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Excited to win (Posted on 2012-06-06) Difficulty: 3 of 5
When referring to the record of team, let x,y mean a team has won x games and lost y games out of a total of x+y games played.

Suppose a team currently has a 1,1 record, having won one of its first two games. From there on the team's probability of winning its game is equal to its current proportion of wins. (For example if after 8 games their record is 5,3 they have a 5/8 chance of winning the next game.)

[1] Find the probability that the team will have a record of a,b after it has played a total of a+b games.

[2] If we know they lost their third game so they are now 1,2, again find the probability of having a record of a,b after a+b games.

No Solution Yet Submitted by Jer    
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Solution solution Comment 1 of 1

The probability of a given state, p(x,y), will occur equals p(x-1,y)*(x-1)/(x-1+y) + p(x,y-1)*(1-(x/(x+y-1)). The following program uses this recursion to produce two tables:one for part 1 and one for part 2:


DEFDBL A-Z
CLS
DIM p(10, 10)
p(1, 1) = 1
FOR wins = 1 TO 10
FOR losses = 1 TO 10
  IF wins > 1 OR losses > 1 THEN
    p(wins, losses) = p(wins - 1, losses) * (wins - 1) / (wins - 1 + losses) + p(wins, losses - 1) * (1 - wins / (wins + losses - 1))
    PRINT USING " .######"; p(wins, losses);
  ELSE
    PRINT USING "#.######"; p(wins, losses);
  END IF
NEXT
PRINT
NEXT

ERASE p
PRINT
p(1, 1) = 1
p(1, 2) = 1
FOR wins = 1 TO 10
FOR losses = 2 TO 10
  IF wins > 1 OR losses > 2 THEN
    p(wins, losses) = p(wins - 1, losses) * (wins - 1) / (wins - 1 + losses) + p(wins, losses - 1) * (1 - wins / (wins + losses - 1))
    PRINT USING " .######"; p(wins, losses);
  ELSE
    PRINT USING "#.######"; p(wins, losses);
  END IF
NEXT
PRINT
NEXT

I've added column and row headings manually.

For part 1:
                              losses  
wins    1        2       3       4       5       6       7       8       9      10  
  1 1.000000 .500000 .333333 .250000 .200000 .166667 .142857 .125000 .111111 .100000  
  2  .500000 .333333 .250000 .200000 .166667 .142857 .125000 .111111 .100000 .090909  
  3  .333333 .250000 .200000 .166667 .142857 .125000 .111111 .100000 .090909 .083333  
  4  .250000 .200000 .166667 .142857 .125000 .111111 .100000 .090909 .083333 .076923  
  5  .200000 .166667 .142857 .125000 .111111 .100000 .090909 .083333 .076923 .071429  
  6  .166667 .142857 .125000 .111111 .100000 .090909 .083333 .076923 .071429 .066667  
  7  .142857 .125000 .111111 .100000 .090909 .083333 .076923 .071429 .066667 .062500  
  8  .125000 .111111 .100000 .090909 .083333 .076923 .071429 .066667 .062500 .058824  
  9  .111111 .100000 .090909 .083333 .076923 .071429 .066667 .062500 .058824 .055556  
 10  .100000 .090909 .083333 .076923 .071429 .066667 .062500 .058824 .055556 .052632  

It seems apparent that for any given number of games played, any (x,y) where each of x and y is at least 1 is equally likely. So p(a,b) = 1/(a+b-1) when a+b games have been played.

For part 2:
                              losses
wins            2        3       4       5       6       7       8       9      10
  1         1.000000 .666667 .500000 .400000 .333333 .285714 .250000 .222222 .200000
  2          .333333 .333333 .300000 .266667 .238095 .214286 .194444 .177778 .163636
  3          .166667 .200000 .200000 .190476 .178571 .166667 .155556 .145455 .136364
  4          .100000 .133333 .142857 .142857 .138889 .133333 .127273 .121212 .115385
  5          .066667 .095238 .107143 .111111 .111111 .109091 .106061 .102564 .098901
  6          .047619 .071429 .083333 .088889 .090909 .090909 .089744 .087912 .085714
  7          .035714 .055556 .066667 .072727 .075758 .076923 .076923 .076190 .075000
  8          .027778 .044444 .054545 .060606 .064103 .065934 .066667 .066667 .066176
  9          .022222 .036364 .045455 .051282 .054945 .057143 .058333 .058824 .058824
 10          .018182 .030303 .038462 .043956 .047619 .050000 .051471 .052288 .052632

Here, it seems that for any given number of games played, the probability increases arithmetically with y, the number of losses. Starting at the original two losses, the ratio of the probabilities for the given number of games played is 1:2:3:...:x+y-2 (for example, for 5 games played, it's just 1:2:3). The denominator, then, for each probability is therefore the sum of the integers from 1 to x+y-2, or (x+y-2)*(x+y-1)/2. So p(a,b) = 2*(b-1)/((a+b-2)*(a+b-1)) when a+b games have been played.


  Posted by Charlie on 2012-06-06 14:59:30
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