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 Rolling to score (Posted on 2010-11-15)
We roll five standard dice (sides numbered 1 to 6) and write down the sum of the top three i.e. of the 3 highest values.
What is the probability to get 15 ?

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 computer-aided solution | Comment 4 of 17 |

The following program examines each possible combination of numbers of the various numbers of pips, using each one's probability to add to the corresponding total's probability of occurring:

`  100       Ndice=5:Ntop=3  110       Remaining=Ndice:CurrProb=1  120       dim TProb(Ndice*6)  130       dim Hist(6)  140    150       gosub *SetQuant(6)  160       for I=1 to Ndice*6  170          CProb=CProb+TProb(I)  180          print I,TProb(I),CProb/1,1-CProb/1  190       next I  200    210       end  220   1100           *SetQuant(Den) 1110           local Nbr,P,Strt 1120           if Den=1 then Strt=Remaining:else Strt=0 1130           for Nbr=Strt to Remaining 1135             if Den=1 then 1136               :P=1 1137             :else 1140               :P=(1//Den)^Nbr*((Den-1)//Den)^(Remaining-Nbr)*combi(Remaining,Nbr) 1150            1160             CurrProb=CurrProb*P 1170             Remaining=Remaining-Nbr 1180             Hist(Den)=Nbr 1190            1200             if Remaining=0 then 1210              :gosub *CountIt(Den) 1220             :else 1230              :gosub *SetQuant(Den-1) 1240            1250             Remaining=Remaining+Nbr 1260             CurrProb=CurrProb//P 1270           next Nbr 1280           return 1290          2100           *CountIt(D) 2110           local Pips,Totv,Totn 2120           for Pips=6 to D step -1 2130             if Hist(Pips)>=Ntop-Totn then 2140              :Totv=Totv+Pips*(Ntop-Totn) 2150              :Totn=Ntop 2160             :else 2170              :Totv=Totv+Pips*Hist(Pips) 2180              :Totn=Totn+Hist(Pips) 2190           next Pips 2200           TProb(Totv)=TProb(Totv)+CurrProb 2210           return  `

The result is a display that shows all the possible totals:

` score   p(score)        p(<= this score)        p(> this score) 1      0               0                       1 2      0               0                       1 3      1/7776         0.0001286008230452674   0.9998713991769547325 4      5/7776         0.0007716049382716048   0.9992283950617283951 5      5/2592         0.0027006172839506172   0.9972993827160493827 6      41/7776        0.0079732510288065843   0.9920267489711934156 7      5/432          0.0195473251028806584   0.9804526748971193415 8      85/3888        0.0414094650205761316   0.9585905349794238683 9      37/972         0.0794753086419753086   0.920524691358024691310      235/3888       0.1399176954732510287   0.860082304526748971211      665/7776       0.2254372427983539094   0.774562757201646090512      881/7776       0.3387345679012345678   0.661265432098765432113      1055/7776      0.4744084362139917695   0.525591563786008230414      385/2592       0.6229423868312757201   0.377057613168724279815      1111/7776      0.7658179012345679012   0.234182098765432098716      935/7776       0.886059670781893004    0.113940329218106995917      305/3888       0.9645061728395061728   0.035493827160493827118      23/648         1.0                     019      0               1.0                     020      0               1.0                     021      0               1.0                     022      0               1.0                     023      0               1.0                     024      0               1.0                     025      0               1.0                     026      0               1.0                     027      0               1.0                     028      0               1.0                     029      0               1.0                     030      0               1.0                     0`

So the probability of a score of exactly 15 is 1111/7776 ~= .1428755144032922. The probability of getting at least 15 is the same as the probability of getting > 14, which is shown above as being approximately 0.3770576131687242798. Adding the individual probabilities for 15 through 18 gives this as exactly 733/1944.

As all the denominators can be expanded to 6^5 = 7776, the following table can be made by a slight modification of the program, in which the last two columns have been multiplied by 7776, so as to provide the numerators for a fraction with that number in the denominator:

`value    p(score=value) p(score<=value) p(score > value)                         -------numerators only ------- 2       0              0                 7776 3       1/7776         1                 7775 4       5/7776         6                 7770 5       5/2592         21                7755 6       41/7776        62                7714 7       5/432          152               7624 8       85/3888        322               7454 9       37/972         618               7158 10      235/3888       1088              6688 11      665/7776       1753              6023 12      881/7776       2634              5142 13      1055/7776      3689              4087 14      385/2592       4844              2932 15      1111/7776      5955              1821 16      935/7776       6890              886 17      305/3888       7500              276 18      23/648         7776              0  `
` `

As the specific interest in this problem is for a score of 15, the following details the derivation of probability 1111/7776 for that score. The sets of showing numbers of pips are all those that score 15. The first probability, for example, is 5/3888 = 10/7776, as 3 5's and two 1's can occur in C(5,3) = C(5,2) = 10 ways.

`     number of  6's  5's  4's  3's  2's  1's     probability  0    3    0    0    0    2        5/3888  0    3    0    0    1    1        5/1944  0    3    0    0    2             5/3888  0    3    0    1    0    1        5/1944  0    3    0    1    1             5/1944  0    3    0    2                  5/3888  0    3    1    0    0    1        5/1944  0    3    1    0    1             5/1944  0    3    1    1                  5/1944  0    3    2                       5/3888  0    4    0    0    0    1        5/7776  0    4    0    0    1             5/7776  0    4    0    1                  5/7776  0    4    1                       5/7776  0    5                            1/7776  1    1    1    0    0    2        5/648  1    1    1    0    1    1        5/324  1    1    1    0    2             5/648  1    1    1    1    0    1        5/324  1    1    1    1    1             5/324  1    1    1    2                  5/648  1    1    2    0    0    1        5/648  1    1    2    0    1             5/648  1    1    2    1                  5/648  1    1    3                       5/1944  2    0    0    1    0    2        5/1296  2    0    0    1    1    1        5/648  2    0    0    1    2             5/1296  2    0    0    2    0    1        5/1296  2    0    0    2    1             5/1296  2    0    0    3                  5/3888  which do indeed add up to 1111/7776.  `

Simulation verification:

DEFDBL A-Z
CLS
DIM valCt(20)
FOR trial = 1 TO 1000000
tot = 0
FOR die = 1 TO 5
pips(die) = INT(RND(1) * 6) + 1
NEXT die
DO
done = 1
FOR i = 1 TO 4
IF pips(i) < pips(i + 1) THEN SWAP pips(i), pips(i + 1): done = 0
NEXT
LOOP UNTIL done = 1
FOR i = 1 TO 3
tot = tot + pips(i)
NEXT
valCt(tot) = valCt(tot) + 1
NEXT trial

FOR i = 3 TO 18
c = c + valCt(i) / (trial - 1)
PRINT USING "## #.####### #.#######"; i; valCt(i) / (trial - 1); c
NEXT

goes through a million trials and gets the following statistics:

`val  fraction =val   fraction <= val 3   0.000142           0.000142 4   0.000625           0.000767 5   0.001896           0.002663 6   0.005248           0.007911 7   0.011541           0.019452 8   0.021726           0.041178 9   0.038271           0.07944910   0.060519           0.13996811   0.085300           0.22526812   0.113187           0.33845513   0.135842           0.47429714   0.148759           0.62305615   0.142818           0.76587416   0.120569           0.88644317   0.077907           0.96435018   0.035650           1.000000`

 Posted by Charlie on 2010-11-15 18:33:10

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