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# Probability analysis of sachin's batting

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An application of probability theories on Sachin Tendulkar's cricketing records. A combination of statistics with cricket . Have fun while you learn :-)

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### Transcript of "Probability analysis of sachin's batting"

1. 1. PROBABILITY ANALYSIS : Sachin Tendulkar’s Test Cricket Records Presented by : Sandipan Maiti Lal Bahadur Shastri Institute of Management
2. 2. FLOWLINE Introduction Data Summarization Century Analysis Half century analysis Not out Analysis Series Total Key Learnings
3. 3. INTRODUCTION Statistics: Game of data. Probability: Game of chances. Cricket: Game close to Indian hearts. Attempt to link the 3 games to reach to some meaningful conclusions.
4. 4. WHY SACHIN ! Most number of test played: Better applicability of statistics. Huge achievements: Solid base for discussion on probability techniques. Variety of data: Scope to cover different probability techniques. Finally: Because he is “THE SACHIN – GOD OF CRICKET”
5. 5. DATA SUMMARIZATION
6. 6. DATA SUMMARIZATION Data Source: http://www.cricketarchive.com/ Raw data : Details of matches played against each team in each season from 1989 to 2011. (Excluding ongoing Test Series). Using cross tabulations to structure the data in organized manner.
7. 7. OPPONENT-WISE RECORDS Half Caught Team Matches Innings Not out Runs HS Avg Century Century outPakistan 18 27 2 1057 194 42.28 2 7 8Newzeal 22 36 5 1532 217 49.42 4 8 10 andEngland 24 39 4 2150 193 61.43 7 10 19Srilanka 25 36 3 1995 203 60.45 9 6 12Australia 31 59 7 3151 241 60.60 11 13 19Banglade 7 9 3 820 248 136.67 5 0 6 shZimbabw 9 14 2 918 201 76.50 3 3 5 e West 16 25 2 1328 179 57.74 3 7 14 Indies South 25 45 4 1741 169 42.46 7 5 13 Africa Total 177 290 32 1469 248 56.95 51 59 106
8. 8. SEASON-WISE RECORDS Half CaughtSeason Matches Innings Not out Runs HS Average Century Century out1989-90 7 10 0 332 88 33.20 0 3 21990-91 4 6 1 256 119 51.20 1 1 31991-92 5 9 1 368 148 46.00 2 0 51992-93 9 12 1 566 165 51.45 2 4 81993-94 7 8 2 501 142 83.50 2 2 51994-95 3 6 0 402 179 67.00 1 2 51995-96 3 4 2 58 52 29.00 0 1 31996-97 15 25 1 1134 177 47.25 3 5 111997-98 8 12 1 935 177 85.00 5 1 51998-99 7 13 1 625 227 52.08 3 2 11999-00 8 16 2 859 217 61.36 3 3 4 .. Contd
9. 9. SEASON-WISE RECORDS…cntd.. Half CaughtSeason Matches Innings Not out Runs HS Average Century Century out2000-01 6 10 1 684 201 76.00 3 2 42001-02 14 23 2 1284 176 61.14 4 6 82002-03 9 15 1 807 193 57.64 2 3 42003-04 9 15 3 659 241 54.92 2 2 42004-05 9 14 2 664 248 55.33 1 4 52005-06 9 13 1 335 189 27.92 1 0 52006-07 3 6 0 199 64 33.17 0 2 32007-08 12 21 3 1114 154 61.89 4 6 132008-09 12 23 2 991 160 47.19 3 4 42009-10 7 10 2 674 143 84.25 5 1 22010-11 11 19 3 1245 214 77.81 4 5 2 Total 177 290 32 14692 248 56.95 51 59 106
10. 10. CENTURY ANALYSIS
11. 11. PROBABILITY OF CENTURY IN A MATCH MATCHES CENTURIES PROBABILITY OF TEAM PLAYED SCORED CENTURY Pakistan 18 2 0.111Newzealand 22 4 0.182 England 24 7 0.292 Srilanka 25 9 0.360 Australia 31 11 0.355Bangladesh 7 5 0.714Zimbabwe 9 3 0.333West Indies 16 3 0.188South Africa 25 7 0.280OVERALL 177 51 0.288
12. 12. PROBABILITY OF CENTURY IN A TEST MATCH0.80.70.60.50.40.3 PROBABILITY OF CENTURY0.20.1 0
13. 13. IF A CENTURY, PROBABLE TEAM TEAM CENTURIES SCORED PROBABILITY Pakistan 2 0.039Bangladesh 4 0.078 England 7 0.137 Srilanka 9 0.176 Australia 11 0.216Bangladesh 5 0.098Zimbabwe 3 0.059West Indies 3 0.059South Africa 7 0.137 TOTAL 51 1.000
14. 14. IF A CENTURY, PROBABLE TEAM0.25 0.20.15 PROBABILITY 0.10.05 0 Pakistan Bangladesh England Srilanka Australia Bangladesh Zimbabwe West Indies South Africa
15. 15. OBSERVATIONS If Sachin plays a test match, then he is most likely to score a century when opponent is Bangladesh. If Sachin scores a century in a test match it is most likely that the opponent is Australia. In both type of above situations Sachin is least likely to score a century against Pakistan.
16. 16. HALF CENTURY ANALYSIS:DISCRETE PROBABILITY
17. 17. Fifties Number of series f(x) X 0 29 0 0.433 1 19 1 0.284 2 17 2 0.254 3 2 3 0.030Total 67 ∑f(x) 1.000
18. 18. Probability distribution for half Cumulative probability centuries distribution 1.200 0.450 0.400 1.000 0.350 0.800 0.300Probability 0.250 0.600 0.200 f(x) f(x) 0.150 0.400 0.100 0.050 0.200 0.000 0 1 2 3 Number of fifties in a series 0.000 0 1 2 3
19. 19. VARIATION AND STD. DEVIATIONX x-µ (x-µ)^2 f(x) f(x)*(x-µ)^2  Variation of random variable x (fifties in a0 -0.88 0.775451 0.433 0.3356 series) is 0.7917 squared fifties1 0.12 0.014257 0.284 0.0040  Standard deviation in the2 1.12 1.253063 0.254 0.3179 number of fifties in a series (σ) is 0.88983 2.12 4.491869 0.030 0.1341 fifties 0.7917 =σ^2
20. 20. EXPECTED FIFTIES IN A SERIESX f(x) xf(x)  Thus the expected value E(x) for Sachin scoring a0 0.433 0.00 half century in a series is1 0.284 0.28 0.88 or almost 1  In every test series he2 0.254 0.51 plays, he is expected to score a half century3 0.030 0.09 1.000 0.88
21. 21. SERIES TOTAL ANALYSIS:NORMAL DISTRIBUTION
22. 22. TOTAL RUNS IN A SERIES Histogram 14Bin(Total in a series) Frequency 12 0 2 10 60 7 120 8 8 Frequency 180 7 6 240 13 Frequency 300 11 4 360 7 2 420 8 0 More than 420 4 0 60 120 180 240 300 360 420 More than 420 Bin
23. 23. TOTAL RUNS IN A SERIES.. Statistical Summary  Slightly skewed towards Mean 219.2836 right. Median 213 Mode 199  Skewness is just 0.07Standard Deviation 126.9055 approximately. Sample Variance 16104.99  Data can be considered Skewness 0.068918 Range 493 to be normally Minimum 0 distributed for analysis Maximum 493 purpose. Sum 14692 Count 67
24. 24. PROBABILITY OF A TOTAL OF 250 IN A SERIES Std Deviation σ = 127 Mean µ = 219 x = 400 f(x) = 0.00314* exp-(250-219)^2)/(2*127^2) = 0.003 = 0.3 %
25. 25. PROBABILITY OF A TOTAL OF 100 IN A SERIES Std Deviation σ = 127 Mean µ = 219 x = 400 f(x) = 0.00314* exp-(100-219)^2)/(2*127^2) = 0.002 = 0.2 %
26. 26. PROBABILITY OF A TOTAL OF 400 IN A SERIES Std Deviation σ = 127 Mean µ = 219 x = 400 f(x) = 0.00314* exp-(400-219)^2)/(2*127^2) = 0.001 = 0.1 %
27. 27. NOT OUT INNINGS :BINOMIAL DISTRIBUTION
28. 28. NOT OUT INNINGS: BINOMIAL DISTRIBUTION Only two possibilities in an innings - out or not out. Remaining not out in any match is independent of being out or not out in any other match. Probability of remaining not out = 1- Probability of being out. => Binomial Probability distribution n! n x P( X )  p (1  p) x x !(n  x )!
29. 29. NOT OUT INNINGS.. Probability of being not out in a match P(N) = Total not out innings/ Total matches = 32/290 = 0.11 Probability of being out in a match P(O) = 1 - Probability of being not out in a match = 1 – 0.11 = 0.89
30. 30. PROBABILLITY OF SINGLE NOT OUT IN 10 MATCHES X=1 n = 10 p = 0.11 1 – p = 0.89 Thus, n! P( X )  p x (1  p)n  x x !(n  x )! = (10!/(1! * 9!)) * (0.11^1)*(0.89^9) = 0.39
31. 31. PROBABILLITY OF TWO NOT OUTS IN 20 MATCHES X=2 n = 20 p = 0.11 1 – p = 0.89 Thus, n! P( X )  p x (1  p)n  x x !(n  x )! = (20!/(2! * 18!)) * (0.11^2)*(0.89^18) = 0. 28
32. 32. KEY LEARNINGS1. Trends contrasting to the preconceived notions. (Binomial Probability Distribution)2. Proves the general statements.(Half Century every match)3. Great tool for analysis.4. Easy to use, apply and understand.
33. 33. Thank You…