Guessing game: Tossing one die


 

(At the end of this unit, there is a TI-84 Plus C calculator simulation for playing the game.)

 

One die is tossed over and over again until each number from one to six shows up at least once. Statistics are kept, and the total number of tosses needed is recorded.

 

Before play starts, all participants guess the total number of tosses. The lowest reasonable guess is 6, but there is no upper limit.

 

If you guess correctly, you earn 5 points; if you are one off, you earn 3 points; if you are 2 off, you earn 2 points, and if you are 3 off, you earn 1 point.

 

Example.

you toss:        Mark on your score card:

 

1

2

3

4

5

6

6

 

 

 

 

 

x

5

 

 

 

 

x

x

5

 

 

 

 

xx

x

6

 

 

 

 

xx

xx

2

 

x

 

 

xx

xx

1

x

x

 

 

xx

xx

1

xx

x

 

 

xx

xx

5

xx

x

 

 

xxx

xx

4

xx

x

 

x

xxx

xx

2

xx

xx

 

x

xxx

xx

3

xx

xx

x

x

xxx

xx

 

Thus, after the game is over, the score card looks as follows:

number on die: 1       2       3       4       5       6      
tally: xx xx x x xxx xx
no. of times tossed: 2 2 1 1 3 2 Total = 11

 

 

So, for example, guesses of 9 and 13 are two off, and each would earn 2 points.

 

Question.

What is the mean value, m, of the number of tosses for this game?

 

Remark.

Students must be told that mean value estimates the average that is computed from very many examples. It is also called the "expected value", which is very misleading, because in most situations we do not expect that any particular result will match the mean or will even be close to the mean.

 

Finding the mean.

Let’s call a toss successful if it returns a new value, and unsuccessful if it returns a value seen before. So the game ends when we score 6 successful tosses.

But the probability of a successful toss depends on how many numbers have already been seen. And this value varies from 0 at the beginning of the game to 5 when only one number is left unseen.

 

Number of numbers seen so far:      Probability of a successful toss:
   0    6/6 = 1
   1    5/6
   2    4/6
   3    3/6
   4    2/6
   5    1/6

 

There are two facts about probability that are needed to answer the original question.

 

(1) If the probability of success in one toss is p, then the mean value of the number of tosses until the first success is 1/p (the reciprocal of the probability of success).

 

Therefore we have:

 

Number of numbers seen so far:      Mean value of number of tosses until there is a successful one:
   0    6/6 = 1
   1    6/5 = 1.2
   2    6/4 = 1.5
   3    6/3 = 2
   4    6/2 = 3
   5    6/1 = 6

 

(2) The mean value of the sum of any random variables is the sum of their mean values.

Here, the number of tosses until the next success is one random variable. The first one is constant because the success in the first toss is assured, but the remaining five are not.)

 

Therefore, the mean value of the number of tosses until the game is finished is,

      m = 1 + 1.2 + 1.5 + 2 + 3 + 6 = 14.7 tosses.

 

Remarks.

Of course you do not expect 14.7 tosses in any game. The actual number of tosses is a whole number. But you expect that the average over very many games would be close to 14.7, provided that the die is not "loaded".

Three to seven students sitting around one table can play one game. They should take turns tossing a die, even if it is not relevant for the outcome.

Each group has to keep score. They may compute the average of their games, but the average of all games played in the class should also be computed. (The average of all games can be different from the average of the averages from different tables.)

Another way to play the game, besides throwing a die, is to use the simulated program ONEDIE shown below. We first give calculator screen dumps showing the program, and then we give an example of a run. Finally, we give an explanation of the program code.

 

 

An example of running the program

 

 

 

Let’s see what the output means.

Roll 1 yielded a 2, and five other numbers (1, 3, 4, 5, and 6) were still needed.

Roll 2 yielded a 6, and four other numbers (1, 3, 4, and 5) were still needed.

Roll 3 yielded a 5, and three other numbers (1, 4, and 5) were still needed.

In roll 4, we got a second copy of 2, so the count of the number of twos was increased, and the three other numbers, 1, 4, and 5 were still needed.

...

Roll 17 yielded a 3, so we are done. It took 17 rolls to get the numbers 1-6.

In order from most to fewest, there were 5 twos and 5 sixes, 3 ones, 2 fives, 1 three, and 1 four that were rolled.

The “wait times”, or numbers of rolls between new numbers, were 5, 4, 5, 1, and 1. So, for example, the third roll was a five, and it took five more rolls to get the next new number, a 1.

 

Program explanation.

 

PROGRAM:ONEDIE

 

:6→dim(LD):Fill (0, LD)

List D holds the number of occurrences for each outcome.

:5→dim(LS):Fill (0,LS)

List S is the five-element list shown at the end. It holds

the no. of times you wait for the next zero to be filled.

:Repeat sum(LD=0)=0

Repeat the process below until the game ends.

:randInt(1,6)→I

A random integer between 1 and 6 is stored in variable I.

It simulates a throw of a die.

:LD(I)+1→LD(I)

The count of the Ith element in list D increases by 1.

:sum(LD=0)→J

The no. of elements equal to zero in list D are put into J.

: If J ≠0 :LS(J+1)→LS(J)

Only when J≠0, which indicates the game is not over,

is the corresponding element in list S increased by one.

:Disp LD :Disp {sum(LD), sum(LD=0)}

List D is displayed. Both the number of tosses so far,

and the count of how many new numbers are still

needed, are displayed.

:Pause:End

The game ends.

:SortD(LD); Disp " ",LD, LS

List D is sorted, a blank line is inserted, and lists

D and S are displayed.

 


Webpage Maintained by Owen Ramsey
Lesson Index