What is the experimental probability of rolling two dice 100 times and getting a sum divisible by three?

Answer 1

The successfule outcomes that fit your problem are a 3,6,9 and 12. There are 36 combinations, of which 3 can be the outcome of 2 events (1,2) and (2,1), 6 can be the outcome of 5 events, (1,5), (2,4), (3,3), (4,2), (5,1), 9 can be the outcome of 4 events (3,6),(4,5),(5,4), (6,3) and 12 is the outcome of 1 event (6,6). So out of the 36 combinations, we have 2+5+4+1 or 12 events, so 12/36 = 0.33. Now, if you throw two dice 100 times, and on the average this experiment should have 33.33 successes. Of course, some times you might have 41 successes, some times 35, but the long term average of 100 throws is 33.33 successes. This I would call the expected number of occurrences, not experimental probability. -- A bit extra to my answer: In experiments that involve chance, the results are never known. I might throw the dice 100 times and calculate a 40% of the time have success or a 25% of the time have success. These estimates are called proportions and are, I think, your "experimental probabilities" or sample estimates of success probability of your population. Of course, as given we know the success probability of the population (0.33). The binomial distribution can provide the probability of tossing two dice, "n" times and obtaining "x" successes, where the probability is 0.33. For example, I can state that 80% of the time, the number of successes will be between 27 and 39, when the dice are thrown 100 times using the binomial distribution. This is calculated by calculating the probability of 39 or fewer successes occur minus the probability of 27 or fewer sucesses occur. I have to use the cumulative distribution function (CDF). In Excel, I calculated: +binom(33-a1,100,0.33,TRUE)-binom(33-a1,100,0.33,TRUE) and varied a1 (whole numbers). When I tried a1=6, I obtained the 80%. At a1=8 (25 to 41 successes) I have the 91% confidence interval. The TRUE parameter means that I am using the CDF of the binomial distribution.