Binary possibilities

volospian · 02-11-2011, 17:08

Guys,

I'm having a mental infarction today. I should be able to solve this puzzle, but I just can't get my head in gear...

If I had a 32 bit binary stream, but I only read the first 26 bits... what percentage of the final numbers *could" be the same as any other number... I thought it was simply the highest 32 bit number minus the highest 26 bit number and then finding the percentage between those two numbers... but I'm not sure it is...

it may help if I explain... we have some 32 bit card readers on our printers and we use the same cards on the door access systems, which, for reasons known only to estates, have 26 bit readers. Now it wasn't a problem until now. They now want to segregate some of the doors off, but we have identified some 32 bit cards that are duplicated in the door system as they have different 32 bit numbers where the difference lies only in the last (or potentially first, depending on how the reader works) 6 bits... so the reader only gets the 26 bits that are the same... does this make sense

so... for example (using 14 bits as it's easier to see) if I had two 14 bit cards with the decimal numbers of 12345 and 14393 the binary streams would look like this

11000000111001
11100000111001

however, if we only read the first 10 (or last, depending on how you read it

) bits they would look like..

0000111001
0000111001

which both add up to 57, so the reader would think they were the same card...

I don't need to know how many are going to be actually identical, just how many could actually report the same as a number already used...

02-11-2011, 17:08	#1
volospian Vodka Martini Join Date: May 2009 Posts: 786	Binary possibilities Guys, I'm having a mental infarction today. I should be able to solve this puzzle, but I just can't get my head in gear... If I had a 32 bit binary stream, but I only read the first 26 bits... what percentage of the final numbers *could" be the same as any other number... I thought it was simply the highest 32 bit number minus the highest 26 bit number and then finding the percentage between those two numbers... but I'm not sure it is... it may help if I explain... we have some 32 bit card readers on our printers and we use the same cards on the door access systems, which, for reasons known only to estates, have 26 bit readers. Now it wasn't a problem until now. They now want to segregate some of the doors off, but we have identified some 32 bit cards that are duplicated in the door system as they have different 32 bit numbers where the difference lies only in the last (or potentially first, depending on how the reader works) 6 bits... so the reader only gets the 26 bits that are the same... does this make sense so... for example (using 14 bits as it's easier to see) if I had two 14 bit cards with the decimal numbers of 12345 and 14393 the binary streams would look like this 11000000111001 11100000111001 however, if we only read the first 10 (or last, depending on how you read it) bits they would look like.. 0000111001 0000111001 which both add up to 57, so the reader would think they were the same card... I don't need to know how many are going to be actually identical, just how many could actually report the same as a number already used...