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 Replies: 12 / Views: 4,276 |
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Pillar of the Community
Thailand
1509 Posts |
Is there anybody out there who can correct me if (and probably) I'm wrong with my maths.
In an eight digit series (like US$) I get:
00011000 to 00099000 = 9
00100100 to 00900900 = 9 x 9 = 81 (to include the above)
01000010 to 09000090 = 9 x 81 = 729
10000001 to 90000009 = 9 x 729 = 6561 = 0.006561%
1 in 15,241 is a radar.
In a seven digit series (like the Thai notes I handle):
0001000 to 0009000 = 9
0010100 to 0090900 = 9 x 9 = 81
0100010 to 0900090 = 9 x 81 = 729
1000001 to 9000009 = 9 x 729 = 6561 = 0.06561%
1 in 1,524 is a radar (so why haven't I found one yet after months and months of looking?)
In the old days I would have written a quick progam to do the donkey work but it's been so long since I messed with loops and IF...THENs.
Vic
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Moderator
 Australia
16826 Posts |
I would reckon it this way.
There are 10,000 numbers in between 00010000 and 00019999. Out of those 10,000 numbers, only one of them is a radar number - 00011000. (assuming all 10,000 of those notes were actually printed). The same truth holds for 0002xxxx, 0003xxxx and so on. Now, there are 10,000 sets of 10,000 notes between 0000xxxx and 9999xxxx, and the only set of notes that wouldn't have a radar amongst them is 0000xxxx, because no countries ordinarily produce a "zero note". That would give a total of 9999 notes out of 99,999,999 or a probability of 1 in 10001 across an entire eight-digit series.
Similar reasoning for an odd number of serial numbers would run like this: There are 1,000 numbers in between 0001000 and 0001999. Out of those 1,000 numbers, only one is a radar number - 0001000. The same truth holds for 0002xxx, 0003xxx and so on. There are 10,000 sets of notes between 0000xxx and 9999xxx, and only one set has no radar: 0000xxx. Therefore there are 9999 notes out of 9,999,999 or a probability of just over 1 in 1000.
Mathematically, if the number of digits in the serial number is n, then the probability of a radar is approximately 1 in 10 to the (n/2)th power, plus one; if n/2 is not a whole number then the fraction is rounded down.
Thus, your odds of finding a radar in the seven-digit Thai series are actually slightly better than finding one in the six-digit Australian series, though both are approximately 1 in 1,000. And I've only ever found one Australian radar note.
Don't say "infinitely" when you mean "very"; otherwise, you'll have no word left when you want to talk about something really infinite. - C. S. Lewis
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Pillar of the Community
United States
1285 Posts |
Thai / Sap, Interesting topic - I wanted to do something similar to this topic but gave up. Thai, I believe that most folks when they see a radar pull it and that's why you never find one. Chances are good they do not know it is radar but rather think the number is cool / neat. As it relates US notes a "true true" radar would have the ending ALPHA same as the Beginning alpha. I search quite a few US $1 and only have found 2 circulated radars so far. But, I have found quite a few un circulated radars. In the long run you are better off staying with UNC's for fancy numbers such as radars. Further, some folks eliminate the leading zero's (0014 3341 - example) and will call this a "mini radar". As it relates to me, technically the first radar would be 0001 1000 and then every 11k till you hit 99 and after 1k for 00 (every 10K mathematically) and repeat it self. Also the pricing on radars varies greatly. Below is a link to my analysis and another topic on binary radars.. Please take a look and chime in with your thoughts. These prices have come down further for 4 digit radars. What you want to find in a radar is also a binary, repeater etc meaning have some other fancy number combo going for it. https://goccf.com/t/62788https://goccf.com/t/61320The best one I have is 4411 1144 (binary, 2 digit bookend radar). The pic of the brick is in the second link. I view repeaters the same as radars (mathematically every 10k) Peace |
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Pillar of the Community
United States
2605 Posts |
For me it works this way:
There are exactly 9999 numbers from 0001 to 9999 (since we're eliminating "zero notes"). Now, for each of this numbers abcd there is exactly one radar, dcbaabcd. Ta-duh! We got 9999 radars out of 99999999 with eight-digit notes.
With seven-digit notes for each number abc between 001 and 999 there are exactly 10 radars, cba0abc, cba1abc, ..., cba9abc. Plus 9 radars for 000 (again we're eliminating "zero notes"). Therefore we have 9990+9=9999 radars out of 9999999.
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Pillar of the Community
Canada
3692 Posts |
IF
Seems to me that "mini-radars" wouldn't be as collectable.
THEN
They don't have that novelty of being a pure radar, and thus rarer.. (?)
STOP
Maybe a mathematician can correct me there if it's rarer.
GO TO
I'd say that anything into the zeros is scarce enough.
Edited by Libertad
05/25/2010 7:53 pm
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Pillar of the Community
United States
2605 Posts |
Quote:
Maybe a mathematician can correct me there if it's rarer. Let's assume we're dealing with eight-digit numbers. Since there are 9999 true radars for 7-digit numbers then we'll have 9999 mini-radars with the first 0. Actually, it's not how many zeros in front, but rather where it is centered. I'll provide the chart (for the figures see Sap's speculations). Red character is the "mirror". True radars: Quantity: #### |#### , 9999 Mini-radars: Quantity: 0### ####, 9999 00### |###, 999 000## ###, 999 0000## |##, 99 00000# ##, 99 000000# |#, 9 So, we have 9999+999+999+99+99+9 mini-radars while only 9999 true ones. |
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Pillar of the Community
United States
1285 Posts |
Just a FYI
The Federal Reserve does NOT go up to 9999 9999.
For 1, 5, 10 and 20 - 9620 0000
For 50 and 100 - 9920 0000
And they roll over to the next ending alpha. The high numbers you see are from collector sets and un cut sheets they sell which have been cut up in some instances. They did this starting in 80's.
Svslav, If I may - what are the odds of a "super repeater"?
0101 0101 (would be the very first one I think). Any combo of 2 numbers repeating them selves such as 7474 7474.
Thanks
Peace
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Pillar of the Community
United States
2605 Posts |
Quote:
Svslav, If I may - what are the odds of a "super repeater"? That's easy. There are only 90 numbers (from 01 to 98 with numbers like 11, 22, etc. are removed) that can "super repeat". So, it's 90 in 99999999 (or whatever smaller number of SN's is there). And there are only 9 "extra super repeaters" like 55555555. |
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Pillar of the Community
United States
1397 Posts |
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Pillar of the Community
 Thailand
1509 Posts |
Wow Nickelman that is some find! Congrats.
I'll also add my thanks now to everyone above who pointed out the error of my (mathematical) ways. Well I did say right at the start that I was probably wrong. Too many experts in too many fields here, for which I'm thankful.
Vic
Edited by thai-vic
06/09/2010 9:45 pm
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Pillar of the Community
United States
1285 Posts |
It depends on how much is printed. For instance Boston's first print block would be true radars and NY's 2nd block, etc Richmond (E) at 5th block etc till san fran (L) at the 12th block. So for Richmond (E) they would have to print 96 Mil X 4 = 384 Mil notes b4 you would hit the E block. Also some fed districts may NOT even hit the matching ALPHA block. In some series ending alpha's never get that many print runs and end around F or G. So for instance I = Minneapolis will not have a print block ending in "I" as they do not print much. Ergo, there is no simple answer to your question.
This is the best way I can think of explaining this to you. Not sure if someone else has a better way of doing it.
Peace
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Pillar of the Community
United States
1285 Posts |
Edited by Ceylon62
07/28/2010 08:18 am
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Valued Member
Canada
290 Posts |
Thought I would revive this thread since probabilities of encountering a radar recently peaked my interest. I agree with your calculations fellows, but these probabilities would only be realistic if there were no variables, such as the fact that collectors will remove radars from the circulation.
Just for fun, I calculated the probabilities of getting a Canadian dollar radar if the notes go up to 9,600,000 instead of 9,999,999. It came up pretty similar. From 9,600,000 to 9,999,999 there are 440 radars that I could count:
9601069 = 10
9610169 = 100
9701079 = 10
9710179 = 100
9801089 = 10
9810189 = 100
9901099 = 10
9910199 = 100
Total = 440
So instead of 9999 radars out of 9,999,999 (1/1000.09990999) there are 9559 out of 9,600,000 which it a probability of 1/1004.28915158. So, I don't think the maximum number of notes printed per prefix really affect your chance of randomly getting a radar. But If you buy straps of UNC notes and you have the choice, you may want to get the first 90 straps because you'll get 9 radars(0001000 to 0009000) and the straps after will only give you a radar each 1100 notes instead of each 1000. I haven't calculated past that, but I guess it oscillates between one radar every 1000 and one radar every 1100 and this all the way up to 9,999,999 ?
Also, I calculated the probabilities of encountering a EUR radar note:
00000100000 = 9
00001010000 = 90
00010001000 = 900
00100000100 = 9000
01000000010 = 90000
10000000001 = 900000
999999 / 99999999999
So 1 chance out of 100000.099999 to find one. Makes me want to hunt for one ^^ Problem is... I live in Canada :(
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Replies: 12 / Views: 4,276 |
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Posted 05/25/2010 06:08 am
