NOTE: Most of the tests in DIEHARD return a p-value, which should be uniform on [0,1) if the input file contains truly independent random bits. Those p-values are obtained by p=F(X), where F is the assumed distribution of the sample random variable X---often normal. But that assumed F is just an asymptotic approximation, for which the fit will be worst in the tails. Thus you should not be surprised with occasional p-values near 0 or 1, such as .0012 or .9983. When a bit stream really FAILS BIG, you will get p's of 0 or 1 to six or more places. By all means, do not, as a Statistician might, think that a p < .025 or p> .975 means that the RNG has "failed the test at the .05 level". Such p's happen among the hundreds that DIEHARD produces, even with good RNG's. So keep in mind that " p happens". ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BIRTHDAY SPACINGS TEST :: :: Choose m birthdays in a year of n days. List the spacings :: :: between the birthdays. If j is the number of values that :: :: occur more than once in that list, then j is asymptotically :: :: Poisson distributed with mean m^3/(4n). Experience shows n :: :: must be quite large, say n>=2^18, for comparing the results :: :: to the Poisson distribution with that mean. This test uses :: :: n=2^24 and m=2^9, so that the underlying distribution for j :: :: is taken to be Poisson with lambda=2^27/(2^26)=2. A sample :: :: of 500 j's is taken, and a chi-square goodness of fit test :: :: provides a p value. The first test uses bits 1-24 (counting :: :: from the left) from integers in the specified file. :: :: Then the file is closed and reopened. Next, bits 2-25 are :: :: used to provide birthdays, then 3-26 and so on to bits 9-32. :: :: Each set of bits provides a p-value, and the nine p-values :: :: provide a sample for a KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: BIRTHDAY SPACINGS TEST, M= 512 N=2**24 LAMBDA= 2.0000 Results for o:\tmp\f49.raw For a sample of size 500: mean o:\tmp\f49.raw using bits 1 to 24 1.930 duplicate number number spacings observed expected 0 71. 67.668 1 144. 135.335 2 142. 135.335 3 67. 90.224 4 53. 45.112 5 18. 18.045 6 to INF 5. 8.282 Chisquare with 6 d.o.f. = 9.70 p-value= .862350 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f49.raw using bits 2 to 25 1.934 duplicate number number spacings observed expected 0 61. 67.668 1 152. 135.335 2 132. 135.335 3 93. 90.224 4 46. 45.112 5 10. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 7.11 p-value= .689146 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f49.raw using bits 3 to 26 1.954 duplicate number number spacings observed expected 0 70. 67.668 1 146. 135.335 2 125. 135.335 3 91. 90.224 4 41. 45.112 5 21. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 3.20 p-value= .217188 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f49.raw using bits 4 to 27 1.934 duplicate number number spacings observed expected 0 64. 67.668 1 148. 135.335 2 133. 135.335 3 93. 90.224 4 42. 45.112 5 15. 18.045 6 to INF 5. 8.282 Chisquare with 6 d.o.f. = 3.54 p-value= .261166 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f49.raw using bits 5 to 28 2.092 duplicate number number spacings observed expected 0 55. 67.668 1 139. 135.335 2 136. 135.335 3 88. 90.224 4 51. 45.112 5 22. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 4.23 p-value= .353953 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f49.raw using bits 6 to 29 2.026 duplicate number number spacings observed expected 0 67. 67.668 1 129. 135.335 2 142. 135.335 3 91. 90.224 4 43. 45.112 5 18. 18.045 6 to INF 10. 8.282 Chisquare with 6 d.o.f. = 1.09 p-value= .018181 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f49.raw using bits 7 to 30 2.118 duplicate number number spacings observed expected 0 56. 67.668 1 132. 135.335 2 140. 135.335 3 96. 90.224 4 40. 45.112 5 22. 18.045 6 to INF 14. 8.282 Chisquare with 6 d.o.f. = 8.02 p-value= .763282 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f49.raw using bits 8 to 31 2.170 duplicate number number spacings observed expected 0 51. 67.668 1 131. 135.335 2 131. 135.335 3 104. 90.224 4 46. 45.112 5 28. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 12.06 p-value= .939334 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f49.raw using bits 9 to 32 2.124 duplicate number number spacings observed expected 0 57. 67.668 1 123. 135.335 2 144. 135.335 3 103. 90.224 4 38. 45.112 5 25. 18.045 6 to INF 10. 8.282 Chisquare with 6 d.o.f. = 9.33 p-value= .844075 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were .862350 .689146 .217188 .261166 .353953 .018181 .763282 .939334 .844075 A KSTEST for the 9 p-values yields .281800 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE OVERLAPPING 5-PERMUTATION TEST :: :: This is the OPERM5 test. It looks at a sequence of one mill- :: :: ion 32-bit random integers. Each set of five consecutive :: :: integers can be in one of 120 states, for the 5! possible or- :: :: derings of five numbers. Thus the 5th, 6th, 7th,...numbers :: :: each provide a state. As many thousands of state transitions :: :: are observed, cumulative counts are made of the number of :: :: occurences of each state. Then the quadratic form in the :: :: weak inverse of the 120x120 covariance matrix yields a test :: :: equivalent to the likelihood ratio test that the 120 cell :: :: counts came from the specified (asymptotically) normal dis- :: :: tribution with the specified 120x120 covariance matrix (with :: :: rank 99). This version uses 1,000,000 integers, twice. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: OPERM5 test for file o:\tmp\f49.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom=102.826; p-value= .623930 OPERM5 test for file o:\tmp\f49.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 95.649; p-value= .423282 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 31x31 matrices. The leftmost :: :: 31 bits of 31 random integers from the test sequence are used :: :: to form a 31x31 binary matrix over the field {0,1}. The rank :: :: is determined. That rank can be from 0 to 31, but ranks< 28 :: :: are rare, and their counts are pooled with those for rank 28. :: :: Ranks are found for 40,000 such random matrices and a chisqua-:: :: re test is performed on counts for ranks 31,30,29 and <=28. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary rank test for o:\tmp\f49.raw Rank test for 31x31 binary matrices: rows from leftmost 31 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 28 204 211.4 .260276 .260 29 5063 5134.0 .982167 1.242 30 23315 23103.0 1.944511 3.187 31 11418 11551.5 1.543413 4.730 chisquare= 4.730 for 3 d. of f.; p-value= .822178 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 32x32 matrices. A random 32x :: :: 32 binary matrix is formed, each row a 32-bit random integer. :: :: The rank is determined. That rank can be from 0 to 32, ranks :: :: less than 29 are rare, and their counts are pooled with those :: :: for rank 29. Ranks are found for 40,000 such random matrices :: :: and a chisquare test is performed on counts for ranks 32,31, :: :: 30 and <=29. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary rank test for o:\tmp\f49.raw Rank test for 32x32 binary matrices: rows from leftmost 32 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 29 222 211.4 .529654 .530 30 5194 5134.0 .700967 1.231 31 22942 23103.0 1.122627 2.353 32 11642 11551.5 .708636 3.062 chisquare= 3.062 for 3 d. of f.; p-value= .657893 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 6x8 matrices. From each of :: :: six random 32-bit integers from the generator under test, a :: :: specified byte is chosen, and the resulting six bytes form a :: :: 6x8 binary matrix whose rank is determined. That rank can be :: :: from 0 to 6, but ranks 0,1,2,3 are rare; their counts are :: :: pooled with those for rank 4. Ranks are found for 100,000 :: :: random matrices, and a chi-square test is performed on :: :: counts for ranks 6,5 and <=4. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary Rank Test for o:\tmp\f49.raw Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 965 944.3 .454 .454 r =5 21803 21743.9 .161 .614 r =6 77232 77311.8 .082 .697 p=1-exp(-SUM/2)= .29415 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 934 944.3 .112 .112 r =5 21801 21743.9 .150 .262 r =6 77265 77311.8 .028 .291 p=1-exp(-SUM/2)= .13526 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 919 944.3 .678 .678 r =5 21941 21743.9 1.787 2.465 r =6 77140 77311.8 .382 2.846 p=1-exp(-SUM/2)= .75905 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 962 944.3 .332 .332 r =5 21710 21743.9 .053 .385 r =6 77328 77311.8 .003 .388 p=1-exp(-SUM/2)= .17633 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 888 944.3 3.357 3.357 r =5 21868 21743.9 .708 4.065 r =6 77244 77311.8 .059 4.125 p=1-exp(-SUM/2)= .87284 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 933 944.3 .135 .135 r =5 21536 21743.9 1.988 2.123 r =6 77531 77311.8 .621 2.745 p=1-exp(-SUM/2)= .74647 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 957 944.3 .171 .171 r =5 21705 21743.9 .070 .240 r =6 77338 77311.8 .009 .249 p=1-exp(-SUM/2)= .11717 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 934 944.3 .112 .112 r =5 21719 21743.9 .029 .141 r =6 77347 77311.8 .016 .157 p=1-exp(-SUM/2)= .07546 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 964 944.3 .411 .411 r =5 21676 21743.9 .212 .623 r =6 77360 77311.8 .030 .653 p=1-exp(-SUM/2)= .27856 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 914 944.3 .972 .972 r =5 21908 21743.9 1.238 2.211 r =6 77178 77311.8 .232 2.442 p=1-exp(-SUM/2)= .70512 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 906 944.3 1.554 1.554 r =5 21807 21743.9 .183 1.737 r =6 77287 77311.8 .008 1.745 p=1-exp(-SUM/2)= .58201 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 997 944.3 2.941 2.941 r =5 21582 21743.9 1.205 4.146 r =6 77421 77311.8 .154 4.301 p=1-exp(-SUM/2)= .88355 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 965 944.3 .454 .454 r =5 21778 21743.9 .053 .507 r =6 77257 77311.8 .039 .546 p=1-exp(-SUM/2)= .23892 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 866 944.3 6.493 6.493 r =5 21676 21743.9 .212 6.705 r =6 77458 77311.8 .276 6.981 p=1-exp(-SUM/2)= .96952 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 901 944.3 1.986 1.986 r =5 21702 21743.9 .081 2.066 r =6 77397 77311.8 .094 2.160 p=1-exp(-SUM/2)= .66044 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 980 944.3 1.350 1.350 r =5 21850 21743.9 .518 1.867 r =6 77170 77311.8 .260 2.127 p=1-exp(-SUM/2)= .65482 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 911 944.3 1.174 1.174 r =5 21811 21743.9 .207 1.381 r =6 77278 77311.8 .015 1.396 p=1-exp(-SUM/2)= .50248 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1042 944.3 10.108 10.108 r =5 21527 21743.9 2.164 12.272 r =6 77431 77311.8 .184 12.455 p=1-exp(-SUM/2)= .99803 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 964 944.3 .411 .411 r =5 21676 21743.9 .212 .623 r =6 77360 77311.8 .030 .653 p=1-exp(-SUM/2)= .27856 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 908 944.3 1.396 1.396 r =5 21670 21743.9 .251 1.647 r =6 77422 77311.8 .157 1.804 p=1-exp(-SUM/2)= .59419 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 963 944.3 .370 .370 r =5 21680 21743.9 .188 .558 r =6 77357 77311.8 .026 .584 p=1-exp(-SUM/2)= .25341 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 951 944.3 .048 .048 r =5 21834 21743.9 .373 .421 r =6 77215 77311.8 .121 .542 p=1-exp(-SUM/2)= .23741 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 957 944.3 .171 .171 r =5 21762 21743.9 .015 .186 r =6 77281 77311.8 .012 .198 p=1-exp(-SUM/2)= .09431 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 966 944.3 .499 .499 r =5 21782 21743.9 .067 .565 r =6 77252 77311.8 .046 .612 p=1-exp(-SUM/2)= .26348 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f49.raw b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 963 944.3 .370 .370 r =5 21804 21743.9 .166 .536 r =6 77233 77311.8 .080 .617 p=1-exp(-SUM/2)= .26534 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: .294155 .135261 .759050 .176329 .872835 .746467 .117167 .075458 .278558 .705116 .582008 .883555 .238920 .969518 .660442 .654820 .502478 .998026 .278558 .594190 .253409 .237412 .094307 .263476 .265344 brank test summary for o:\tmp\f49.raw The KS test for those 25 supposed UNI's yields KS p-value= .506476 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE BITSTREAM TEST :: :: The file under test is viewed as a stream of bits. Call them :: :: b1,b2,... . Consider an alphabet with two "letters", 0 and 1 :: :: and think of the stream of bits as a succession of 20-letter :: :: "words", overlapping. Thus the first word is b1b2...b20, the :: :: second is b2b3...b21, and so on. The bitstream test counts :: :: the number of missing 20-letter (20-bit) words in a string of :: :: 2^21 overlapping 20-letter words. There are 2^20 possible 20 :: :: letter words. For a truly random string of 2^21+19 bits, the :: :: number of missing words j should be (very close to) normally :: :: distributed with mean 141,909 and sigma 428. Thus :: :: (j-141909)/428 should be a standard normal variate (z score) :: :: that leads to a uniform [0,1) p value. The test is repeated :: :: twenty times. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: THE OVERLAPPING 20-tuples BITSTREAM TEST, 20 BITS PER WORD, N words This test uses N=2^21 and samples the bitstream 20 times. No. missing words should average 141909. with sigma=428. --------------------------------------------------------- tst no 1: 142224 missing words, .74 sigmas from mean, p-value= .76890 tst no 2: 142221 missing words, .73 sigmas from mean, p-value= .76676 tst no 3: 141812 missing words, -.23 sigmas from mean, p-value= .41006 tst no 4: 142042 missing words, .31 sigmas from mean, p-value= .62171 tst no 5: 141633 missing words, -.65 sigmas from mean, p-value= .25926 tst no 6: 141893 missing words, -.04 sigmas from mean, p-value= .48478 tst no 7: 142233 missing words, .76 sigmas from mean, p-value= .77525 tst no 8: 142206 missing words, .69 sigmas from mean, p-value= .75589 tst no 9: 141567 missing words, -.80 sigmas from mean, p-value= .21190 tst no 10: 141817 missing words, -.22 sigmas from mean, p-value= .41460 tst no 11: 141844 missing words, -.15 sigmas from mean, p-value= .43934 tst no 12: 141858 missing words, -.12 sigmas from mean, p-value= .45227 tst no 13: 142420 missing words, 1.19 sigmas from mean, p-value= .88360 tst no 14: 141638 missing words, -.63 sigmas from mean, p-value= .26306 tst no 15: 142652 missing words, 1.74 sigmas from mean, p-value= .95865 tst no 16: 142341 missing words, 1.01 sigmas from mean, p-value= .84341 tst no 17: 142014 missing words, .24 sigmas from mean, p-value= .59660 tst no 18: 142213 missing words, .71 sigmas from mean, p-value= .76100 tst no 19: 142000 missing words, .21 sigmas from mean, p-value= .58389 tst no 20: 141286 missing words, -1.46 sigmas from mean, p-value= .07264 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The tests OPSO, OQSO and DNA :: :: OPSO means Overlapping-Pairs-Sparse-Occupancy :: :: The OPSO test considers 2-letter words from an alphabet of :: :: 1024 letters. Each letter is determined by a specified ten :: :: bits from a 32-bit integer in the sequence to be tested. OPSO :: :: generates 2^21 (overlapping) 2-letter words (from 2^21+1 :: :: "keystrokes") and counts the number of missing words---that :: :: is 2-letter words which do not appear in the entire sequence. :: :: That count should be very close to normally distributed with :: :: mean 141,909, sigma 290. Thus (missingwrds-141909)/290 should :: :: be a standard normal variable. The OPSO test takes 32 bits at :: :: a time from the test file and uses a designated set of ten :: :: consecutive bits. It then restarts the file for the next de- :: :: signated 10 bits, and so on. :: :: :: :: OQSO means Overlapping-Quadruples-Sparse-Occupancy :: :: The test OQSO is similar, except that it considers 4-letter :: :: words from an alphabet of 32 letters, each letter determined :: :: by a designated string of 5 consecutive bits from the test :: :: file, elements of which are assumed 32-bit random integers. :: :: The mean number of missing words in a sequence of 2^21 four- :: :: letter words, (2^21+3 "keystrokes"), is again 141909, with :: :: sigma = 295. The mean is based on theory; sigma comes from :: :: extensive simulation. :: :: :: :: The DNA test considers an alphabet of 4 letters:: C,G,A,T,:: :: determined by two designated bits in the sequence of random :: :: integers being tested. It considers 10-letter words, so that :: :: as in OPSO and OQSO, there are 2^20 possible words, and the :: :: mean number of missing words from a string of 2^21 (over- :: :: lapping) 10-letter words (2^21+9 "keystrokes") is 141909. :: :: The standard deviation sigma=339 was determined as for OQSO :: :: by simulation. (Sigma for OPSO, 290, is the true value (to :: :: three places), not determined by simulation. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: OPSO test for generator o:\tmp\f49.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for o:\tmp\f49.raw using bits 23 to 32 142049 .482 .6850 OPSO for o:\tmp\f49.raw using bits 22 to 31 142048 .478 .6837 OPSO for o:\tmp\f49.raw using bits 21 to 30 141745 -.567 .2855 OPSO for o:\tmp\f49.raw using bits 20 to 29 142136 .782 .7828 OPSO for o:\tmp\f49.raw using bits 19 to 28 141808 -.349 .3634 OPSO for o:\tmp\f49.raw using bits 18 to 27 141631 -.960 .1686 OPSO for o:\tmp\f49.raw using bits 17 to 26 142294 1.326 .9077 OPSO for o:\tmp\f49.raw using bits 16 to 25 142345 1.502 .9335 OPSO for o:\tmp\f49.raw using bits 15 to 24 141986 .264 .6043 OPSO for o:\tmp\f49.raw using bits 14 to 23 142488 1.995 .9770 OPSO for o:\tmp\f49.raw using bits 13 to 22 142006 .333 .6306 OPSO for o:\tmp\f49.raw using bits 12 to 21 142350 1.520 .9357 OPSO for o:\tmp\f49.raw using bits 11 to 20 142270 1.244 .8932 OPSO for o:\tmp\f49.raw using bits 10 to 19 141751 -.546 .2925 OPSO for o:\tmp\f49.raw using bits 9 to 18 141260 -2.239 .0126 OPSO for o:\tmp\f49.raw using bits 8 to 17 142506 2.057 .9802 OPSO for o:\tmp\f49.raw using bits 7 to 16 141227 -2.353 .0093 OPSO for o:\tmp\f49.raw using bits 6 to 15 142197 .992 .8394 OPSO for o:\tmp\f49.raw using bits 5 to 14 141869 -.139 .4447 OPSO for o:\tmp\f49.raw using bits 4 to 13 142015 .364 .6422 OPSO for o:\tmp\f49.raw using bits 3 to 12 141808 -.349 .3634 OPSO for o:\tmp\f49.raw using bits 2 to 11 142207 1.026 .8477 OPSO for o:\tmp\f49.raw using bits 1 to 10 141981 .247 .5976 OQSO test for generator o:\tmp\f49.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for o:\tmp\f49.raw using bits 28 to 32 142193 .962 .8319 OQSO for o:\tmp\f49.raw using bits 27 to 31 141587 -1.093 .1373 OQSO for o:\tmp\f49.raw using bits 26 to 30 142051 .480 .6845 OQSO for o:\tmp\f49.raw using bits 25 to 29 141425 -1.642 .0503 OQSO for o:\tmp\f49.raw using bits 24 to 28 142316 1.379 .9160 OQSO for o:\tmp\f49.raw using bits 23 to 27 141678 -.784 .2165 OQSO for o:\tmp\f49.raw using bits 22 to 26 141421 -1.655 .0489 OQSO for o:\tmp\f49.raw using bits 21 to 25 141801 -.367 .3567 OQSO for o:\tmp\f49.raw using bits 20 to 24 141832 -.262 .3966 OQSO for o:\tmp\f49.raw using bits 19 to 23 142184 .931 .8241 OQSO for o:\tmp\f49.raw using bits 18 to 22 142088 .606 .7276 OQSO for o:\tmp\f49.raw using bits 17 to 21 141941 .107 .5427 OQSO for o:\tmp\f49.raw using bits 16 to 20 141734 -.594 .2761 OQSO for o:\tmp\f49.raw using bits 15 to 19 142178 .911 .8188 OQSO for o:\tmp\f49.raw using bits 14 to 18 141902 -.025 .4901 OQSO for o:\tmp\f49.raw using bits 13 to 17 141676 -.791 .2145 OQSO for o:\tmp\f49.raw using bits 12 to 16 142170 .884 .8116 OQSO for o:\tmp\f49.raw using bits 11 to 15 142173 .894 .8143 OQSO for o:\tmp\f49.raw using bits 10 to 14 142213 1.029 .8484 OQSO for o:\tmp\f49.raw using bits 9 to 13 142259 1.185 .8821 OQSO for o:\tmp\f49.raw using bits 8 to 12 141580 -1.116 .1321 OQSO for o:\tmp\f49.raw using bits 7 to 11 142035 .426 .6649 OQSO for o:\tmp\f49.raw using bits 6 to 10 141813 -.327 .3720 OQSO for o:\tmp\f49.raw using bits 5 to 9 141934 .084 .5333 OQSO for o:\tmp\f49.raw using bits 4 to 8 142158 .843 .8004 OQSO for o:\tmp\f49.raw using bits 3 to 7 142310 1.358 .9128 OQSO for o:\tmp\f49.raw using bits 2 to 6 141602 -1.042 .1488 OQSO for o:\tmp\f49.raw using bits 1 to 5 142395 1.646 .9502 DNA test for generator o:\tmp\f49.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for o:\tmp\f49.raw using bits 31 to 32 141921 .034 .5137 DNA for o:\tmp\f49.raw using bits 30 to 31 142060 .444 .6716 DNA for o:\tmp\f49.raw using bits 29 to 30 141813 -.284 .3881 DNA for o:\tmp\f49.raw using bits 28 to 29 141473 -1.287 .0990 DNA for o:\tmp\f49.raw using bits 27 to 28 141309 -1.771 .0383 DNA for o:\tmp\f49.raw using bits 26 to 27 141675 -.691 .2447 DNA for o:\tmp\f49.raw using bits 25 to 26 141810 -.293 .3848 DNA for o:\tmp\f49.raw using bits 24 to 25 141971 .182 .5722 DNA for o:\tmp\f49.raw using bits 23 to 24 141900 -.028 .4890 DNA for o:\tmp\f49.raw using bits 22 to 23 142283 1.102 .8648 DNA for o:\tmp\f49.raw using bits 21 to 22 141870 -.116 .4538 DNA for o:\tmp\f49.raw using bits 20 to 21 141376 -1.573 .0578 DNA for o:\tmp\f49.raw using bits 19 to 20 142000 .267 .6054 DNA for o:\tmp\f49.raw using bits 18 to 19 142521 1.804 .9644 DNA for o:\tmp\f49.raw using bits 17 to 18 141529 -1.122 .1309 DNA for o:\tmp\f49.raw using bits 16 to 17 141226 -2.016 .0219 DNA for o:\tmp\f49.raw using bits 15 to 16 142408 1.471 .9294 DNA for o:\tmp\f49.raw using bits 14 to 15 142027 .347 .6357 DNA for o:\tmp\f49.raw using bits 13 to 14 142003 .276 .6088 DNA for o:\tmp\f49.raw using bits 12 to 13 141997 .259 .6020 DNA for o:\tmp\f49.raw using bits 11 to 12 142309 1.179 .8808 DNA for o:\tmp\f49.raw using bits 10 to 11 141151 -2.237 .0126 DNA for o:\tmp\f49.raw using bits 9 to 10 141842 -.199 .4213 DNA for o:\tmp\f49.raw using bits 8 to 9 141750 -.470 .3192 DNA for o:\tmp\f49.raw using bits 7 to 8 141895 -.042 .4831 DNA for o:\tmp\f49.raw using bits 6 to 7 141805 -.308 .3791 DNA for o:\tmp\f49.raw using bits 5 to 6 141414 -1.461 .0720 DNA for o:\tmp\f49.raw using bits 4 to 5 142044 .397 .6544 DNA for o:\tmp\f49.raw using bits 3 to 4 141724 -.547 .2923 DNA for o:\tmp\f49.raw using bits 2 to 3 141338 -1.685 .0460 DNA for o:\tmp\f49.raw using bits 1 to 2 141793 -.343 .3657 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for o:\tmp\f49.raw Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for o:\tmp\f49.raw 2501.02 .014 .505734 byte stream for o:\tmp\f49.raw 2546.11 .652 .742852 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST for specific bytes. :: :: Consider the file under test as a stream of 32-bit integers. :: :: From each integer, a specific byte is chosen , say the left- :: :: most:: bits 1 to 8. Each byte can contain from 0 to 8 1's, :: :: with probabilitie 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the specified bytes from successive integers provide a string :: :: of (overlapping) 5-letter words, each "letter" taking values :: :: A,B,C,D,E. The letters are determined by the number of 1's, :: :: in that byte:: 0,1,or 2 ---> A, 3 ---> B, 4 ---> C, 5 ---> D,:: :: and 6,7 or 8 ---> E. Thus we have a monkey at a typewriter :: :: hitting five keys with with various probabilities:: 37,56,70,:: :: 56,37 over 256. There are 5^5 possible 5-letter words, and :: :: from a string of 256,000 (overlapping) 5-letter words, counts :: :: are made on the frequencies for each word. The quadratic form :: :: in the weak inverse of the covariance matrix of the cell :: :: counts provides a chisquare test:: Q5-Q4, the difference of :: :: the naive Pearson sums of (OBS-EXP)^2/EXP on counts for 5- :: :: and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Chi-square with 5^5-5^4=2500 d.of f. for sample size: 256000 chisquare equiv normal p value Results for COUNT-THE-1's in specified bytes: bits 1 to 8 2383.57 -1.647 .049830 bits 2 to 9 2520.58 .291 .614475 bits 3 to 10 2341.40 -2.243 .012448 bits 4 to 11 2338.46 -2.284 .011171 bits 5 to 12 2457.73 -.598 .274998 bits 6 to 13 2453.49 -.658 .255352 bits 7 to 14 2515.52 .220 .586872 bits 8 to 15 2410.71 -1.263 .103334 bits 9 to 16 2523.33 .330 .629287 bits 10 to 17 2516.79 .237 .593846 bits 11 to 18 2557.41 .812 .791574 bits 12 to 19 2514.32 .202 .580236 bits 13 to 20 2471.06 -.409 .341159 bits 14 to 21 2543.94 .621 .732817 bits 15 to 22 2545.33 .641 .739238 bits 16 to 23 2460.23 -.562 .286934 bits 17 to 24 2618.87 1.681 .953620 bits 18 to 25 2527.54 .389 .651544 bits 19 to 26 2516.09 .228 .590013 bits 20 to 27 2498.88 -.016 .493663 bits 21 to 28 2325.31 -2.471 .006745 bits 22 to 29 2531.19 .441 .670411 bits 23 to 30 2493.23 -.096 .461862 bits 24 to 31 2525.20 .356 .639199 bits 25 to 32 2578.47 1.110 .866435 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THIS IS A PARKING LOT TEST :: :: In a square of side 100, randomly "park" a car---a circle of :: :: radius 1. Then try to park a 2nd, a 3rd, and so on, each :: :: time parking "by ear". That is, if an attempt to park a car :: :: causes a crash with one already parked, try again at a new :: :: random location. (To avoid path problems, consider parking :: :: helicopters rather than cars.) Each attempt leads to either :: :: a crash or a success, the latter followed by an increment to :: :: the list of cars already parked. If we plot n: the number of :: :: attempts, versus k:: the number successfully parked, we get a:: :: curve that should be similar to those provided by a perfect :: :: random number generator. Theory for the behavior of such a :: :: random curve seems beyond reach, and as graphics displays are :: :: not available for this battery of tests, a simple characteriz :: :: ation of the random experiment is used: k, the number of cars :: :: successfully parked after n=12,000 attempts. Simulation shows :: :: that k should average 3523 with sigma 21.9 and is very close :: :: to normally distributed. Thus (k-3523)/21.9 should be a st- :: :: andard normal variable, which, converted to a uniform varia- :: :: ble, provides input to a KSTEST based on a sample of 10. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: CDPARK: result of ten tests on file o:\tmp\f49.raw Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 3520 z-score: -.137 p-value: .445521 Successes: 3508 z-score: -.685 p-value: .246694 Successes: 3501 z-score: -1.005 p-value: .157553 Successes: 3509 z-score: -.639 p-value: .261324 Successes: 3509 z-score: -.639 p-value: .261324 Successes: 3550 z-score: 1.233 p-value: .891189 Successes: 3504 z-score: -.868 p-value: .192812 Successes: 3527 z-score: .183 p-value: .572463 Successes: 3532 z-score: .411 p-value: .659449 Successes: 3520 z-score: -.137 p-value: .445521 square size avg. no. parked sample sigma 100. 3518.000 14.339 KSTEST for the above 10: p= .634605 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file o:\tmp\f49.raw Sample no. d^2 avg equiv uni 5 1.0562 .7520 .654079 10 .3951 .7221 .327716 15 .6566 .8513 .483103 20 .1690 1.2502 .156242 25 .3954 1.2311 .327900 30 .9473 1.0742 .614056 35 .4063 1.0474 .335247 40 1.5054 1.0196 .779737 45 .1042 1.0328 .099444 50 2.3451 1.0761 .905283 55 .0644 1.0731 .062681 60 .8897 1.1341 .591044 65 .3186 1.1181 .274010 70 .2596 1.0984 .229643 75 .2604 1.0756 .230257 80 1.4486 1.0888 .766812 85 .2596 1.0546 .229627 90 4.2536 1.0804 .986087 95 .5968 1.0566 .451056 100 .1689 1.0769 .156134 MINIMUM DISTANCE TEST for o:\tmp\f49.raw Result of KS test on 20 transformed mindist^2's: p-value= .645808 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file o:\tmp\f49.raw sample no: 1 r^3= 6.501 p-value= .19482 sample no: 2 r^3= 52.647 p-value= .82708 sample no: 3 r^3= 45.081 p-value= .77747 sample no: 4 r^3= 7.569 p-value= .22298 sample no: 5 r^3= 19.850 p-value= .48402 sample no: 6 r^3= 15.810 p-value= .40962 sample no: 7 r^3= 9.243 p-value= .26515 sample no: 8 r^3= 32.330 p-value= .65962 sample no: 9 r^3= 36.630 p-value= .70506 sample no: 10 r^3= 4.727 p-value= .14577 sample no: 11 r^3= 95.263 p-value= .95822 sample no: 12 r^3= .004 p-value= .00012 sample no: 13 r^3= 37.558 p-value= .71405 sample no: 14 r^3= 9.582 p-value= .27342 sample no: 15 r^3= 27.324 p-value= .59780 sample no: 16 r^3= 15.104 p-value= .39558 sample no: 17 r^3= 78.917 p-value= .92796 sample no: 18 r^3= 5.701 p-value= .17307 sample no: 19 r^3= 16.218 p-value= .41761 sample no: 20 r^3= 69.350 p-value= .90090 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file o:\tmp\f49.raw p-value= .144430 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: RESULTS OF SQUEEZE TEST FOR o:\tmp\f49.raw Table of standardized frequency counts ( (obs-exp)/sqrt(exp) )^2 for j taking values <=6,7,8,...,47,>=48: 1.3 -.3 .1 .3 .6 .8 -.2 2.0 -.7 1.2 .6 -.7 -.4 .1 .0 3.0 -.7 -.7 -.4 -.1 -2.3 .5 .3 -.2 -.7 -.4 .4 -.9 1.0 -1.5 .6 .2 1.0 .9 .3 -.3 -.7 1.1 -1.2 .4 .1 3.0 -.1 Chi-square with 42 degrees of freedom: 45.917 z-score= .427 p-value= .687042 ______________________________________________________________ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The OVERLAPPING SUMS test :: :: Integers are floated to get a sequence U(1),U(2),... of uni- :: :: form [0,1) variables. Then overlapping sums, :: :: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. :: :: The S's are virtually normal with a certain covariance mat- :: :: rix. A linear transformation of the S's converts them to a :: :: sequence of independent standard normals, which are converted :: :: to uniform variables for a KSTEST. The p-values from ten :: :: KSTESTs are given still another KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test no. 1 p-value .147451 Test no. 2 p-value .930882 Test no. 3 p-value .519466 Test no. 4 p-value .775626 Test no. 5 p-value .473977 Test no. 6 p-value .363304 Test no. 7 p-value .095702 Test no. 8 p-value .574041 Test no. 9 p-value .210696 Test no. 10 p-value .109100 Results of the OSUM test for o:\tmp\f49.raw KSTEST on the above 10 p-values: .347831 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the RUNS test. It counts runs up, and runs down, :: :: in a sequence of uniform [0,1) variables, obtained by float- :: :: ing the 32-bit integers in the specified file. This example :: :: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95:: :: contains an up-run of length 3, a down-run of length 2 and an :: :: up-run of (at least) 2, depending on the next values. The :: :: covariance matrices for the runs-up and runs-down are well :: :: known, leading to chisquare tests for quadratic forms in the :: :: weak inverses of the covariance matrices. Runs are counted :: :: for sequences of length 10,000. This is done ten times. Then :: :: repeated. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The RUNS test for file o:\tmp\f49.raw Up and down runs in a sample of 10000 _________________________________________________ Run test for o:\tmp\f49.raw : runs up; ks test for 10 p's: .061870 runs down; ks test for 10 p's: .741429 Run test for o:\tmp\f49.raw : runs up; ks test for 10 p's: .057787 runs down; ks test for 10 p's: .058657 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the CRAPS TEST. It plays 200,000 games of craps, finds:: :: the number of wins and the number of throws necessary to end :: :: each game. The number of wins should be (very close to) a :: :: normal with mean 200000p and variance 200000p(1-p), with :: :: p=244/495. Throws necessary to complete the game can vary :: :: from 1 to infinity, but counts for all>21 are lumped with 21. :: :: A chi-square test is made on the no.-of-throws cell counts. :: :: Each 32-bit integer from the test file provides the value for :: :: the throw of a die, by floating to [0,1), multiplying by 6 :: :: and taking 1 plus the integer part of the result. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Results of craps test for o:\tmp\f49.raw No. of wins: Observed Expected 98162 98585.86 98162= No. of wins, z-score=-1.896 pvalue= .02900 Analysis of Throws-per-Game: Chisq= 23.90 for 20 degrees of freedom, p= .75325 Throws Observed Expected Chisq Sum 1 66195 66666.7 3.337 3.337 2 37739 37654.3 .190 3.528 3 27286 26954.7 4.071 7.599 4 19420 19313.5 .588 8.186 5 13861 13851.4 .007 8.193 6 10111 9943.5 2.820 11.013 7 7130 7145.0 .032 11.045 8 5088 5139.1 .508 11.552 9 3599 3699.9 2.750 14.302 10 2698 2666.3 .377 14.679 11 1862 1923.3 1.956 16.635 12 1373 1388.7 .178 16.813 13 1025 1003.7 .451 17.264 14 720 726.1 .052 17.316 15 494 525.8 1.927 19.244 16 394 381.2 .433 19.677 17 296 276.5 1.369 21.046 18 187 200.8 .952 21.999 19 140 146.0 .245 22.244 20 93 106.2 1.644 23.888 21 289 287.1 .012 23.901 SUMMARY FOR o:\tmp\f49.raw p-value for no. of wins: .028997 p-value for throws/game: .753246 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Results of DIEHARD battery of tests sent to file o:\tmp\f49.txt