Mixing
Probability of attaining same prime factor [on hold]
Clash Royale CLAN TAG#URR8PPP
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EDITED:
Two coloums of numbers are presented, there are 234 entries in both columns. Column A, values are not multiples of 19, natural numbers increasing by a random value with mean of 30.
Column B, all numbers are multiples of 19, multiplied by a random integer, from 2 to 300. Summing column B, will give a prime factor of 19 x 55049 and column A, prime factor of 17 x 55049.
Question, what are the chances for obtaining the same, prime factor 55049?
A = [5, 11, 24, 58, 106, 123, 243, 244, 249, 292, 293, 306, 321, 336, 354, 374, 417, 421, 496, 532, 545, 566, 583, 587, 641, 659, 675, 711, 728, 744, 749, 757, 784, 810, 833, 869, 879, 929, 978, 1016, 1019, 1047, 1050, 1055, 1077, 1099, 1102, 1117, 1156, 1167, 1177, 1220, 1244, 1249, 1254, 1255, 1272, 1288, 1326, 1338, 1348, 1394, 1430, 1439, 1440, 1446, 1484, 1486, 1573, 1591, 1621, 1622, 1635, 1636, 1646, 1748, 1752, 1762, 1766, 1788, 1820, 1839, 1848, 1852, 1874, 1897, 1934, 1941, 1950, 1969, 1970, 1979, 2018, 2081, 2085, 2147, 2176, 2202, 2203, 2252, 2257, 2285, 2300, 2316, 2319, 2320, 2322, 2332, 2345, 2400, 2406, 2425, 2428, 2466, 2473, 2507, 2528, 2532, 2551, 2561, 2571, 2602, 2603, 2610, 2614, 2657, 2665, 2717, 2747, 2752, 2767, 2807, 2811, 2816, 2818, 2825, 2846, 2862, 2896, 2913, 2952, 2961, 3005, 3020, 3022, 3027, 3047, 3087, 3113, 3119, 3138, 3171, 3220, 3255, 3270, 3285, 3426, 3433, 3465, 3491, 3509, 3522, 3531, 3544, 3563, 3641, 3651, 3694, 3698, 3714, 3715, 3728, 3743, 3761, 3763, 3780, 3824, 3828, 3841, 3852, 3862, 3874, 3904, 3916, 3948, 3971, 4008, 4050, 4109, 4122, 4154, 4158, 4167, 4183, 4192, 4193, 4209, 4215, 4238, 4259, 4268, 4307, 4328, 4330, 4347, 4365, 4382, 4398, 4418, 4463, 4480, 4522, 4527, 4618, 4632, 4672, 4676, 4699, 4739, 4762, 4775, 4784, 4788, 4844, 4888, 4890, 4896, 4906, 4907, 4913, 4928, 4957, 4993, 4998, 5015, 5052, 5084, 5106, 5135, 5167, 5174, 5207, 5227, 5230, 5272, 5308, 5317, 5324, 5353, 5395, 5408, 5460, 5464, 5509, 5536, 5565, 5594, 5597, 5602, 5616, 5638, 5648, 5658, 5664, 5690, 5692, 5710, 5716, 5753, 5763, 5793, 5854, 5863, 5874, 5880, 5895, 5903, 5939, 5945, 5947, 5963, 5990, 6033, 6037, 6043, 6064, 6087, 6125, 6130, 6149, 6161, 6189, 6195, 6208]
B = [836, 2641, 7809, 5871, 1938, 4541, 8189, 11894, 2261, 6042, 12863, 8018, 7923, 1995, 5510, 13148, 3743, 3439, 8778, 3116, 2394, 4826, 8569, 14915, 2280, 1938, 16074, 7923, 3363, 6954, 6650, 2679, 4104, 5206, 7999, 6517, 3515, 7030, 4712, 3211, 3762, 5035, 6270, 5643, 6289, 7391, 6897, 10241, 3021, 7334, 5358, 8816, 3040, 3838, 5092, 3477, 5377, 2527, 7448, 4351, 5130, 3857, 6973, 4104, 1900, 1463, 3743, 6289, 1976, 2261, 4009, 4180, 2660, 5719, 3553, 3477, 4332, 4294, 836, 4598, 2052, 1653, 931, 1767, 1482, 836, 7809, 2090, 3059, 4104, 8436, 4560, 1615, 3914, 4085, 1976, 3496, 3990, 7448, 2109, 2679, 4636, 1615, 2527, 2603, 1843, 4579, 2109, 1045, 2375, 3572, 5339, 2090, 1425, 2375, 8265, 3534, 3933, 2109, 5358, 2983, 2204, 2698, 7942, 6517, 2033, 4750, 7049, 2489, 3857, 2204, 5795, 2546, 1843, 9443, 5358, 13281, 3648, 5206, 3762, 2489, 4123, 1406, 513, 1273, 342, 1197, 2375, 1786, 1691, 855, 6441, 4921, 1767, 7144, 3059, 1520, 5757, 6783, 2831, 2698, 2755, 2698, 1254, 5472, 2128, 5681, 4104, 4484, 3363, 3325, 7315, 3686, 2052, 817, 2318, 2166, 1026, 3135, 2071, 1026, 950, 1767, 1026, 1026, 2926, 1615, 1653, 3819, 2413, 9576, 4446, 7980, 4332, 2166, 5016, 3002, 9025, 2622, 2831, 12046, 1957, 1691, 3591, 1919, 3059, 3116, 1634, 855, 1748, 779, 6460, 8360, 4142, 3040, 2223, 3040, 2527, 836, 1520, 3002, 665, 152, 1805, 4864, 1254, 817, 931, 798, 1330, 1463, 2318, 1159, 627, 513, 2375, 5396, 6422, 6593, 2546, 4902, 3249, 4256, 6726, 1425, 1368, 3002, 931, 2432, 1425, 1805, 2793, 4237, 950, 494, 836, 1178, 1843, 3363, 1615, 304, 475, 2394, 969, 2242, 1558, 2831, 836, 323, 1653, 76, 817, 1425, 3439, 2109, 1026, 1045, 779, 133, 76, 1444, 874, 228, 2090, 551, 361, 1292, 1349, 1026, 1862, 1995, 2223, 1349, 1330]
Sum A = 17 x 55049 = 935833
Sum B = 19 x 55049 = 1045931
Just attaining the same factor is, by two assumption, "random" numbers, the probability, mathematically speaking is 1/f * 1/f, very small.
But this does not take into account, the number series in list a and b, producing the same factor. I want to quantify this probability.
I am thinking to quantify this probability, if we assume, for list A to increase by gaussian mean with added noise of about 5, multiply this probability with a two digit random number and floor the integer to the lowest int. Having a probability distribution describing the increase in verse values with a gaussian mean, offset of about 5. Next for list B, the numerical values shifts sometimes with 1000's and 100. Therefor we could 19 * X, approach X, with a randomly produced digit number varying from 1 to 3 digits, X is produced the same way as for A, with multiplying with a gaussian probability distribution.
So programming wise it would look something like
r.uniform() x random.randint(1,200) = X
X * 19 = entries in B.
Assume that the two lists are generated as described above.
Now that we have roughly estimated how to describe this process randomly. We can then apply other statistical methods in assuming, what is the probability for the null hypothesis, i.e. gaining the same 19 x 55049, 17 x 55049, factor on both list is within bounds of non-significant p values. I am still not sure how to do this.
probability probability-distributions prime-factorization prime-twins
asked Aug 3 at 17:52
Asael
11
put on hold as off-topic by Andrés E. Caicedo, Adrian Keister, max_zorn, Leucippus, Somos Aug 4 at 3:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." РAndr̩s E. Caicedo, Adrian Keister, max_zorn, Leucippus, Somos
 |Â
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up vote
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EDITED:
Two coloums of numbers are presented, there are 234 entries in both columns. Column A, values are not multiples of 19, natural numbers increasing by a random value with mean of 30.
Column B, all numbers are multiples of 19, multiplied by a random integer, from 2 to 300. Summing column B, will give a prime factor of 19 x 55049 and column A, prime factor of 17 x 55049.
Question, what are the chances for obtaining the same, prime factor 55049?
A = [5, 11, 24, 58, 106, 123, 243, 244, 249, 292, 293, 306, 321, 336, 354, 374, 417, 421, 496, 532, 545, 566, 583, 587, 641, 659, 675, 711, 728, 744, 749, 757, 784, 810, 833, 869, 879, 929, 978, 1016, 1019, 1047, 1050, 1055, 1077, 1099, 1102, 1117, 1156, 1167, 1177, 1220, 1244, 1249, 1254, 1255, 1272, 1288, 1326, 1338, 1348, 1394, 1430, 1439, 1440, 1446, 1484, 1486, 1573, 1591, 1621, 1622, 1635, 1636, 1646, 1748, 1752, 1762, 1766, 1788, 1820, 1839, 1848, 1852, 1874, 1897, 1934, 1941, 1950, 1969, 1970, 1979, 2018, 2081, 2085, 2147, 2176, 2202, 2203, 2252, 2257, 2285, 2300, 2316, 2319, 2320, 2322, 2332, 2345, 2400, 2406, 2425, 2428, 2466, 2473, 2507, 2528, 2532, 2551, 2561, 2571, 2602, 2603, 2610, 2614, 2657, 2665, 2717, 2747, 2752, 2767, 2807, 2811, 2816, 2818, 2825, 2846, 2862, 2896, 2913, 2952, 2961, 3005, 3020, 3022, 3027, 3047, 3087, 3113, 3119, 3138, 3171, 3220, 3255, 3270, 3285, 3426, 3433, 3465, 3491, 3509, 3522, 3531, 3544, 3563, 3641, 3651, 3694, 3698, 3714, 3715, 3728, 3743, 3761, 3763, 3780, 3824, 3828, 3841, 3852, 3862, 3874, 3904, 3916, 3948, 3971, 4008, 4050, 4109, 4122, 4154, 4158, 4167, 4183, 4192, 4193, 4209, 4215, 4238, 4259, 4268, 4307, 4328, 4330, 4347, 4365, 4382, 4398, 4418, 4463, 4480, 4522, 4527, 4618, 4632, 4672, 4676, 4699, 4739, 4762, 4775, 4784, 4788, 4844, 4888, 4890, 4896, 4906, 4907, 4913, 4928, 4957, 4993, 4998, 5015, 5052, 5084, 5106, 5135, 5167, 5174, 5207, 5227, 5230, 5272, 5308, 5317, 5324, 5353, 5395, 5408, 5460, 5464, 5509, 5536, 5565, 5594, 5597, 5602, 5616, 5638, 5648, 5658, 5664, 5690, 5692, 5710, 5716, 5753, 5763, 5793, 5854, 5863, 5874, 5880, 5895, 5903, 5939, 5945, 5947, 5963, 5990, 6033, 6037, 6043, 6064, 6087, 6125, 6130, 6149, 6161, 6189, 6195, 6208]
B = [836, 2641, 7809, 5871, 1938, 4541, 8189, 11894, 2261, 6042, 12863, 8018, 7923, 1995, 5510, 13148, 3743, 3439, 8778, 3116, 2394, 4826, 8569, 14915, 2280, 1938, 16074, 7923, 3363, 6954, 6650, 2679, 4104, 5206, 7999, 6517, 3515, 7030, 4712, 3211, 3762, 5035, 6270, 5643, 6289, 7391, 6897, 10241, 3021, 7334, 5358, 8816, 3040, 3838, 5092, 3477, 5377, 2527, 7448, 4351, 5130, 3857, 6973, 4104, 1900, 1463, 3743, 6289, 1976, 2261, 4009, 4180, 2660, 5719, 3553, 3477, 4332, 4294, 836, 4598, 2052, 1653, 931, 1767, 1482, 836, 7809, 2090, 3059, 4104, 8436, 4560, 1615, 3914, 4085, 1976, 3496, 3990, 7448, 2109, 2679, 4636, 1615, 2527, 2603, 1843, 4579, 2109, 1045, 2375, 3572, 5339, 2090, 1425, 2375, 8265, 3534, 3933, 2109, 5358, 2983, 2204, 2698, 7942, 6517, 2033, 4750, 7049, 2489, 3857, 2204, 5795, 2546, 1843, 9443, 5358, 13281, 3648, 5206, 3762, 2489, 4123, 1406, 513, 1273, 342, 1197, 2375, 1786, 1691, 855, 6441, 4921, 1767, 7144, 3059, 1520, 5757, 6783, 2831, 2698, 2755, 2698, 1254, 5472, 2128, 5681, 4104, 4484, 3363, 3325, 7315, 3686, 2052, 817, 2318, 2166, 1026, 3135, 2071, 1026, 950, 1767, 1026, 1026, 2926, 1615, 1653, 3819, 2413, 9576, 4446, 7980, 4332, 2166, 5016, 3002, 9025, 2622, 2831, 12046, 1957, 1691, 3591, 1919, 3059, 3116, 1634, 855, 1748, 779, 6460, 8360, 4142, 3040, 2223, 3040, 2527, 836, 1520, 3002, 665, 152, 1805, 4864, 1254, 817, 931, 798, 1330, 1463, 2318, 1159, 627, 513, 2375, 5396, 6422, 6593, 2546, 4902, 3249, 4256, 6726, 1425, 1368, 3002, 931, 2432, 1425, 1805, 2793, 4237, 950, 494, 836, 1178, 1843, 3363, 1615, 304, 475, 2394, 969, 2242, 1558, 2831, 836, 323, 1653, 76, 817, 1425, 3439, 2109, 1026, 1045, 779, 133, 76, 1444, 874, 228, 2090, 551, 361, 1292, 1349, 1026, 1862, 1995, 2223, 1349, 1330]
Sum A = 17 x 55049 = 935833
Sum B = 19 x 55049 = 1045931
Just attaining the same factor is, by two assumption, "random" numbers, the probability, mathematically speaking is 1/f * 1/f, very small.
But this does not take into account, the number series in list a and b, producing the same factor. I want to quantify this probability.
I am thinking to quantify this probability, if we assume, for list A to increase by gaussian mean with added noise of about 5, multiply this probability with a two digit random number and floor the integer to the lowest int. Having a probability distribution describing the increase in verse values with a gaussian mean, offset of about 5. Next for list B, the numerical values shifts sometimes with 1000's and 100. Therefor we could 19 * X, approach X, with a randomly produced digit number varying from 1 to 3 digits, X is produced the same way as for A, with multiplying with a gaussian probability distribution.
So programming wise it would look something like
r.uniform() x random.randint(1,200) = X
X * 19 = entries in B.
Assume that the two lists are generated as described above.
Now that we have roughly estimated how to describe this process randomly. We can then apply other statistical methods in assuming, what is the probability for the null hypothesis, i.e. gaining the same 19 x 55049, 17 x 55049, factor on both list is within bounds of non-significant p values. I am still not sure how to do this.
probability probability-distributions prime-factorization prime-twins
asked Aug 3 at 17:52
Asael
11
put on hold as off-topic by Andrés E. Caicedo, Adrian Keister, max_zorn, Leucippus, Somos Aug 4 at 3:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." РAndr̩s E. Caicedo, Adrian Keister, max_zorn, Leucippus, Somos
why is this down voted?
– Asael
Aug 3 at 18:01
This question is very unclear. It might help if you name a variable or two, and clarify what you're asking for.
– G Tony Jacobs
Aug 3 at 18:01
Tried to do my best at explaining, try to see the edited version. Thanks.
– Asael
Aug 3 at 18:06
There is no such thing as a random integer until you tell us how they are determined. There is no uniform probability measure on a countable infinite space.
– Thomas Andrews
Aug 3 at 18:37
Thomas Andrews, I am currently at the process at writing the reason as of why I am asking this question. And how the integers are picked, in both columns.
– Asael
Aug 3 at 18:39
 |Â
show 2 more comments
up vote
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down vote
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up vote
-3
down vote
favorite
EDITED:
Two coloums of numbers are presented, there are 234 entries in both columns. Column A, values are not multiples of 19, natural numbers increasing by a random value with mean of 30.
Column B, all numbers are multiples of 19, multiplied by a random integer, from 2 to 300. Summing column B, will give a prime factor of 19 x 55049 and column A, prime factor of 17 x 55049.
Question, what are the chances for obtaining the same, prime factor 55049?
A = [5, 11, 24, 58, 106, 123, 243, 244, 249, 292, 293, 306, 321, 336, 354, 374, 417, 421, 496, 532, 545, 566, 583, 587, 641, 659, 675, 711, 728, 744, 749, 757, 784, 810, 833, 869, 879, 929, 978, 1016, 1019, 1047, 1050, 1055, 1077, 1099, 1102, 1117, 1156, 1167, 1177, 1220, 1244, 1249, 1254, 1255, 1272, 1288, 1326, 1338, 1348, 1394, 1430, 1439, 1440, 1446, 1484, 1486, 1573, 1591, 1621, 1622, 1635, 1636, 1646, 1748, 1752, 1762, 1766, 1788, 1820, 1839, 1848, 1852, 1874, 1897, 1934, 1941, 1950, 1969, 1970, 1979, 2018, 2081, 2085, 2147, 2176, 2202, 2203, 2252, 2257, 2285, 2300, 2316, 2319, 2320, 2322, 2332, 2345, 2400, 2406, 2425, 2428, 2466, 2473, 2507, 2528, 2532, 2551, 2561, 2571, 2602, 2603, 2610, 2614, 2657, 2665, 2717, 2747, 2752, 2767, 2807, 2811, 2816, 2818, 2825, 2846, 2862, 2896, 2913, 2952, 2961, 3005, 3020, 3022, 3027, 3047, 3087, 3113, 3119, 3138, 3171, 3220, 3255, 3270, 3285, 3426, 3433, 3465, 3491, 3509, 3522, 3531, 3544, 3563, 3641, 3651, 3694, 3698, 3714, 3715, 3728, 3743, 3761, 3763, 3780, 3824, 3828, 3841, 3852, 3862, 3874, 3904, 3916, 3948, 3971, 4008, 4050, 4109, 4122, 4154, 4158, 4167, 4183, 4192, 4193, 4209, 4215, 4238, 4259, 4268, 4307, 4328, 4330, 4347, 4365, 4382, 4398, 4418, 4463, 4480, 4522, 4527, 4618, 4632, 4672, 4676, 4699, 4739, 4762, 4775, 4784, 4788, 4844, 4888, 4890, 4896, 4906, 4907, 4913, 4928, 4957, 4993, 4998, 5015, 5052, 5084, 5106, 5135, 5167, 5174, 5207, 5227, 5230, 5272, 5308, 5317, 5324, 5353, 5395, 5408, 5460, 5464, 5509, 5536, 5565, 5594, 5597, 5602, 5616, 5638, 5648, 5658, 5664, 5690, 5692, 5710, 5716, 5753, 5763, 5793, 5854, 5863, 5874, 5880, 5895, 5903, 5939, 5945, 5947, 5963, 5990, 6033, 6037, 6043, 6064, 6087, 6125, 6130, 6149, 6161, 6189, 6195, 6208]
B = [836, 2641, 7809, 5871, 1938, 4541, 8189, 11894, 2261, 6042, 12863, 8018, 7923, 1995, 5510, 13148, 3743, 3439, 8778, 3116, 2394, 4826, 8569, 14915, 2280, 1938, 16074, 7923, 3363, 6954, 6650, 2679, 4104, 5206, 7999, 6517, 3515, 7030, 4712, 3211, 3762, 5035, 6270, 5643, 6289, 7391, 6897, 10241, 3021, 7334, 5358, 8816, 3040, 3838, 5092, 3477, 5377, 2527, 7448, 4351, 5130, 3857, 6973, 4104, 1900, 1463, 3743, 6289, 1976, 2261, 4009, 4180, 2660, 5719, 3553, 3477, 4332, 4294, 836, 4598, 2052, 1653, 931, 1767, 1482, 836, 7809, 2090, 3059, 4104, 8436, 4560, 1615, 3914, 4085, 1976, 3496, 3990, 7448, 2109, 2679, 4636, 1615, 2527, 2603, 1843, 4579, 2109, 1045, 2375, 3572, 5339, 2090, 1425, 2375, 8265, 3534, 3933, 2109, 5358, 2983, 2204, 2698, 7942, 6517, 2033, 4750, 7049, 2489, 3857, 2204, 5795, 2546, 1843, 9443, 5358, 13281, 3648, 5206, 3762, 2489, 4123, 1406, 513, 1273, 342, 1197, 2375, 1786, 1691, 855, 6441, 4921, 1767, 7144, 3059, 1520, 5757, 6783, 2831, 2698, 2755, 2698, 1254, 5472, 2128, 5681, 4104, 4484, 3363, 3325, 7315, 3686, 2052, 817, 2318, 2166, 1026, 3135, 2071, 1026, 950, 1767, 1026, 1026, 2926, 1615, 1653, 3819, 2413, 9576, 4446, 7980, 4332, 2166, 5016, 3002, 9025, 2622, 2831, 12046, 1957, 1691, 3591, 1919, 3059, 3116, 1634, 855, 1748, 779, 6460, 8360, 4142, 3040, 2223, 3040, 2527, 836, 1520, 3002, 665, 152, 1805, 4864, 1254, 817, 931, 798, 1330, 1463, 2318, 1159, 627, 513, 2375, 5396, 6422, 6593, 2546, 4902, 3249, 4256, 6726, 1425, 1368, 3002, 931, 2432, 1425, 1805, 2793, 4237, 950, 494, 836, 1178, 1843, 3363, 1615, 304, 475, 2394, 969, 2242, 1558, 2831, 836, 323, 1653, 76, 817, 1425, 3439, 2109, 1026, 1045, 779, 133, 76, 1444, 874, 228, 2090, 551, 361, 1292, 1349, 1026, 1862, 1995, 2223, 1349, 1330]
Sum A = 17 x 55049 = 935833
Sum B = 19 x 55049 = 1045931
Just attaining the same factor is, by two assumption, "random" numbers, the probability, mathematically speaking is 1/f * 1/f, very small.
But this does not take into account, the number series in list a and b, producing the same factor. I want to quantify this probability.
I am thinking to quantify this probability, if we assume, for list A to increase by gaussian mean with added noise of about 5, multiply this probability with a two digit random number and floor the integer to the lowest int. Having a probability distribution describing the increase in verse values with a gaussian mean, offset of about 5. Next for list B, the numerical values shifts sometimes with 1000's and 100. Therefor we could 19 * X, approach X, with a randomly produced digit number varying from 1 to 3 digits, X is produced the same way as for A, with multiplying with a gaussian probability distribution.
So programming wise it would look something like
r.uniform() x random.randint(1,200) = X
X * 19 = entries in B.
Assume that the two lists are generated as described above.
Now that we have roughly estimated how to describe this process randomly. We can then apply other statistical methods in assuming, what is the probability for the null hypothesis, i.e. gaining the same 19 x 55049, 17 x 55049, factor on both list is within bounds of non-significant p values. I am still not sure how to do this.
probability probability-distributions prime-factorization prime-twins
asked Aug 3 at 17:52
Asael
11
EDITED:
Two coloums of numbers are presented, there are 234 entries in both columns. Column A, values are not multiples of 19, natural numbers increasing by a random value with mean of 30.
Column B, all numbers are multiples of 19, multiplied by a random integer, from 2 to 300. Summing column B, will give a prime factor of 19 x 55049 and column A, prime factor of 17 x 55049.
Question, what are the chances for obtaining the same, prime factor 55049?
A = [5, 11, 24, 58, 106, 123, 243, 244, 249, 292, 293, 306, 321, 336, 354, 374, 417, 421, 496, 532, 545, 566, 583, 587, 641, 659, 675, 711, 728, 744, 749, 757, 784, 810, 833, 869, 879, 929, 978, 1016, 1019, 1047, 1050, 1055, 1077, 1099, 1102, 1117, 1156, 1167, 1177, 1220, 1244, 1249, 1254, 1255, 1272, 1288, 1326, 1338, 1348, 1394, 1430, 1439, 1440, 1446, 1484, 1486, 1573, 1591, 1621, 1622, 1635, 1636, 1646, 1748, 1752, 1762, 1766, 1788, 1820, 1839, 1848, 1852, 1874, 1897, 1934, 1941, 1950, 1969, 1970, 1979, 2018, 2081, 2085, 2147, 2176, 2202, 2203, 2252, 2257, 2285, 2300, 2316, 2319, 2320, 2322, 2332, 2345, 2400, 2406, 2425, 2428, 2466, 2473, 2507, 2528, 2532, 2551, 2561, 2571, 2602, 2603, 2610, 2614, 2657, 2665, 2717, 2747, 2752, 2767, 2807, 2811, 2816, 2818, 2825, 2846, 2862, 2896, 2913, 2952, 2961, 3005, 3020, 3022, 3027, 3047, 3087, 3113, 3119, 3138, 3171, 3220, 3255, 3270, 3285, 3426, 3433, 3465, 3491, 3509, 3522, 3531, 3544, 3563, 3641, 3651, 3694, 3698, 3714, 3715, 3728, 3743, 3761, 3763, 3780, 3824, 3828, 3841, 3852, 3862, 3874, 3904, 3916, 3948, 3971, 4008, 4050, 4109, 4122, 4154, 4158, 4167, 4183, 4192, 4193, 4209, 4215, 4238, 4259, 4268, 4307, 4328, 4330, 4347, 4365, 4382, 4398, 4418, 4463, 4480, 4522, 4527, 4618, 4632, 4672, 4676, 4699, 4739, 4762, 4775, 4784, 4788, 4844, 4888, 4890, 4896, 4906, 4907, 4913, 4928, 4957, 4993, 4998, 5015, 5052, 5084, 5106, 5135, 5167, 5174, 5207, 5227, 5230, 5272, 5308, 5317, 5324, 5353, 5395, 5408, 5460, 5464, 5509, 5536, 5565, 5594, 5597, 5602, 5616, 5638, 5648, 5658, 5664, 5690, 5692, 5710, 5716, 5753, 5763, 5793, 5854, 5863, 5874, 5880, 5895, 5903, 5939, 5945, 5947, 5963, 5990, 6033, 6037, 6043, 6064, 6087, 6125, 6130, 6149, 6161, 6189, 6195, 6208]
B = [836, 2641, 7809, 5871, 1938, 4541, 8189, 11894, 2261, 6042, 12863, 8018, 7923, 1995, 5510, 13148, 3743, 3439, 8778, 3116, 2394, 4826, 8569, 14915, 2280, 1938, 16074, 7923, 3363, 6954, 6650, 2679, 4104, 5206, 7999, 6517, 3515, 7030, 4712, 3211, 3762, 5035, 6270, 5643, 6289, 7391, 6897, 10241, 3021, 7334, 5358, 8816, 3040, 3838, 5092, 3477, 5377, 2527, 7448, 4351, 5130, 3857, 6973, 4104, 1900, 1463, 3743, 6289, 1976, 2261, 4009, 4180, 2660, 5719, 3553, 3477, 4332, 4294, 836, 4598, 2052, 1653, 931, 1767, 1482, 836, 7809, 2090, 3059, 4104, 8436, 4560, 1615, 3914, 4085, 1976, 3496, 3990, 7448, 2109, 2679, 4636, 1615, 2527, 2603, 1843, 4579, 2109, 1045, 2375, 3572, 5339, 2090, 1425, 2375, 8265, 3534, 3933, 2109, 5358, 2983, 2204, 2698, 7942, 6517, 2033, 4750, 7049, 2489, 3857, 2204, 5795, 2546, 1843, 9443, 5358, 13281, 3648, 5206, 3762, 2489, 4123, 1406, 513, 1273, 342, 1197, 2375, 1786, 1691, 855, 6441, 4921, 1767, 7144, 3059, 1520, 5757, 6783, 2831, 2698, 2755, 2698, 1254, 5472, 2128, 5681, 4104, 4484, 3363, 3325, 7315, 3686, 2052, 817, 2318, 2166, 1026, 3135, 2071, 1026, 950, 1767, 1026, 1026, 2926, 1615, 1653, 3819, 2413, 9576, 4446, 7980, 4332, 2166, 5016, 3002, 9025, 2622, 2831, 12046, 1957, 1691, 3591, 1919, 3059, 3116, 1634, 855, 1748, 779, 6460, 8360, 4142, 3040, 2223, 3040, 2527, 836, 1520, 3002, 665, 152, 1805, 4864, 1254, 817, 931, 798, 1330, 1463, 2318, 1159, 627, 513, 2375, 5396, 6422, 6593, 2546, 4902, 3249, 4256, 6726, 1425, 1368, 3002, 931, 2432, 1425, 1805, 2793, 4237, 950, 494, 836, 1178, 1843, 3363, 1615, 304, 475, 2394, 969, 2242, 1558, 2831, 836, 323, 1653, 76, 817, 1425, 3439, 2109, 1026, 1045, 779, 133, 76, 1444, 874, 228, 2090, 551, 361, 1292, 1349, 1026, 1862, 1995, 2223, 1349, 1330]
Sum A = 17 x 55049 = 935833
Sum B = 19 x 55049 = 1045931
Just attaining the same factor is, by two assumption, "random" numbers, the probability, mathematically speaking is 1/f * 1/f, very small.
But this does not take into account, the number series in list a and b, producing the same factor. I want to quantify this probability.
I am thinking to quantify this probability, if we assume, for list A to increase by gaussian mean with added noise of about 5, multiply this probability with a two digit random number and floor the integer to the lowest int. Having a probability distribution describing the increase in verse values with a gaussian mean, offset of about 5. Next for list B, the numerical values shifts sometimes with 1000's and 100. Therefor we could 19 * X, approach X, with a randomly produced digit number varying from 1 to 3 digits, X is produced the same way as for A, with multiplying with a gaussian probability distribution.
So programming wise it would look something like
r.uniform() x random.randint(1,200) = X
X * 19 = entries in B.
Assume that the two lists are generated as described above.
Now that we have roughly estimated how to describe this process randomly. We can then apply other statistical methods in assuming, what is the probability for the null hypothesis, i.e. gaining the same 19 x 55049, 17 x 55049, factor on both list is within bounds of non-significant p values. I am still not sure how to do this.
probability probability-distributions prime-factorization prime-twins
asked Aug 3 at 17:52
Asael
11
edited 2 days ago
asked Aug 3 at 17:52
Asael
11
asked Aug 3 at 17:52
Asael
11
asked Aug 3 at 17:52
Asael
11
11
put on hold as off-topic by Andrés E. Caicedo, Adrian Keister, max_zorn, Leucippus, Somos Aug 4 at 3:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." РAndr̩s E. Caicedo, Adrian Keister, max_zorn, Leucippus, Somos
put on hold as off-topic by Andrés E. Caicedo, Adrian Keister, max_zorn, Leucippus, Somos Aug 4 at 3:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." РAndr̩s E. Caicedo, Adrian Keister, max_zorn, Leucippus, Somos
why is this down voted?
– Asael
Aug 3 at 18:01
This question is very unclear. It might help if you name a variable or two, and clarify what you're asking for.
– G Tony Jacobs
Aug 3 at 18:01
Tried to do my best at explaining, try to see the edited version. Thanks.
– Asael
Aug 3 at 18:06
There is no such thing as a random integer until you tell us how they are determined. There is no uniform probability measure on a countable infinite space.
– Thomas Andrews
Aug 3 at 18:37
Thomas Andrews, I am currently at the process at writing the reason as of why I am asking this question. And how the integers are picked, in both columns.
– Asael
Aug 3 at 18:39
 |Â
show 2 more comments
why is this down voted?
– Asael
Aug 3 at 18:01
This question is very unclear. It might help if you name a variable or two, and clarify what you're asking for.
– G Tony Jacobs
Aug 3 at 18:01
Tried to do my best at explaining, try to see the edited version. Thanks.
– Asael
Aug 3 at 18:06
There is no such thing as a random integer until you tell us how they are determined. There is no uniform probability measure on a countable infinite space.
– Thomas Andrews
Aug 3 at 18:37
Thomas Andrews, I am currently at the process at writing the reason as of why I am asking this question. And how the integers are picked, in both columns.
– Asael
Aug 3 at 18:39
why is this down voted?
– Asael
Aug 3 at 18:01
why is this down voted?
– Asael
Aug 3 at 18:01
This question is very unclear. It might help if you name a variable or two, and clarify what you're asking for.
– G Tony Jacobs
Aug 3 at 18:01
This question is very unclear. It might help if you name a variable or two, and clarify what you're asking for.
– G Tony Jacobs
Aug 3 at 18:01
Tried to do my best at explaining, try to see the edited version. Thanks.
– Asael
Aug 3 at 18:06
Tried to do my best at explaining, try to see the edited version. Thanks.
– Asael
Aug 3 at 18:06
There is no such thing as a random integer until you tell us how they are determined. There is no uniform probability measure on a countable infinite space.
– Thomas Andrews
Aug 3 at 18:37
There is no such thing as a random integer until you tell us how they are determined. There is no uniform probability measure on a countable infinite space.
– Thomas Andrews
Aug 3 at 18:37
Thomas Andrews, I am currently at the process at writing the reason as of why I am asking this question. And how the integers are picked, in both columns.
– Asael
Aug 3 at 18:39
Thomas Andrews, I am currently at the process at writing the reason as of why I am asking this question. And how the integers are picked, in both columns.
– Asael
Aug 3 at 18:39
 |Â
show 2 more comments
1 Answer
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up vote
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down vote
To answer a probability question, we need to have a precise experiment in mind. I think this is the experiment you're describing: We add $234$ numbers chosen "randomly" in column $A$. We add $234$ numbers, chosen "randomly" from among multiples of $19$, in column $B$. We wish to know the probability that both sums are multiples of the prime number $55049$.
The problem here is the idea of choosing numbers "randomly". We can't choose "random" natural numbers, unless we define a distribution that governs our selection, and there is no simple uniform distribution on an infinite discrete set.
Under certain assumptions, the probability that a "random" number being a multiple of $55049$ is precisely $frac155049$. Without those assumptions, though, it could really be anything.
With more information about how these numbers are chosen randomly, we could give an answer to your question, but without that, there's really nothing we can say.
answered Aug 3 at 18:20
G Tony Jacobs
25.5k43483
I understand your point. There is no uniform distribution of randomness that describes infinite series. What if we just implement a random gaussian distribution and multiply, that probability distribution with numbers varrying from 2, to 4 digt numbers? This way it can accurately, mimick the natural numbers in column A and B. Mean of both probability distribution is in mean of A and B. That is how these numbers are chosen randomly. Just by random generators in python.
– Asael
Aug 3 at 18:30
is there a specific formula to estimate the probability of the prime factor of 55049? for above comment.
– Asael
Aug 3 at 18:35
i edited my question
– Asael
Aug 3 at 19:12
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
To answer a probability question, we need to have a precise experiment in mind. I think this is the experiment you're describing: We add $234$ numbers chosen "randomly" in column $A$. We add $234$ numbers, chosen "randomly" from among multiples of $19$, in column $B$. We wish to know the probability that both sums are multiples of the prime number $55049$.
The problem here is the idea of choosing numbers "randomly". We can't choose "random" natural numbers, unless we define a distribution that governs our selection, and there is no simple uniform distribution on an infinite discrete set.
Under certain assumptions, the probability that a "random" number being a multiple of $55049$ is precisely $frac155049$. Without those assumptions, though, it could really be anything.
With more information about how these numbers are chosen randomly, we could give an answer to your question, but without that, there's really nothing we can say.
answered Aug 3 at 18:20
G Tony Jacobs
25.5k43483
I understand your point. There is no uniform distribution of randomness that describes infinite series. What if we just implement a random gaussian distribution and multiply, that probability distribution with numbers varrying from 2, to 4 digt numbers? This way it can accurately, mimick the natural numbers in column A and B. Mean of both probability distribution is in mean of A and B. That is how these numbers are chosen randomly. Just by random generators in python.
– Asael
Aug 3 at 18:30
is there a specific formula to estimate the probability of the prime factor of 55049? for above comment.
– Asael
Aug 3 at 18:35
i edited my question
– Asael
Aug 3 at 19:12
add a comment |Â
up vote
0
down vote
To answer a probability question, we need to have a precise experiment in mind. I think this is the experiment you're describing: We add $234$ numbers chosen "randomly" in column $A$. We add $234$ numbers, chosen "randomly" from among multiples of $19$, in column $B$. We wish to know the probability that both sums are multiples of the prime number $55049$.
The problem here is the idea of choosing numbers "randomly". We can't choose "random" natural numbers, unless we define a distribution that governs our selection, and there is no simple uniform distribution on an infinite discrete set.
Under certain assumptions, the probability that a "random" number being a multiple of $55049$ is precisely $frac155049$. Without those assumptions, though, it could really be anything.
With more information about how these numbers are chosen randomly, we could give an answer to your question, but without that, there's really nothing we can say.
answered Aug 3 at 18:20
G Tony Jacobs
25.5k43483
I understand your point. There is no uniform distribution of randomness that describes infinite series. What if we just implement a random gaussian distribution and multiply, that probability distribution with numbers varrying from 2, to 4 digt numbers? This way it can accurately, mimick the natural numbers in column A and B. Mean of both probability distribution is in mean of A and B. That is how these numbers are chosen randomly. Just by random generators in python.
– Asael
Aug 3 at 18:30
is there a specific formula to estimate the probability of the prime factor of 55049? for above comment.
– Asael
Aug 3 at 18:35
i edited my question
– Asael
Aug 3 at 19:12
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To answer a probability question, we need to have a precise experiment in mind. I think this is the experiment you're describing: We add $234$ numbers chosen "randomly" in column $A$. We add $234$ numbers, chosen "randomly" from among multiples of $19$, in column $B$. We wish to know the probability that both sums are multiples of the prime number $55049$.
The problem here is the idea of choosing numbers "randomly". We can't choose "random" natural numbers, unless we define a distribution that governs our selection, and there is no simple uniform distribution on an infinite discrete set.
Under certain assumptions, the probability that a "random" number being a multiple of $55049$ is precisely $frac155049$. Without those assumptions, though, it could really be anything.
With more information about how these numbers are chosen randomly, we could give an answer to your question, but without that, there's really nothing we can say.
answered Aug 3 at 18:20
G Tony Jacobs
25.5k43483
To answer a probability question, we need to have a precise experiment in mind. I think this is the experiment you're describing: We add $234$ numbers chosen "randomly" in column $A$. We add $234$ numbers, chosen "randomly" from among multiples of $19$, in column $B$. We wish to know the probability that both sums are multiples of the prime number $55049$.
The problem here is the idea of choosing numbers "randomly". We can't choose "random" natural numbers, unless we define a distribution that governs our selection, and there is no simple uniform distribution on an infinite discrete set.
Under certain assumptions, the probability that a "random" number being a multiple of $55049$ is precisely $frac155049$. Without those assumptions, though, it could really be anything.
With more information about how these numbers are chosen randomly, we could give an answer to your question, but without that, there's really nothing we can say.
answered Aug 3 at 18:20
G Tony Jacobs
25.5k43483
answered Aug 3 at 18:20
G Tony Jacobs
25.5k43483
answered Aug 3 at 18:20
G Tony Jacobs
25.5k43483
answered Aug 3 at 18:20
G Tony Jacobs
25.5k43483
25.5k43483
I understand your point. There is no uniform distribution of randomness that describes infinite series. What if we just implement a random gaussian distribution and multiply, that probability distribution with numbers varrying from 2, to 4 digt numbers? This way it can accurately, mimick the natural numbers in column A and B. Mean of both probability distribution is in mean of A and B. That is how these numbers are chosen randomly. Just by random generators in python.
– Asael
Aug 3 at 18:30
is there a specific formula to estimate the probability of the prime factor of 55049? for above comment.
– Asael
Aug 3 at 18:35
i edited my question
– Asael
Aug 3 at 19:12
add a comment |Â
I understand your point. There is no uniform distribution of randomness that describes infinite series. What if we just implement a random gaussian distribution and multiply, that probability distribution with numbers varrying from 2, to 4 digt numbers? This way it can accurately, mimick the natural numbers in column A and B. Mean of both probability distribution is in mean of A and B. That is how these numbers are chosen randomly. Just by random generators in python.
– Asael
Aug 3 at 18:30
is there a specific formula to estimate the probability of the prime factor of 55049? for above comment.
– Asael
Aug 3 at 18:35
i edited my question
– Asael
Aug 3 at 19:12
I understand your point. There is no uniform distribution of randomness that describes infinite series. What if we just implement a random gaussian distribution and multiply, that probability distribution with numbers varrying from 2, to 4 digt numbers? This way it can accurately, mimick the natural numbers in column A and B. Mean of both probability distribution is in mean of A and B. That is how these numbers are chosen randomly. Just by random generators in python.
– Asael
Aug 3 at 18:30
I understand your point. There is no uniform distribution of randomness that describes infinite series. What if we just implement a random gaussian distribution and multiply, that probability distribution with numbers varrying from 2, to 4 digt numbers? This way it can accurately, mimick the natural numbers in column A and B. Mean of both probability distribution is in mean of A and B. That is how these numbers are chosen randomly. Just by random generators in python.
– Asael
Aug 3 at 18:30
is there a specific formula to estimate the probability of the prime factor of 55049? for above comment.
– Asael
Aug 3 at 18:35
is there a specific formula to estimate the probability of the prime factor of 55049? for above comment.
– Asael
Aug 3 at 18:35
i edited my question
– Asael
Aug 3 at 19:12
i edited my question
– Asael
Aug 3 at 19:12
add a comment |Â
why is this down voted?
– Asael
Aug 3 at 18:01
This question is very unclear. It might help if you name a variable or two, and clarify what you're asking for.
– G Tony Jacobs
Aug 3 at 18:01
Tried to do my best at explaining, try to see the edited version. Thanks.
– Asael
Aug 3 at 18:06
There is no such thing as a random integer until you tell us how they are determined. There is no uniform probability measure on a countable infinite space.
– Thomas Andrews
Aug 3 at 18:37
Thomas Andrews, I am currently at the process at writing the reason as of why I am asking this question. And how the integers are picked, in both columns.
– Asael
Aug 3 at 18:39