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Discard Dice (Posted on 2012-07-27) Difficulty: 3 of 5

In a dice game, 2 players each simultaneously cast 4 fair 6-sided dice, to form a 'hand'.

Each player then:-
(i) discards the highest-scoring die from his hand, and
(ii) scores the value of the next-highest (or highest-equal) die in his hand.

If my opponent has cast {6,4,4,3}, discarding the 6 to score 4, what is the probability that my score exceeds his?

See The Solution Submitted by broll    
Rating: 3.0000 (1 votes)

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Solution solution | Comment 2 of 4 |

The two tosses of 4 dice each are of course independent, so it doesn't matter how the opponent scored 4, whether by {6,4,4,3} or some other 4-scoring hand, like {5,4,2,2} or {4,4,4,1}. The goal is to find the probability of exceeding 4, which is to say, scoring 5 or 6.

It's easiest to analyze the probability of any given highest-2 pair by considering separately the case where the top two scoring dice have the same pip-count from the case where the two are different.

If both are the same there are 2, 3 or 4 of the dice with that number showing. If that number is called n then there are these three sub-cases, in which there are C(4,n) in each instance of n, ways of choosing which 2, 3 or 4 are the high value and in each of these ways, the probability of getting the given arrangement is (1/6)^n * ((score-1)/6)^(4-n), where "score" is the duplicated high value in question. So the overall probability of (s,s) being the top two dice is Sigma{n=2 to 4}C(4,n) * (1/6)^n * ((s-1)/6)^(4-n). The reason for the (s-1) is that that's how many possibilities there are that are less than s, the score.

Similar reasoning applies to the case where the top two faces are different, except here there is only one of the highest scoring die and possibly 1, 2 or 3 of the second highest value. There are 4 ways of choosing which is the high-scoring die and C(3,n) ways of choosing the n=1,2 or 3. of the second highest, for Sigma{n=1 to 3} 4*C(3,n)*(1/6) * (1/6)^n * ((s-1)/6)^(3-n), where s is the second highest value, and therefore the score.

This program calculates these values for every pair of highest two values:

   10     DIM rslt(6, 6), scTot(6)
   20
   30     CLS
   40
   50     FOR top = 6 TO 1 STEP -1
   60     FOR second = top TO 1 STEP -1
   70       ct = ct + 1
   80
   90       IF top = second THEN
  100        :tot = 0
  110        :FOR n = 2 TO 4
  120          :p = (1 // 6) ^ n * ((top - 1) // 6) ^ (4 - n) * combi(4, n)
  130          :tot = tot + p
  140        :NEXT n
  150        :PRINT top; second,
  160        :PRINT  tot,tot/1
  170      :ELSE
  180        :tot = 0
  190        :FOR n = 1 TO 3
  200          :p = (1 // 6) * (1 // 6) ^ n * ((second - 1) // 6) ^ (3 - n) * combi(3, n) * 4
  210          :tot = tot + p
  220        :NEXT n
  230        :PRINT top; second,
  240        :PRINT  tot,tot/1
  250      :END IF
  260       cumulative = cumulative + tot
  270       rslt(top, second) = tot
  280     NEXT
  290     NEXT
  300     PRINT ct,
  310     PRINT  cumulative: PRINT
  320
  330     FOR score = 1 TO 6
  340       tot = 0
  350       FOR first = 1 TO 6
  360         tot = tot + rslt(first, score)
  370       NEXT
  380       PRINT score, tot,tot/1
  390       c = c + tot
  400       scTot(score) = tot
  410     NEXT
  420     PRINT "", c,c/1
  430
  440     PRINT scTot(5) + scTot(6)

The resulting table:

 high    prob.        prob. expressed as decimal
 pair
 6  6    19/144         0.1319444444444444444
 6  5    61/324         0.1882716049382716048
 6  4    37/324         0.1141975308641975308
 6  3    19/324         0.0586419753086419753
 6  2    7/324          0.0216049382716049382
 6  1    1/324          0.0030864197530864197
 5  5    113/1296       0.087191358024691358
 5  4    37/324         0.1141975308641975308
 5  3    19/324         0.0586419753086419753
 5  2    7/324          0.0216049382716049382
 5  1    1/324          0.0030864197530864197
 4  4    67/1296        0.0516975308641975308
 4  3    19/324         0.0586419753086419753
 4  2    7/324          0.0216049382716049382
 4  1    1/324          0.0030864197530864197
 3  3    11/432         0.0254629629629629629
 3  2    7/324          0.0216049382716049382
 3  1    1/324          0.0030864197530864197
 2  2    11/1296        0.0084876543209876542
 2  1    1/324          0.0030864197530864197
 1  1    1/1296         0.0007716049382716048
 21      1
 


so there are 21 such possible pairs, and the total probability does in fact check out as 1.
 
The program also tabulated the total probability for each score (second number):

 
score    prob.       prob. expressed as decimal
 1       7/432          0.0162037037037037036
 2       41/432         0.0949074074074074073
 3       29/144         0.2013888888888888888
 4       121/432        0.2800925925925925925
 5       119/432        0.2754629629629629629
 6       19/144         0.1319444444444444444
Total    1       1.0
        


Again, it adds up to 1, reassuringly.

The total of score 5 and score 6 probabilities is then given as 119/432 + 19/144 =

 11/27 ~= .4074074074074074
 
which would then be the answer to the puzzle.


  Posted by Charlie on 2012-07-27 16:53:43
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