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2nd grade math (Posted on 2017-01-12) Difficulty: 3 of 5
My son is doing a math worksheet. He is practicing the concept of 'carrying' when doing sums. The sheet has 16 problems, each is summing two three-digit numbers. What struck me as interesting was that the creator of the problems made every problem have at least two carries and most have three.

What is the probability distribution for the number of carries in finding the sum of two randomly selected three-digit numbers?

Feel free to generalize.

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Solution solution | Comment 1 of 4
What pairs of digits will cause a carry? Consider the following table:

  0 1 2 3 4 5 6 7 8 9
0 . . . . . . . . . +
1 . . . . . . . . + x 
2 . . . . . . . + x x
3 . . . . . . + x x x
4 . . . . . + x x x x
5 . . . . + x x x x x
6 . . . + x x x x x x
7 . . + x x x x x x x
8 . + x x x x x x x x
9 + x x x x x x x x x

An x indicates that the given pair of digits will always cause a carry. A + indicates that the pair will cause a carry if and only if there was a carry coming into the addition of the digit pair. A dot indicates no carry coming out of the addition of this pair.

The first (rightmost) digit pair can never have a carry coming into it.  The leftmost digit pair cannot contain a zero. The remaining digit(s) can have either. For larger than 3-digit numbers the middle digits can have varying probabilities of having a carry coming into them depending on how far from the right they are.

Starting at the right ends of the pair of 3-digit numbers: 

The probability of a carry is 45/100 = 9/20 (or technically, 45/81 if in the general case we're talking about 1-digit numbers, so as to exclude zeros).

Going to the 10's digit, there's a 9/20 probability it will have a carry coming into it, so we must give a 9/20 weight to the probability of a carry going out of it given that a carry came in, and an 11/20 (i.e. 55/100) weight to the probability given that a carry did not come in. But actually we're interested in counting the carries, so it's pretty useless to consider the overall probability of sending a carry to the 100's digit pair. So:

prob of no carry, no carry: (11/20)^2
prob of no carry, carry: (11/20)*(9/20)
prob of carry, no carry: (9/20)*(9/20)
prob of carry, carry: (9/20)*(11/20)

So, going into the 100's digit there's:

(i)   prob of carry count = 0, no carry going in now:  (11/20)^2 = 121/400
(ii)  prob of carry count = 1, no carry going in now:  (9/20)^2 = 81/400
(iii) prob of carry count = 1, carry going in now: 99/400
(iv)  prob of carry count = 2, carry going in now: 99/400

For the calculation of that last digit (100's digit), we must consider that if there is no carry going into it there is a 45/81 probability there will be a carry coming out; if there is a carry going in there's a 53/81 probability of a carry going out (making the sum a 4-digit number from two 3-digit numbers).

(i)   0 -> 0   (11/20)^2 * 36/81       =   121/900
(i)   0 -> 1   (11/20)^2 * 45/81       =   121/720
(ii)  1 -> 1   (9/20)^2 * 36/81        =     9/100
(ii)  1 -> 2   (9/20)^2 * 45/81        =     9/80
(iii) 1 -> 1   99/400 * 28/81          =    77/900
(iii) 1 -> 2   99/400 * 53/81          =   583/3600
(iv)  2 -> 2   99/400 * 28/81          =    77/900
(iv)  2 -> 3   99/400 * 53/81          =   583/3600

p(0) = 121/900                    =  121/900    ~=  .134444444444444
p(1) = 121/720 + 9/100 + 77/900   = 1237/3600   ~=  .343611111111112
p(2) = 9/80 + 583/3600 + 77/900   =    9/25     ~=  .36
p(3) = 583/3600                   =  583/3600   ~=  .161944444444444

A program for more generalized case:

   10    dim ProbCtOv(10,1),NewProbCtOv(10,1)
   20    kill "2grdmath.txt": open "2grdmath.txt" for output as #2
   30    for N=1 to 6
   40      erase ProbCtOv():dim ProbCtOv(10,1)
   50      ProbCtOv(0,0)=1
   60      ProbCtOv(0,1)=0
   70      for K=1 to N
   80        erase NewProbCtOv():dim NewProbCtOv(10,1)
   90        if K<N then POvOv=11//20:PNOvOv=9//20
  100          :else POvOv=53//81:PNOvOv=45//81
  110        POvNOv=1-POvOv:PNOvNOv=1-PNOvOv
  115        for I=0 to N
  120          NewProbCtOv(I,0)=NewProbCtOv(I,0)+ProbCtOv(I,0)*PNOvNOv
  130          NewProbCtOv(I+1,1)=NewProbCtOv(I+1,1)+ProbCtOv(I,0)*PNOvOv
  140          NewProbCtOv(I,0)=NewProbCtOv(I,0)+ProbCtOv(I,1)*POvNOv
  150          NewProbCtOv(I+1,1)=NewProbCtOv(I+1,1)+ProbCtOv(I,1)*POvOv
  155        next I
  160        for I=0 to N
  170          ProbCtOv(I,0)=NewProbCtOv(I,0)
  180          ProbCtOv(I,1)=NewProbCtOv(I,1)
  190        next
  200      next K
  210      print N
  220      for I=0 to N
  230        print ProbCtOv(I,0)+ProbCtOv(I,1),
  235        print #2, ProbCtOv(I,0)+ProbCtOv(I,1),
  240      next I
  250      print:print #2,
  260    next N
  270    close #2

produces the following info for numbers of carries from zero to n (annotated by hand with double /'s replaced by single):

digits           number of carries probabilities
n                0     1         2           3           ...

1          4/9   5/9  
2          11/45   83/180   53/180  
3          121/900   1237/3600   9/25   583/3600  
4          1331/18000   17171/72000   2727/8000   2061/8000   6413/72000  
5          14641/360000   45617/288000   22743/80000   11727/40000   13959/80000   70543/1440000  
6          161051/7200000   2940179/28800000   702801/3200000   453627/1600000   370323/1600000   72963/640000   775973/28800000 

With decimals shown:

      0         1         2         3        4          5        6         7        8          9
 1 
 0.4444444 0.5555556

 2 
 0.2444444 0.4611111 0.2944444

 3 
 0.1344444 0.3436111 0.3600000 0.1619444

 4 
 0.0739444 0.2384861 0.3408750 0.2576250 0.0890694

 5 
 0.0406694 0.1583924 0.2842875 0.2931750 0.1744875 0.0489882

 6 
 0.0223682 0.1020895 0.2196253 0.2835169 0.2314519 0.1140047 0.0269435

 7 
 0.0123025 0.0643848 0.1611039 0.2482995 0.2539153 0.1725524 0.0726227 0.0148189

 8 
 0.0067664 0.0399412 0.1138099 0.2032093 0.2478664 0.2114120 0.1234458 0.0453985 0.0081504

 9 
 0.0037215 0.0244589 0.0781246 0.1582502 0.2232617 0.2272116 0.1669166 0.0856021 0.0279700 0.0044827

Edited on January 12, 2017, 2:49 pm
  Posted by Charlie on 2017-01-12 14:48:35

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