Thursday, July 18, 2013

Applying Simple Arithmetic and Pobability for Everyday Train Journeys!

Last week during my journey back from New Delhi to Bathinda in my favorite “Punjab Mail” I was eavesdropping conversation between a family and one lonely traveler.  The family want to exchange a seat of theirs, which is 5 compartments away, with this person so that they can all sit together. The lonely traveler is ready to help out, but with a condition that he don’t mind any berths other than Upper or Side Upper or Side Lower. Seat number the family want to exchange is “42” and they have no clue which berth this is. All were totally clueless to my utter surprise! I had been routinely confronted with this birth allocation problem, especially when tickets in “RAC” gets confirmed while checking the “PNR status”. I could help them out almost instantaneously (which is middle berth!) with a very simple solution.
This is the berth allocation for Seat # 1 to #8
1 L
5 M
2 M
6 U
3 U
7 SL
4 L
0 SU

Whatever your seat number, divide it with 8 and look-up the remainder in the above table to get the answer. 

  • Let's take seat# 42, 42/8= 5 with remainder 2, and 2 is M.
  • Let’s take seat# 76, 76/8 = 9 with 4 as remainder, 4 is L.
  • Let’s take seat# 71, 71/8=9 with 0 as remainder, 0 is SU

This always works for SL (Sleeper Class) and 3AC (AC 3 Tier) classes, as Indian Railway have uniformity in (at least) numbering of the seats. For 2AC, refer the following table:
1 L
4 U
2 U
5 SL
3 L
0 SU

And for this, you should divide by 6 (instead of 8) and look up the remainder the above table.

  • Let's take seat# 53; 53/6 = 8 with remainder 5, 5 is SL
  • Let's take seat# 54; 54/6 = 9 with remainder 0, 0 is SU

Simple, isn’t it? But the problem is that most of us do not apply these simple mathematical logic in our lives! I’m sure there are engineers and even math teachers who is new to this trick!

On a related note, what is the probability for getting a preferred seat (in Sleeper Class or 3AC) for the person described in the beginning? We all know berth allocation during ticket reservation is quite random even if we request for a specific seat. And we know that If there are ‘n’ likely outcomes of an experiment of which one is called a success ‘s’, then

the probability of success P(A)= s/n

(s=Number of ways (A) can occur; n=Total number of possible outcome)

  • Probability of getting any long births (side births are shorter and many tall persons don’t prefer)= 6/8 = 0.75. (Probability is always expressed in decimals; 0.75 means "75% chance" in our everyday language) That means quite a good probability!
  • Probability of getting long-lower berth (long-lower births are lower births excluding side lower) = 2/8 (only two seats out of 8 possible allocations!) = 0.25.
  • Probability of getting Side berths (well, probably introverts prefer side births; no one will disturb!) = 2/8= 0.25
  • Probability of getting any berths other than Upper or Side Upper or Side Lower= 4/8 = 0.5 (as we say, “50-50 chance”)
 Hope you'll start applying these simple arithmetic and probabilities next time when you are confronted with similar situations!

Author disclosure: I'm a ferroequinophile and married to a ferroequinoengineer :-)

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