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# Oracle & Counterintuitive results

In the world of Oracle tuning, many professionals do not realize they we live in a world of probabilities, a realm devoid of absolutes.

In many areas of Oracle tuning we often see counterintuitive results, leading many respected professional to exclaim "it should not work that way".

This has led to many Oracle myths, often precipitated by counterintuitive results.  For example, Oracle tuning activities do not jive with the official documentation, yet are proven empirically:

• The benefits of different blocksizes - Naysayers claim that it cannot be true (despite empirical evidence) because the OS read size is always the same.

• Read Only Tablespaces - Naysayers and luddites say that it's not worthwhile because there is only a 6% benefit, not realizing that a 6% improvement can reduce I/O by millions of operations per minute.

• Multiple blocksizes - Naysayers claim that they cannot work because they were designed for transportable tablespaces, not understand how multiple blocksizes have been used successfully on IBM mainframes since the 1980's.

• Ratio's as a tuning tool - Some neophytes notes that ratio's can be manipulated and incorrectly conclude that Oracle tuning ratio's (i.e. the buffer hit ratio) and entirely meaningless.

Here is a real world example of counterintuitive results that illustrates the problem when people without a proper formal education get in "over their head" when performing Oracle tuning activities.

The Monty Hall problem

A favorite Google job interview question is "develop an algorithm where you can enter one of three doors with an equal probability by tossing a fair coin". See other Google job interview questions here.

In non-Oracle areas, we see the same issue with the "Monty Hall" problem.  In this probability analysis:

1 - Contestants are shown three doors.

2 - Behind one door is a new car, while the other two doors have a "booby prize" (a goat).

3 - After the contestant picks a door, Monty Hall reveals a door that he knows to contain a goat

4 - Monty then asks the contestant if they should switch doors of stick to their original choice.

To fully understand the underlying probabilities, see this Youtube description of the Monty Hall Problem.

Upon its face, the "switch" or "stick" choice offers a 50/50 probability, and a great many debates have taken place over this counterintuitive result.

To play the game and see the outcomes, see this UCSD simulation of the Monty Hall problem reveals that a player has a 66% change of winning if they switch.

Oracle ACE Steve Karam explains:

Yes, once the host opens a door, it /seems /there's a 50/50 chance of you having a goat or a car.  However, it's saying you must bank on the original statistics (33/33/33) when making your decision, because the host's choice is rigged.

*Swapping scenarios:*

33% chance:  The door you pick has a car.  The host reveals a goat.  You swap.  You get a goat.

66% chance:  The door you pick has a goat.  The host reveals the other goat.  You swap.  You get a car.

*Conclusion:* Swapping gave you a car 66% of the time.

*Non-swap scenarios:*

33% chance:  The door you pick has a car.  The host reveals a goat.  You don't swap.  You get a car.

66% chance:  The door you pick has a goat.  The host reveals the other goat.  You don't swap.  You get a goat.

*Conclusion:*  Not swapping gives you a car 33% of the time.

Really it's a mindscrew.  Regardless of the door the host opens, you have a 1 in 3 chance of choosing the car and a 2 in 3 chance of choosing a goat.  Because the host KNOWS which door has a goat, they are not changing the odds at all.  It's still a 1 in 3 chance you chose the car, and a 2 in 3 chance you chose a goat.  Because the host MUST open a door with a goat (whether you chose a car or not), it doesn't change the odds at all.

2 in 3 times, you chose a goat, 100% of the time the host will open a door with a goat, and therefore 2 in 3 times, switching gives you a car.

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Burleson is the American Team

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