1000 REM Probable.bas 1020 ' MJR 15-03-98 1030 ' To test the theory of CARGOAT 1040 ' Theory: There are three boxes and you can choose to open any one box. 1050 ' One box contains a new car, and each other box an old goat. 1060 ' The probability of selecting a new car is approximately 33% 1070 ' After you have chosen, you are shown the contents of one of the 1072 ' two remaining boxes. It is a goat. 1074 ' You are offered the opportunity of abandoning your first 1076 ' choice and taking the unopened remaining box. Ought you to 1078 ' change your mind and take this opportunity? 1080 ' It is claimed via statistical theory that the probability that 1082 ' this other box contains the car is approx 66%, not the 50% 1084 ' you might assume if there is now only a choice of two boxes. 1090 ' In other words, if you change your selection then the chance 1092 ' of your winning is 66% instead of 33% if you do not change. 1094 ' You might be forgiven in thinking that your chances are 50% 1096 ' on either box. 1200 ' Comments: It is claimed that, having already made the first choice, 1202 ' the event has passed and the probability cannot be altered, 1204 ' so it remains a 33% probability that the car is in the 1206 ' first-chosen box, and 66% that it is in the unopened 1208 ' remaining box. 1220 ' Try to formulate a common sense explanation of this result 1230 ' and illustrate it with a Basic program not involving 1232 ' formal Statistical Theory. 1300 ' Argument: A: I have chosen Box X, leaving Box Y and Box Z 1302 ' B: Box X has been shifted into a Green Area 1304 ' C: Boxes Y and Z have been shifted to a Red Area 1306 ' D: All the boxes have an equal chance of containing the car 1308 ' E: So the car is more likely to be in the Red Area 1310 ' F: If I could choose an area it would be the Red Area because 1312 ' it contains two boxes. 1314 ' G: The car can only be in one box so one of the boxes in the 1316 ' Red Area must contain a goat, and if that box is removed 1318 ' then my prospects cannot be affected. 1320 ' H: The box most likely to contain the car is the remaining 1322 ' box in the Red Area. 1324 ' I: The probability that the car is in the red area is 2 in 3 1326 ' J: If the car is in the Red Area then it is in the remaining 1328 ' box. 1330 ' K: So, the probability that the car is in the remaining box 1332 ' in the Red Area is 2 in 3. 1334 ' L: The probability that the car is in Box in the Green Area 1336 ' is 1 in 3. 1338 ' M: So, if I am given the opportunity I should change my 1340 ' selection to the remaining box in the Red Area. 1350 ' 2000 PRINT "Program name: Probable.bas" 2010 RANDOMIZE 2020 INPUT "Enter reqd number of trials: ", AA% 2030 FOR IA% = 1 TO AA% 2032 PRINT IA%; 2040 BA% = INT(RND * 3) + 1' Set winning box number 2050 PRINT BA%; 2100 BB% = INT(RND * 3) + 1' Punter's original choice (from 3 boxes) 2110 PRINT BB%; 2112 BE% = BC% 2120 IF BB% = BA% THEN BC% = BC% + 1' Accum score 2122 IF BC% = BE% THEN BF% = 0 ELSE BF% = 1' 1 indicates win 2124 PRINT BF%; 2130 PRINT BC%; 2200 BD% = INT(BC% / IA% * 100)' Percent 2210 PRINT BD%; 2310 IF BB% = 1 THEN CA% = 2: CB% = 3' If choice = 1 then residue are 2 & 3 2312 IF BB% = 2 THEN CA% = 1: CB% = 3 2314 IF BB% = 3 THEN CA% = 1: CB% = 2 2320 CC% = 0' Reset number of the box to be eliminated. 2330 IF CA% <> BA% THEN CC% = CB%' If residue #1 is goat then choice = #2 2332 IF CB% <> BA% THEN CC% = CA%' If #2 is goat then choice = #1 2333 PRINT CC%; ' Box number of residual (non-eliminated) box being offered. 2334 CE% = CD% 2340 IF CC% = BA% THEN CD% = CD% + 1 2341 IF CD% = CE% THEN CG% = 0 ELSE CG% = 1' 1 indicates win 2342 PRINT CG%; ' Win / Lose 2343 PRINT CD%; ' Accum score 2344 CF% = INT(CD% / IA% * 100)' Percent 2346 PRINT CF% 2900 NEXT IA% 3000 PRINT "Trial #; Goal; Try_1; Y/N; Wins; Percent; Try_2; Y/N; Wins; Percent" 3010 INPUT "Repeat ? Y/N :", DA$ 3020 IF DA$ = "N" OR DA$ = "n" THEN 9900 ELSE RUN 9900 PRINT "Done" 9999 END