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Probability Investigation

Introduction

    My aim in this investigation is to investigate three questions:

'font-size:14.0pt; '>1.     The probabilities of each of A, B or C winning the game.

'font-size:14.0pt; '>2.     Who will be the most likely winner?

'font-size:14.0pt; '>3.     The most likely length of the game in terms of the number of rolls of the dice to produce a winner.

Predictions'font-size:14.0pt; '>

    Before I actually do the game in the proper way I’m going to make a few predictions. This will help me to understand possible results better.

     I predict that C will be the most likely winner as it has the highest probability of winning the game when it’s his turn. I say this even though it takes its turn last. I believe B will be the second most likely winner as it too has the second highest probability; and the same goes for A- least likely for the same reason.

    As for the most likely length of game in terms of the number of rolls of the dice to produce a winner - I believe that the answer to this question could be 3 as it is the first opportunity for C to win and it has a 50% chance of winning every throw it has (3/6). Obviously A and more so B will win before it even gets to C sometimes but over the whole game the most likely winner will be C on average.

Method

    To do this investigation – first I played the game. We were told by our teacher to play it 30 times and to take down the results with ultimate care - which we did successfully.

Results

    After carrying out the experiment in a careful and strategic way I found some interesting results.

    I found that A won just twice (2/30), B won thirteen time (13/30) and C won a total of fifteen times (15/30). This shows that at this stage that C is the more likely winner.

    Another very important set of results in terms of finding a result for the last question, most likely length of game. I found that B won eight times on the second throw, C won thirteen times on the third throw, A won twice on the fourth throw, B won three times on the fifth throw, C won twice on the sixth throw, B won once on the eighth throw and once on the eleventh throw.

    Below is a table displaying the results – it shows the number of throws it took for someone to win and who actually won.

Information on table of results

    On the table below along the top is the number of throws it took for someone to win.

    And, along the side is the game number (game number 1,2,3,4,5 etc.)

    Where it has the letter X it means that whichever letter was throwing at the time – didn’t win on that particular throw. Therefore when it says another letter, either A, B or C, it means that particular letter won on that specific throw.

Table of Results                   

    After all this I decided it would be a good idea (as did my teacher) to collaborate my results with someone else. I decided this would be a good idea because then I can back up any possible assumptions/rules that may be starting to appear.

    Having collaborated my results with a friend, Laurence, I came to a more definite but not yet firm conclusion. His individual results were: A won four times (4/30), B won twelve times (12/30) and C won fourteen times (14/30).

    Now, collaborating the results together I came to this: A won six times (6/60), B won twenty-five times (25/60) and finally C won twenty-nine times (29/60).

Conclusion 

    Now that I’ve carried out the experiment and I’m happy that I did it in a clear and organised way I can come to a few conclusions. I believe at the present moment C has the highest probability in winning – (29/60) overall confirms this.

    Also another conclusion is that C was the most likely winner, I believe this was because on each throw it took, it had a 50% chance of winning whereas this wasn’t the case with A and B. A had only a 16.6% chance of winning and B had a reasonable 33.3% chance of winning each time it took a throw. I have a good idea that this was the telling factor in the results that I took down. I think that due to the percentage chance of winning C won the most and A the least.

    I am finding that trends are beginning to appear in the results. For a start from my results A never won on its first go. If B won it normally won on its first go as the next person to throw was C and it won frequently on the third throw – in fact it won thirteen out of fifteen times it won, on its first throw (the third throw). This amounted to almost half the number of games played that C won on its first throw. Again I think this was to do with the percentage chance of C winning. Lastly I’ve found that all the games were particularly short in length. As a whole the average length of game was 109/30 = 3.63 throws. However I think the most likely length of game is three throws in which C wins.

    Now I’ve collected some results I can look to find a rule/pattern/formula for n. This is important to try and work out as it might help me discerning a rule for this particular investigation. The formula for finding the probability is Probability = number of successful outcomes / total number of outcomes. Using this A was 6/60 = 0.1, B was 25/60 = 0.42 and C was 29/60 = 0.48. These answers tell us that C was the most likely winner as I predicted, B a close second and A the least likely. So, the formula for finding the probability is s/n. s stands for the no. of successful outcomes and n stands for the total possible outcomes.

    Now I am going to write some new conclusions that I have come to after doing some extensive theory work. Hopefully I will be able to give the right answers at the end of this. Below are all the probabilities that I’ve worked out.

P (A wins on 1^st throw) = 1/6 (3/18)

  (B                          )  = 5/18

   (C                             )  = 5/18

Nobody wins on 1^st throw = 5/18

P (A wins on 2^nd throw) = 15/324

   (B                              ) = 25/324

   (C                              ) = 25/324

Nobody wins on 2^nd throw = 25/324

P (A wins on 3^rd throw) = 75/5832

   (B                             ) = 125/5832

   (C                             ) = 125/5832

Nobody wins on 3^rd throw = 125/5832

P (Of A winning after 1^st and 2^nd throw) = 69/324

   (      B                                                  ) = 115/324

   (      C                                                  ) = 115/324

   (    nobody                                           ) = 25/324

P (Of A winning after 1^st, 2^nd and 3^rd throw) = 1317/5832

   (      B                                                        ) = 2195/5832

   (      C                                                        ) = 2195/5832

   (    nobody                                                 ) = 125/5832

P (A wins on 4^th throw) = 124/104976

   (B                             ) = 625/104976

   (C                             ) = 625/104976

(nobody                    ) = 103602/104976

    I deducted from these results that there was a clear ratio between the results of A compared to B and C. A’s results were in the ratio of 3:5 to B’s and C’s results. This was a clear and very important result. For example the results for the probabilities of each of A, B and C after the 1^st and 2^nd throws were A = 69/324, B and C (the same) 115/324. If you divide 115 by 5 you get 23, this multiplied by three is 69 – this clearly shows the ratio of 3:5.

    From these findings I was then able to work out the answer for the most likely winner. As the probability for each of A, B and C respectively had been worked out to show that A was less likely than B and C – and the fact that B and C had the same probability which was the equal highest of all of them; B and C had to be the most equally likely winner.

    Finally I worked out that the most likely length of game in terms of number of rolls of the dice was three rolls. First of all I thought that as B and C were the most likely winners that possibly 2,3,5 and 6 (which corresponded to B and C’s first two throws) would be the most likely number of rolls before there was a winner. However I looked at my own results and Laurence’s and found that C won the most times out of all the games played (in my results 15/30) and an amazing thirteen times out of that fifteen it won on its first throw (which was the game’s third throw). This was the main reason I decided that three rolls was the most likely.

    So, to sum up my answers are as follows:

'font-size:14.0pt; '>1.     P (Of A winning) = 3/13

        (      B              ) = 5/13

        (      C              ) = 5/13

'font-size:14.0pt; '>2.     B and C were the most likely winners equally.

3. Three rolls                        

Evaluation

    Overall my experiment has been a successful one and I’ve come to a few conclusions that seem to be getting towards firm conclusions, which I can rely on. I believe my experiment to be fairly reliable because I took the results down carefully and accurately. Also I collaborated my results with someone else to give me a more reliable look at my results and confirm any possible patterns. This made it increasingly reliable.

    However saying this I would have liked to done a bit more playing of the game as the more I played the game the more reliable the results would have been. Although one could argue playing the game over and over again may not help or influence any conclusions that one’s already drawn up. Also maybe I could have drawn a bigger probability tree to possibly help improve the reliability of my investigation and give me a clearer idea of a firmer conclusion.

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