’re thinking that since 1 door is being opened, the probability of opening the door with a car behind would be 1/2 for each of the remaining 2 doors? Then think again!

Although we can make use of conditional probability to solve this problem, I will illustrate how to solve by using diagrams and lay-man explanation.

In the 3 situations below, the car is always in Door A, and we select different initial doors in each of the 3 different situations shown. 

Situation 1:

• Car is in Door A.
• You select Door A initially.

• Game host opens Door B to show you the car isn’t in B, and asks if you want to switch your choice.
• You decides to switch from Door A to C.

• Game host opens Door C, and you don’t win the car.

Situation 2:

• Car is in Door A.
• You select Door B initially.

• Game host opens Door C to show you the car isn’t in C, and asks if you want to switch your choice.
• You decides to switch from Door B to A.

• Game host opens Door A, and you win the car!!! 

Situation 3:

• Car is in Door A.
• You select Door C initially.

• Game host opens Door B to show you the car isn’t in B, and asks if you want to switch your choice.
• You decides to switch from Door C to A.

• Game host opens Door A, and you win the car!!!

So, from the illustration, you could see that the player switched in all three situations, and got to win the car in 2 out of 3. So, the probability of winning the car if you switch is 2/3, while the probability of winning the car if you stick to your initial choice is only 1/3.

So, switching is always a better option when you see it in a mathematical perspective. Get it?

This problem is actually called the "Monty Hall Problem", and it could be solved by using Bayes’ Theorem easily.

Extension: If you’re still interested until this point, do work this problem out.

Now, imagine there are 4 doors instead of 3, and the car is hidden behind one of the doors. The player chooses a door. The game host then opens some other door that is a loser. The player then switchs to another door. The game host then opens another as-yet-unopened losing door, different from the player’s current choice. Now, there are only two unopened doors left: the player’s current choice and another one. The player decides to switch again. What is his probability of winning the car?

Well, lets say you’re a lucky contestant in the final round of a game show that gives you the chance of winning a brand new sports car. All you need to do is to pick the right door out of 3 doors, and if the car appears behind the door that you have chosen, you would win. Sounds simple?

Here’s the actual situation:

Suppose the 3 doors are labelled A, B and C. The showhost will ask you to choose a certain door. You name the door of your choice (e.g. A), and the host will open one of the other two doors instead (e.g. C) to show you that there’s no car behind that opened door. He then asks you if you would like to change your mind and select the other unopened door (e.g. B) instead.  

Will you change your choice? Which door has a higher probability of having a car behind, or both have the same probability?  

So I’ve decided to join this upcoming event. It would be my first Biathlon and I’ve decided to register for the Olympic Distance (1.5km swim, 10km run). Having 2 more weeks to go before the competition, I’m happy with my preparation so far. Swam 5 times and ran 3 times this week, clocking a total of 4.5km of swim and 24km run. But doing the two disciplines one after another would be a different ball game altogether.

Well, at least I’ll achieve a personal best no matter what how slow I complete it!