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Each 5-digit group has a small probability of being in a given slot within the large number, but there are 88,889 such slots. This is the condition for which the Poisson distribution was designed. Any given group is expected to occur 88,889/100,000 times. The Poisson distribution with that mean has a distribution as given in the following table, also shown with value multiplied by 100,000, to indicate how many you'd expect to occur 0 times, 1 time, etc.:
prob freq accounting
for
0 0.411111834 41111 0
1 0.365433198 36543 36543
2 0.162414958 16241 32482
3 0.048123011 4812 14436
4 0.010694016 1069 4276
5 0.001901161 190 950
6 0.000281654 28 168
7 0.000035766 4 28
8 0.000003974
9 0.000000392 ______
88883
Rounding errors (including ignoring those with more than 7 occurrences) account for being 6 short of 88889.
So 36,543/88,883 of the groups could be expected to have no matches elsewhere in the number. That's about a 41% probability that a given group has no matches elsewhere in the number.
More precisely, we could use the actual binomial distribution.
0.00001^I*0.99999^(88889-I)*C(88889,I)
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I V
0 0.411110007
1 0.365435228 36544 36544
2 0.162415657 16242 32484
3 0.048122616 4812 14436
4 0.010693674 1069 4276
5 0.001901033 190 950
6 0.000281622 28 168
7 0.000035760 4 28
8 0.000003973
9 0.000000392 _____
88886
The 36544/88886 is still about a 41% probability that a given group of 5 has at no match anywhere else in the overall number, taking group-of-5 boundaries into consideration with no overlap.
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