Suppose we have a random process where an outcome is observed and two things are measured. For example, suppose we toss a fair coin three times and we observe the sequence


where H is a head and T a tail. Suppose we record

In the three tosses above, we observe 2 heads and 3 runs in the sequence.

We are interested in talking about probabilities involving both measurements "number of heads" and "number of runs". These are described as joint probabilities , since they reflect the outcomes of two variables.

To construct this type of probability distribution, we first describe the collection of possible outcomes for the two variables. The number of heads in three tosses could be 0, 1, 2, or 3, and the number of runs in a sequence could be 1, 2, or 3. We represent these outcomes by the following two-way table:

0 1 2 3

Next we have to place probabilities in the above table. If we toss a coin three times, there are 8 possible outcomes. Since the coin is fair, each of the outcomes has the same probability. In the table below, we list the eight outcomes, the number of heads and the number of runs in the outcome and the probability of the outcome.

H H H 3 1 1/8
H H T 2 2 1/8
H T H 2 3 1/8
H T T 1 2 1/8
T H H 2 2 1/8
T H T 1 3 1/8
T T H 1 2 1/8
T T T 0 1 1/8

Now we are ready to fill in the probability table. Start with the box in the first row and first column. What is the probability that the number of runs is equal to 1 and the number of heads is equal to 0? Looking at the outcome table, we see that this happens once (for outcome TTT). So the probability of 1 run and 0 heads is equal to the probability of TTT, which is 1/8. What's the probability of 1 run and 1 head. We see from the outcome table that this never happens, so the probability in this box is 0. Next look at the box in the second row and second column. To find the probability of 2 runs and 1 head, we see that this happens twice in the outcome table (HTT and TTH). So the probability in this box is 2/8. If we continue this for all boxes, we get the following probability table.

0 1 2 3
1 1/8 0 0 1/8
NUMBER OF RUNS 2 0 2/8 2/8 0
3 0 1/8 1/8 0

For the following questions, it might be helpful to convert this probability table to a count table. Suppose that we tossed three coins 800 times. Then we would expect to get 1 run and 0 head 1/8th of the experiments, or 100 times, we expect to get 2 runs and 1 head 2/8th of the time, or 200 experiments, and so on. By converting probabilities to counts, we get the following count table. This represents what we think would happen if we did this coin experiment many times. Note that I have added an extra row and extra column to the table. The TOTAL column gives the number of counts in each row and the TOTAL row gives the number of counts in each column.

0 1 2 3 TOTAL
1 100 0 0 100 200
NUMBER OF RUNS 2 0 200 200 0 400
3 0 100 100 0 200
TOTAL 100 300 300 100 800

Using this count table, we'll ask some questions which explore the connection between the number of heads in this experiment and the number of runs.

  1. If you toss three coins, how many runs are likely to occur? Look at the TOTAL column of the table. This tells us that, out of 800 experiments, we expect to get 200 "one run", 400 "two runs", and 200 "three runs". The most likely possibility is "two runs" which has probability 400/800 = .5.

  2. If we are told that we get two runs in our coin tossing, what does that tell you about the number of heads? Look only at the "two runs" row of the table. Two runs happened 400 times in our hypothetical experiments. Of these 400 counts, only 1 and 2 heads occurred with frequencies 200 and 200. So if we are given that two runs occur, then we know that the only two possibilities are 1 and 2 heads with respective probabilities 200/400 and 200/400

  3. What if we are told that the experiment resulted in exactly 2 heads. Have you learned anything about the number of runs in the sequence? Focus now on the "2 heads" column of the table. The number of 1 runs, 2 runs, and 3 runs in this column are 0, 200, and 100. So if we know that there were 2 heads, then the probabilities of 1 runs, 2 runs, and 3 runs are 0/300, 200/300, and 100/300. So it is most likely in this case that 2 runs will occur and 1 run in the sequence is impossible.


Page Author: Jim Albert (© 1996)
Document: http://www-math.bgsu.edu/~albert/m115/probability/two_way_table.html
Last Modified: November 24, 1996