Monte Carlo Simulation of Western 649

Someone recently expressed disbelief to me that Western 649 is actually a smarter gamble than a typical 50/50 draw, so I decided to run a short Monte Carlo simulation to prove that my previous analysis based on probability theory was correct. I used Excel to generate 500,000 random numbers between 1 and 13,983,816 to simulate one Western 649 draw with 500,000 random combinations played. The criteria for determining which of my simulated numbers were winners were as follows:

A total of 3,000 draws (each with 500,000 random combinations played) were simulated. The tables below summarize the results of my simulation:

Number of winners in the simulated Western 649 draws

Summary of WCLC's profit and house edge, and Western 649 expected value

In the first table, we see that the expected values generally agree well with the averages from the simulation, apart from there being a few more jackpot winners than expected from probability theory. At most I had three winners sharing the $50,000 prize, which from my previous analysis I calculated had about 1 in 753 chance of occurrence in a draw with 500,000 random combinations. I also had two winners sharing the $1,000,000 prize in one of my simulated draws, which I calculated previously to have about 1 in 1,621 chance of occurrence. 

The second table shows that the worst individual draw for the WCLC resulted in a net loss of $920,090. But these losses were generally few and far between. The best draw for the WCLC produced $139,330 in profit. After 3,000 simulated draws, the WCLC earned a total of more than $251 million on the Western 649 game, averaging nearly $84,000 in profits per draw. Note that these numbers ignore operating costs to run the lottery: in real life, the WCLC has to pay for equipment and personnel to run the lottery, sell tickets, etc. However, it's still illustrative of how lucrative running this lottery is (WCLC runs two Western 649 draws every week). The simulation results showed an overall house edge of 33.50%. This corresponds to players losing, on average, $0.335 of every $1 spent playing Western 649. This is only about 3.4% off what the analysis based on probability theory indicated. The expected average profit was $86,665.97 per draw, which corresponds to 34.67% house edge and players losing $0.3467 of every $1 wagered.

Since this is a random number simulation with a finite number of trials, minor differences between the simulation results and the results from probability theory should be expected. The average would approach the expected value as the number of trials approaches infinity. I've plotted the overall house edge against the number of simulated draws in the graph below to show that my simulation results do tend to approach the expected value.

Overall house edge versus number of simulated draws

In summary, a simple Monte Carlo simulation of 3,000 Western 649 draws supports the results of my previous analysis based on probability theory. The expected value of Western 649 is better than the expected value of a typical 50/50 draw.

Guessing on Multiple Choice Tests

You've probably heard someone give advice along the lines of "when in doubt, guess 'C'" when taking multiple choice tests. The alleged reasoning behind it is that there's a statistical advantage in guessing 'C' consistently rather than randomizing your answers.

I decided to investigate this claim. Let's first check the case when the professor distributes the correct answers evenly and randomly, not favouring any one letter. Take for example a test with 30 questions, where each answer is independent of the others and there are four choices per question (A, B,C, or D). Assuming that you really have no clue which is the correct answer, any guess has a 25% probability of being correct. This is true regardless of whether you choose randomly (let's ignore for the moment that humans are notoriously bad at generating random values on their own) or if you decide to choose the same letter consistently. Here's why:

The instructor randomly chooses which letter is correct, without bias, so the probability of any letter happening to be the correct answer on a particular question is 25%. If you also guess randomly without favouring any letter (i.e. each letter is guessed, on average, 25% of the time), you should then expect that 25% of the correct answers are 'A', 25% of your guesses are 'A', thus 6.25% of your guesses (25% of 25%) are correct. The same is true of 'B', 'C', and 'D', so that overall you expect 4*6.25% = 25% of your guesses to be correct. Now if you consistently choose 'C' for the same random test, the probability that 'A' is correct is 25% but the probability that you guess 'A' is 0%. This is also true of 'B' and 'D'. When you choose 'C' 100% of the time, overall you should expect that 3*0.25*0 + 1*0.25*1 = 25% of your guesses are correct.

Since the guess is either right or wrong, this is a problem for the binomial distribution. The mean is n*p and the variance is sqrt[n*p*(1-p)], where n is the number of trials and p is the probability of the desired outcome. In our example multiple-choice test, whether you guess randomly or choose 'C' every time, you would expect, on average, to get 30*0.25 = 7.5 correct answers. To reinforce the point that both approaches to guessing are equal here, I simulated 50,000 of these tests in Excel and generated the following plot.

Probability distribution for correct guesses on tests with random answers to four-choice questions.

The average was 7.49 correct answers with a standard deviation of 2.37 when all guesses were random. When 'C' was guessed consistently, the average was 7.50 with a standard deviation of 2.38. The binomial distribution with n = 30 and p = 0.25 predicts an average of 7.50 and a standard deviation of 2.37. It's pretty clear that both guessing schemes give the same results when the correct answers are random and evenly distributed among the possible choices. 

Now that it's clear that there's no advantage if answers are randomly and evenly distributed, let's investigate the case where the instructor favours one letter over the others. It turns out that no matter how biased the instructor's distribution of correct answers might be, if you randomly guess each letter 25% of the time, your probability of choosing the right answer is still 25%. Let's assign unknown probabilities for each letter being assigned as correct by the instructor: pA, pB, pC, and pD. The sum of these unknown probabilities must be unity. So when the probability of guessing A, B, C, or D is 25% each, the probability of being correct is then 0.25pA+0.25pB+0.25pC+0.25pD = 0.25(pA+pB+pC+pD) = 0.25*1 = 25%. That all changes if you consistently choose 'C' though. In this case, your probability of getting the right answer is 0pA+0pB+1*pC+0pD = pC. I varied pC between the most unfavourable case (where 'C' is never the correct answer) to the most favourable (where 'C' is always the correct answer) to generate the following plot:

Probability distribution for correct answers when 'C' guessed consistently for different values of pC.

Obviously, there's some benefit to choosing 'C' if you know that the instructor favours 'C' over the other letters. But you also get screwed if the instructor doesn't like to use 'C' or simply chooses to favour another letter because he wants to penalize the people who always guess 'C'. While it looks pretty nice that more of the curves I plotted are shifted to the right rather than left of the curve for random guessing, if the choice of which letter gets favoured is random, the chance that 'C' is never the right answer is higher than the chance that it is always correct. To illustrate, I ran a simulation which I think is slightly more realistic than the examples above.

I assumed that pB + pC is 54% on average, which is just a guess on my part but I feel it is reasonable because answers aren't randomly assigned a letter. Numerical answers are often ordered from smallest to largest and statements like "All of the above" are reserved only for 'D' because they'd be confusing if they weren't. I used the normal distribution to generate the random variations in individual tests, so that pB + pC can vary from 0 to 1, but is usually close to the average. I similarly used normally distributed random numbers to split up pB and pC, so that on average pC is half of (pB + pC), but can vary from 0 to 100%. Same idea with pA and pD. Random guesses by the test taker are still equally distributed on average between the four choices. Plotted below are the results of 250,000 simulated tests.

Probability distribution for correct answers based on "realistic" multiple choice tests.

With random guessing we expected to be right 25% of the time on average and to see a binomial distribution after many tests. When 'C' is guessed consistently, things get more complicated, but a simple approximation can be found using the normal distribution. As you can see in the plot, the approximation works fairly well. It is clear that random guessing gives you more consistent results than guessing 'C' all the time. Always guessing 'C' increases your chances of getting 1 in 3 guesses right, but it also increases your chances of doing worse than 1 in 6, simply because your results are influenced by how the distribution of answers was biased by the instructor for the particular test. Overall, consistently guessing 'C' resulted in 2% more correct answers. But of the 250,000 simulated tests, random guessing beat consistently guessing 'C' on a total of 125,568 tests (i.e 50.23% of the time).

To summarize, if you happen to know that the creators of the test favour 'C' for correct answers, guessing 'C' consistently gives you an edge. In the long run, sticking with 'C' probably gives you a very slim advantage over random guessing, though guessing randomly gives you better consistency in the results of your guesses.