Teaser #8

[Softus]

You're asked to give three different characters from a password, for, say, access to an online banking web site.

The password is eight characters long.

The characters are always requested in sequence. For example:
1st, 2nd, 6th
or:
3rd, 4th, 7th
or:
5th, 6th, 8th

...and never in reverse order (for example 6th, 2nd, 1st etc.).

What's the probability of the last character being one of the three requested, on any given instance of a request?

 

[johnny_t]

There's probably a mathematical way of doing this, but ny pure number crunching.....

Sequences can be ...

1 2 3
1 2 4
1 2 5
1 2 6
1 2 7
1 2 8

1 3 4
1 3 5
1 3 6
1 3 7
1 3 8

.
.
.
.
.

1 7 8

or for sequences starting with 1, there are 6 + 5 + 4 + 3 +2 + 1 (21) possibilites.

now for sequences starting two, there are 5 + 4 + 3 + 2 + 1 (15)
starting with three, there are 4 + 3 + 2 + 1 (10)
starting with four, there are 3 + 2 + 1 (6)
starting with five, there are 2 + 1 (3)
and starting with 6, there is just one (6 7

which gives a total of 56 permutations, and the number of those that ends in an 8 is equal to the number of individual numbers I've lsted above, which is 21.

So the probability is 21/56, but I'm going to waste my morning on the mathematial proof now........

 

[empip]

56 3 character combos from 8 characters
6 occurrances with the final position used.

6/56 or 10.7%

I may have misinterpreted this because the question infers the first two characters being adjacent with a reducing number of spaces to the third.
But I will take a flyer here.
[code:1]
10000011
10000101
10001001
10010001
10100001
11000001
[/code:1]
Nothing ventured !

 

[Softus]

I may have misinterpreted this because the question infers the first two characters being adjacent with a reducing number of spaces to the third.

A reasonable inference, but I meant no such implication. 1,3,7, for example, is also a valid sequence.

 

[empip]

Hmmm the flyer just deservedly crashed ...
Misused excel BIN2DEC() forgot 'places' !

Back on the trail !

3 from 8 = (8x7x6 / 3x2x1) = 4x7x2 = 56
Then using binary pattern
21 Instances from 56 sets 21/56 ..... 37.5%

Binary pattern ..
2 bits and lsb

password
00000111
00001011
00001101
00010011
00010101
00011001
00100011
00100101
00101001
00110001
01000011
01000101
01001001
01010001
01100001
10000011
10000101
10001001
10010001
10100001
11000001

 

[vokera80e]

i'l say that in yours you have....I-Y-1.....

OOOOPS

 

[empip]

Oh bollo!
Combinations
3 from 8 = 8x7x6 / 3x2x1 = 56

Then with 1 fixed character there remains 2 from 7 = 7x6 / 2x1 = 21
being 37.5 % of 56.