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If you get an 11/100 on a CS test, but you claim it should be counted as a 'C', they'll probably decide you deserve the upgrade.
Title text: If you get an 11/100 on a CS test, but you claim it should be counted as a 'C', they'll probably decide you deserve the upgrade.


The binary numeral system refers to a counting system in base-2, which uses only the digits 0 and 1, as opposed to the more familiar base-10 decimal system, which uses the digits 0 through 9. In this case, the scale of 1 to 10 is using binary, so in decimal it would be a scale of 1 to 2. Since 4 in binary is "100" it doesn't fit into the range "1" to "10" in a binary system. And Megan doesn't even know the number "4" because she's only working on the binary system, this character does not exist for her.

It is also possible that Megan is using base-3 or base-4, both of which don't have a 4 (base-3 counts 1, 2, 10, etc., and base 4 counts 1, 2, 3, 10 etc.)

It should be noted that if Megan is speaking out loud in such a way that confuses Cueball, she would be saying "ten" out loud; this would automatically indicate she is indeed using base-10 (or higher). The correct pronunciation of "10" in base-2 is "one zero", so Megan is lying to Cueball.

The title text uses a similar joke. Since test scores are usually written as either a letter grade or a percentage, 11 correct questions out of 100 would be a failing score in decimal notation. However, 11/100 in binary translates to 3/4 in decimal, which would be 75%, accepted in most classes as a 'C' grade.

It could also be argued that a score of 11 should count as a "B", as 11 is B in hexadecimal; however, this link is a bit more tenuous, as the whole score would then be interpreted as "B/64".


Megan: On a scale of 1 to 10, how likely is it that this question is using binary?
Cueball: ...4?
Megan: What's a 4?

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One of correct answers is P = 1 + 1 - |sgn(10 - 1 - 1)|

(|x| is absolute value of x, sgn(x) is 1 when x > 0, 0 when x = 0, and -1 when x < 0)

If 10 = 1 + 1, then P = 10 - |sgn(0)| = 10 - |0| = 10

If 10 > 1 + 1, then P = 1 + 1 - |sgn(10 - 1 - 1)| = 1 + 1 - |1| = 1

If 10 < 1 + 1, then P = 1 + 1 - |sgn(10 - 1 - 1)| = 1 + 1 - |-1| = 1

So P is 10 iif the question was is in binary, and 1 iif it was not in binary. 16:26, 6 February 2013 (UTC)

The absolute value is unnecessary. When is 10 ever less than 1+1? 20:28, 3 January 2014 (UTC)
In base -2. 12:32, 26 June 2015 (UTC)

I don't think the explanation is right, I mean i don't know binary but i don't think the joke is that he's saying a 4 as in 100% Lackadaisical (talk) 00:23, 7 November 2013 (UTC)

A 4 is not 100%, but a 3/4 is always 75%. 22:47, 26 January 2014 (UTC)
Actually, my comment was in reference to this: "Since 4 in binary is "100" (one-zero-zero) the joke is that it is 100% likely that the question is binary -- or it could simply be 4 of 10 - which means that the question has evolved into recursive ambiguity. Also, the person asking the question does not know what a 4 is since there is no 4 in binary." The problem I had with it was taken care of in a previous edit (The quote was taken from the 31 December 2013 edit.)--Lackadaisical (talk) 22:03, 26 March 2015 (UTC)

1.(1) is the best answer I've got Halfhat (talk) 11:53, 5 April 2014 (UTC)

"How likely" it is? As everyone knows, "every base is base 10", since every base number in its own numbering system is written as "10" (2 is 10 in binary, 16 is 10 in hex and so on). So that question could be in EVERY number system possible. There are an infinite amount of number systems that use the symbols 1 and 0. I suppose the probability is then 1 over an infinite number of systems, then very unlikely, so I'd say (as 0 is not in the range of possible answers) the answer is 1. Which, incidentally, is also an acceptable answer for every system. If we want instead to take into account that Megan doesn't know what a 4 is, the possibilities are only base 2, 3 and 4. So the likeliness is 1/3, which corresponds anyway to 1 in those number systems. -- 14:05, 3 June 2014 (UTC)

Naw, there's unary. In unary 1 is |, 5 is |||||, ten is ||||||||||, and so on. StillNotOriginal 02:16, 21 May 2018 (UTC)

If we assume the question was auditory to the questioned, the way '10' was articulated is another source of information: the enquirer can pronounce '10' as "ten" or "one zero". In the first case, if he said "ten", this points towards the base system being greater than/equal to 10. For example "ten" would be "10" in decimal (base 10) and "16" in hexadecimal (base 16). The word "ten" stops making sense for bases 2 through 9, which leads us to the conclusion of this case: the probability of the question being base 2 is zero, therefore the closest possible score one can give is 1. In the case that "10" is pronounced "one zero": using the same logic as in the first case, the question can use any base 2 through 9. giving us a probability of 1/8, the nearest score again being 1.

So if the question were: "On a scale from one to one zero, how likely is this question using the decimal system?" Could be answered with full confidence (talk) (please sign your comments with ~~~~)

It seems that the best answer to this question is 1.11111... because it approaches 10 in binary, and is very low in almost any other number system. (talk) (please sign your comments with ~~~~)

I removed the claim that 11 could be interpreted as a (hex) 'B' grade. The grade isn't given as '11', but as '11/100', which would be 'B/64' in hex. Condor70 (talk) 13:28, 21 August 2015 (UTC)

Just pronounce numbers as if they were decimal. If you group the digits every 4, use the -yllion and not the -illion system. It's not that hard to say "one, ten, eleven, one hundred, one hundred [and] one, one hundred [and] ten, one hundred [and] eleven, ten hundred, ten hundred [and] one, ten hundred [and] ten, ten hundred [and] eleven, eleven hundred, eleven hundred [and] one, eleven hundred [and] ten, eleven hundred [and] eleven, one myriad" when you count in binary "1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 1 0000." 17:40, 3 July 2024 (UTC)