Big Board Plinko Probabilities
based on Paths and Tree Diagrams

To investigate the conditional probabilities for a chip dropped from slot E in the big Plinko board, we can use an argument based on paths and tree diagrams.  

To obtain the number of paths to each of the slots using Pascal's Triangle, we display Pascal's Triangle superimposed on the big Plinko game board.  To account for the loss of paths when a chip comes into contact with the side of the board, we must subtract.  The subtractions we must make are shown in RED on the board.  The final path counts for the affected cells are shown in blue.

Plinko Board
        A   B   C   D

 

E

 

F

 

G

 

H

 

I

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

2

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

3

 

3

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

4

 

6

 

4

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

5

 

10

 

10

 

5

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

6

 

15

 

20

 

15

 

6

 

1

 

 

 

 

 

 

 

 

 

 

 

1

 

7

 

21

 

35

 

35

 

21

 

7

 

1

 

 

 

 

 

 

 

 

 

1

 

8

 

28

 

56

 

70

 

56

 

28

 

8

 

1

 

 

 

 

 

 

 

1

 

9

 

36

 

84

 

126

 

126

 

84

 

36

 

9

 

1

 

 

 

 

 

1

 

10
-1
9

 

45

 

120

 

210

 

252

 

210

 

120

 

45

 

10
-1
9

 

1

 

 

 

1

 

11

 

55
-1
54

 

165

 

330

 

462

 

462

 

330

 

165

 

55
-1
54

 

11

 

1

 

1

 

12

 

66
-12
54

 

220
-1
219

 

495

 

792

 

924

 

792

 

495

 

220
-1
219

 

66
-12
54

 

12

 

1

 

   

 

A'

 

B'

 

C'

 

D'

 

E'

 

F'

 

G'

 

H'

 

I'

 

 

 

 

Thus, there are 54 paths from E to A', 219 paths from E to B', etc.

Our board has 13 rows.  If a chip DOES NOT encounter the side of the game board, the probability that it will fall into a particular slot at the bottom is 

.

To understand why this is true, consider a 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 12 branches along each path, the probability of that path is the product of the probabilities along each branch, giving 1/212.

If a chip encounters the side one time, the probability that it will fall into a particular slot is

 .

To understand this, think of a tree diagram and the probabilities along each branch.  If a chip encounters the side one time, there will be a branch along which the chip was forced to make a move to the right or left.  Thus, along that path, there are 11 branches with probability 1/2 and 1 branch with probability 1.  The probability of that path is the product of the probabilities along each branch, giving 1/211 .

If a chip enounters the side twice, the probability with which it will fall into a particular slot is

.

Again, we think of a tree diagram and the probabilities along each branch.  If a chip encounters the side twice, there are two branches along which the chip was forced to make a move to the right or left.  Thus, along that path, there are 10 branches with probability 1/2 and 2 branches with probability 1.  The probability of that path is the product of the probabilities along each branch, giving 1/210. 

In general for our 13-row board, if a chip encounters the side K times along a particular path,  the probability for that path is 1/212-K

Inspecting our 13-row board, we see that maximum number of times a chip falling from E will encounter the side is 2.  Specifically, for a chip falling from slot E, the maximum number of times the chip will encounter the side is 2.  To compute the probability that a chip will fall into slot A', let 

b = number of paths that encounter the side exactly twice,
c = number of paths that encounter the side exactly once, and
d = number of paths that do not encounter the side.

Then the probability that a chip will fall from E into a slot at the bottom is

where  b + c + d = number of paths from E to that slot.   The problem now is to determine how many paths from E to one of the slots at the bottom encounters the side exactly twice and how many paths encounter the side exactly once.

We first consider paths from E to A'.  Examining the board's left half (Figure 1), we see that there is 1 path that encounters the side exactly twice. 

Figure 1.

9 paths encounter the side exactly once (Figure 2).  

Figure 2.

Thus the number of paths that do not encounter the side is 54 - (9+1) = 44.  Note that

is the probability that a chip will fall into slot A' given that it starts in E.  

Now we consider a chip that starts at E and lands in B'.  There is 1 path that encounters the side exactly twice  (Figure 3).

Figure 3.

There are two paths from E to B' that encounter the side exactly once (Figure 4).

Figure 4.

So there are 219 - (10 + 1) = 208 paths that do not encounter the side.  Thus, the probability that a chip will fall into B' given that it starts in E is

.

There are no paths from E to C' that encounter the side exactly twice.  In fact, there is only one path to C' that encounters the side at all (Figure 5).  

Figure 5.

Thus the probability that a chip will fall into C' given that it starts in E is

.

The probability that a chip starting in E will land in G', H', and I' are obtained by considering the symmetry of the board.

Chips starting in E that land in D', E', or F' will not encounter the side of the board, so the conditional probabilities are computed as the number of paths to D', E', or F' multiplied by 1/212 = 1/4096.  So

P(D' | E) = 792/4096
P(E' | E) = 924/4096 and
P(F' | E) = 792/4096.