Binary, base 2, bits, and coin flips

[RabbitWho]

I am studying psychology and have always been awful at maths (I think it's because I don't have much working memory), I am coming at this as a complete beginner. It is relevant to the Information Processing model of psychology.
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The book says: Imagine you throw a coin once.. once it lands you have 1 bit of information. There are two (2) possible results
[log₂(2)] = 1 bit

If you throw a coin twice you have two bits of info, there are four possible results (xx,++,+x,x+) (I don't know why they mention the possible results, is that relevant?)
[log₂(4)] = 2 bits

Am I right in thinking the 4 in that equation corresponds to the four possible results?

The examples in the book end there, and I have been trying to expand it to check that I understand.

[log₂(8)] = how many bits? I guess that question means "How many binary digits do you need to represent 8?" There are 256 possible results.. is that relevant?

[log2(8)] =3 bits if we were talking about coin tosses, and 4 bits on a computer because for some reason they always add one. Am I right?In a computer
zero= 0
one = 1
two = 10
three = 11
four = 100

Why isn't 01 four?
Here I will guessfive = 101
six = 110
seven =111
eight = 1000

log base 2 of 8 is 3 so why do I need 8 digits to represent 8? Why can't I use 011?

A friend suggested that the computer would see 01 as being the same as 10. I am imagining a cent and a euro, they aren't the same coin so a tails with one is not the same as a tails with the other, they aren't interchangeable.. but I would have thought the same went for computer transistors? That is what 1 and 0 represent for computers, right? on and off switches?

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Thanks for any help you can give. I have googled it, but all the tutorials I find impart all the information relevant for IT then very quickly go into more advanced things like bytes etc. and don't deal in detail with the basic concept.

 

[.Scott]

If you throw a coin twice you have two bits of info, there are four possible results (xx,++,+x,x+) (I don't know why they mention the possible results, is that relevant?)
[log₂(4)] = 2 bits

Am I right in thinking the 4 in that equation corresponds to the four possible results?

Yes.

The examples in the book end there, and I have been trying to expand it to check that I understand.

[log₂(8)] = how many bits? I guess that question means "How many binary digits do you need to represent 8?" There are 256 possible results.. is that relevant?

You need 3 bits to represent one of eight possibilities.

[log2(8)] =3 bits if we were talking about coin tosses, and 4 bits on a computer because for some reason they always add one. Am I right?

If you are talking about how many states can be coded, it would be 3 on a computer as well. If you are talking about encoding the number 8, then there is an extra bit - because the lowest number encoded by a bit string is usually zero. So you are counting from 0 to 8, nine different value - so you you ned more than 3 bits.

In a computer
zero= 0
one = 1
two = 10
three = 11
four = 100

Why isn't 01 four?

You're not showing both bits.
In a computer
zero= 000
one = 001
two = 010
three = 011
four = 100

So 01 would be 001 - the code for one.

Here I will guessfive = 101
six = 110
seven =111
eight = 1000

log base 2 of 8 is 3 so why do I need 8 digits to represent 8? Why can't I use 011?

Same as above - this time with four bits.
In a computer
five = 0101
six = 0110
seven =0111
eight = 1000

So 011 would be 0011 - the code for three.

A friend suggested that the computer would see 01 as being the same as 10. I am imagining a cent and a euro, they aren't the same coin so a tails with one is not the same as a tails with the other, they aren't interchangeable.. but I would have thought the same went for computer transistors? That is what 1 and 0 represent for computers, right? on and off switches?

There are codes where a different number of bits (coin tosses) are use for different numbers. But in normal computer coding, there are fixed fields of bits - with a preset size. So if your number is eight and you are using 32-bit arithmetic, then you will see 00000000000000000000000000001000.

 

[RabbitWho]

Thank you so much for your help. I think I understand.. So, for example

[log₂(32)] = 5 bits?

 

[.Scott]

Thank you so much for your help. I think I understand.. So, for example

[log₂(32)] = 5 bits?

Yup!

 

[Mark44]

Thank you so much for your help. I think I understand.. So, for example

[log₂(32)] = 5 bits?

log₂(32) = 5, which is a number -- units should not be included.
Using 5 bits you can represent 32 numbers: 0, 1, 2, 3, ..., 31
The binary bit pattern for 31 (ie, in base-2) is 11111, which means $1 \text{ x } 2^4 + 1 \text{ x } 2^3 + 1 \text{ x } 2^2 + 1 \text{ x } 2^1 + 1 \text{ x } 2^0 = 16 + 8 + 4 + 2 + 1 = 31$

 

[Svein]

Thank you so much for your help. I think I understand.. So, for example

[log₂(32)] = 5 bits?

Going the other way: A simple microcontroller handles information in 8 bit chunks (such a chunk is called a byte). What is the largest value you can represent in a byte?

 

[meBigGuy]

In decimal the positions are weighted as powers of 10. In binary the positions are weighted as powers of 2

In decimal 10 symbols (0-9). The positions are (from right to left) 10^0, 10^1, 10^2 or 1, 10, 100
In binary 2 symbols (0-1). The positions are (from right to left) 2^0, 2^1, 2^2 or 1, 2, 4

Totally analagous