Theoretical Probabilities:  A Simpler Problem

Theoretical Probabilities:  A Simpler Problem

The discussion after the initial Plinko board activity leads to an observation that the experimental probabilities depend upon the number of repetitions in an experiment; the experimental probabilities might not generalize to other experiments in which fewer or more repetitions are completed.  We determine theoretical probabilities to overcome this disadvantage.  Generally, when we speak of probabilities, we are referring to theoretical probabilities.

When all outcomes in an experiment are equally likely, the probability that an event will happen, P(event), is computed by

.

In the Plinko game, however, not all events are equally likely because when the chip hits the side of the board, it is forced to fall in a particular direction.  Thus, we need a different strategy to calculate the probabilities.

The strategy we use is a probability tree diagram.  To see how to construct the tree diagram, we consider a smaller and simpler game board.  Suppose the board is as shown in Figure 2, having only three starting and ending slots.

 Figure 2.  Plinko board with 3 starting and ending slots.

If a chip is dropped from slot B, it can fall left or right.  Thus, the probability it will fall to the left is 1/2 and the probability that it will fall to the right is 1/2.  We can illustrate this using a tree diagram that we construct as in Figure 3.  We label each branch of the tree with the probability that the chip will take a path along that branch.  This is stage 1 of the tree.

 Figure 3.  Stage 1: Tree diagram illustrating the possible paths and probabilities from slot B to 2nd row of slots.

Continuing in the same manner, we see that there are two choices for each of the two possibilities in Step 1, each occurring with probability 1/2.  This is illustrated in Figure 4.

 Figure 4.  Stage 2: Tree diagram illustrating the possible paths and probabilities from slot B to 3nd row of slots.

At this point in the chip's path, there is a possibility that the chip will hit the wall of the game board if it takes one of the paths shown in red in Figure 5.  If the chip encounters the wall of the game board, it will, with probability 1, fall to the right (if it hits the left side) or fall to the left (if it hits the right side).

 Figure 5.  Paths which bring the chip in contact with the wall are shown in red.

Continuing in the manner illustrated in these examples, we can construct the entire tree diagram for the small Plinko board.  The probability that a chip falling from slot B will land in a particular slot along a particular path is computed by multiplying the probabilities along the path.  This is the multiplicative property of probability tree diagrams.

 Figure 6.  Tree diagram for small Plinko board.

The theoretical probability (this is a conditional probability) of a chip landing in one of the slots, given that it starts in B is calculated by adding the probabilities for each path from B to the slot.  This is the additive property of probability tree diagrams.  We denote the conditional probability that the chip will land in slot A' given that it starts in B by P(A' | B).  The probability of the chip falling into slot A' given that it started in B is computed by adding the probabilities associated with each path from B to A',

P(A' | B) = 1/8 + 1/16 + 1/16 = 1/4.

Similarly we compute the conditional probabilities that a chip will land in slot B' or C' given that it starts in B:

P(B' | B) = 1/8 + 1/16 + 1/16 + 1/16 + 1/16 + 1/8 = 1/2, and
P(C' | B) = 1/8 + 1/16 + 1/16 = 1/4.

For the large Plinko board, the problem of finding the probabilities that a chip will fall into a particular slot may seem insurmountable because the tree diagram would be so large.  However, there is an approach we can take that directly relates to the number of paths to a particular slot and Pascal's Triangle. Named for French mathematician Blaise Pascal in 1654,  Pascal's Triangle is a triangular arrangement of numbers corresponding to the probabilities involved in flipping coins; it is often used to investigate probability patterns associated with equally likely events. Students may have seen Pascal's Triangle in connection to the coefficients in the expansions of powers of binomial expressions.  If students have been introduced to Pascal's Triangle, now is a good time to review the method with which it is constructed.  If students have not been introduced to Pascal's Triangle, this is a good time to do so.

It is relatively easy to count the paths from A, B, or C to the three ending slots in the small Plinko board.  Suppose a chip starts in slot B. When it hits the first peg, it can fall to the right or left.  If it falls to the left, at the next peg it can fall to the right or left.  If the chip falls to the left, it encounters the wall so it is forced to fall to the right.  At the last peg, it can fall to the right or left.  If it falls to the left, it will land in slot A' so this constitutes a path to A' from B.   Figure 7 shows that there are 3 paths from B to A'.

 Figure 7.  Counting the paths from B to A'

In a similar way we can determine that there are 6 paths from B to B', and 3 paths from B to C'. Basically, our counting process is by "brute force," and if our board were larger it would be hard to count the paths correctly unless we had a more efficient way to count.  Fortunately there is a more efficient approach and it is closely associated with Pascal's triangle.  To see how Pascal's Triangle comes into the picture, look at the animation in Figure 8.

 Figure 8. Paths from B to A', B', and C'.

Starting at B (we label the starting slot with 1), the chip can fall to the left or right, so there is 1 path from B to each of the two cells in the second row.  Label those cells with 1.  If the chip falls to the left, there are two paths to the next row.  Similarly, if the chip falls to the right, there are two paths to the next row.  So in the third row, the path count (from left to right) is 1 2 1.  Continuing in the same manner, the path count in the fourth row is 3  3 and finally in the last row, the path count is 3 6 3. As we count the paths from B, we can ALMOST see the emergence of Pascal's Triangle and the numbers in successive rows can be generated in much the same way as we construct Pascal's Triangle.

To see a more efficient way to count the paths, superimpose Pascal's Triangle on the game board (illustrated below). The shaded region corresponds to cells in the board.  The red numbers are the numbers in Pascal's Triangle that agree with the numbers in our path count.

 A B C 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 A' B' C'

Looking at the 4th row of Pascal's Triangle, we see that the 1's at each end are outside the board.  Thus, we can assume that the chip that would have come from those cells will not get to the 5th row, so to obtain the correct path count we subtract 1 from the 4's in the 5th row of Pascal's Triangle.

 A B C 1 1 1 1 2 1 1 3 3 1 1 4 -1 3 6 4 -1 3 1 A' B' C'

When we do the subtractions, the path count agrees with the table obtained in Figure 8.

 A B C 1 1 1 1 2 1 3 3 3 6 3 A' B' C'

 A Word of Caution:  At this point it appears that since there are 12 possible paths, the probabilities can be computed as P(A' | B) = 3/12 = 1/4, P(B' | B) = 6/12= 1/2, and P(C' | B) = 3/12 = 1/4, but although we obtain the same numerical values for this example when using this approach, it is a coincidence.

We can also use Pascal's Triangle to help us compute the theoretical probabilities.  The argument is based upon how probabilities are computed using a tree diagram. The end result of the argument is a simple procedure we can use to compute the probabilities.

Consider the tree diagram and the probabilities along the branches.  If a chip does not encounter the side, the probabilities along each branch will be 1/2.  Thus, since there are 4 branches along each path, the probability of that path is the product of the probabilities along each branch, giving 1/24 = 16.  This is the denominator of our probability fractions.

We can compute the numerators using Pascal's Triangle.  Consider the 5th row of Pascal's Triangle and reflect the two outer 1's to the middle as shown below.

The sum of the last row entries gives the denominator of the probability fraction; the entries give the numerators so that

P(A' | B) = 4/16 = 1/4,
P(B' | B) = 8/16 = 1/2, and
P(C' | B) = 4/16 = 1/4.

This technique also works for a chip that is dropped from columns A or C.  Click here to see the details of the computations using both a tree diagram and the approach using Pascal's Triangle.

The table below summarizes the conditional probabilities.

 Column End Begin A' B' C' A 3/8 1/2 1/8 B 1/4 1/2 1/4 C 1/8 1/2 3/8

 Random Starting Slot:  The discussion above dealt with conditional probabilities, for example, the probability of the chip landing in slot B' given that it started in B.  Suppose the player starts from slot that is randomly selected.  How can we compute the probability of the chip falling in a particular ending slot? That question can be answered with the help of the multiplicative and additive properties of tree diagrams.  In this case the tree is somewhat different because when the play begins, we have three choices that can be made (starting slots A, B, or C), each with probability 1/3.  So the tree diagram begins with three branches, one to each of A, B, or C.  The rest of the tree is filled out with the tree diagrams we discussed earlier. The probabilities can now be calculated using the additive property P(A') = 1/3 (P(A' | A) + P(A' | B) + P(A' | C)) = 1/3 (3/8 + 1/4 + 1/8) = 1/4 P(B') = 1/3 (P(B' | A) + P(B' | B) + P(B' | C)) = 1/3 (1/2 + 1/2 + 1/2) = 1/2 P(C') = 1/3 (P(C' | A) + P(C' | B) + P(C' | C)) = 1/3 (1/8 + 1/4 + 3/8) = 1/4.
 Activity:  It is interesting to compare theoretical probabilities to the corresponding experimental probabilities for the small Plinko board.  TI-83 and TI-89 calculator programs and a Javascript routine are available to facilitate simulations for the experimental conditional probabilities as well as the probabilities associated with a random starting slot.  Click on the appropriate buttons below to open the simulation windows.  Instructions are available as the simulation page opens.  Each simulation runs 100 trials.   If you run the simulation in Internet Explorer, only the last path will be shown. If you run the simulation in Netscape, each path will be shown (very quickly--depending on how quickly your screen refreshes, only part of the path may be displayed).   You can see and download a sample activity data collection sheet here.  (Word format) TI-83 calculator program can be viewed and downloaded from here. TI-89 calculator program can be viewed and downloaded from here. Javascript routines: User selects starting slot:      Starting slot selected randomly:    NOTE:  The Javascript routines have been tested in Internet Explorer 5 and Netscape Navigator 4.7.  The browser must be Javascript enabled.  We recommended that your display resolution is  800 x 600.