A simple application of elementary probability shows that lottery odds as reported by the operators are incorrect.

Note: I have delayed posting this for a while since the results I come up with seem incorrect. Hopefully someone can respond and tell me where the problem lies.

Solved?

If I follow the logic expressed at What are the odds of hitting the same number twice on a roulette table? and apply it to the lottery situation then the lottery is computed correctly by the lottery operators. The critical aspect is that the first selection’s probability doesn’t matter to the person taking the risk. The second part of the cited page gives the odds for a specific number being picked twice. I think that case applies to an external observer to a particular lottery game instance. Huh?

About a year ago while speaking with my brother Robert on the phone he casually mentioned a joke he made. He said lotteries are so funny since the winning number has to be picked twice in order to win. First by the lottery company and then by the player. We both laughed.

But, then later I did a double take, huh? That is true, you have to pick the number set, and then a few days later, the lottery company will also pick a set. If the sets match then you win. Ok, that makes sense. But, if seen this way, the probability value they give for winning couldn’t be correct. Could it? They only give the odds of picking any set, not the winning set. It has to be much harder to win, thus, the probability much lower.

**How to compute the Probability?**

In probability theory there are rules for combination of events. If the events in an “*experiment*” are independent, you just multiply the probability values of each: P(A and B) = P(A intersection B) = P(A)P(B).

Further details are on this High School wiki page:

Multiplying probabilities

Probabilities are multiplied together whenever an event occurs in multiple “stages” or “steps.” For example, consider rolling a single die twice; the probability of rolling a 6 both times is calculated by multiplying the probabilities for the individual steps involved. Intuitively, the first step is simply the first roll, and the second step is the second roll. Therefore, the final probability for rolling a 6 twice is as follows:

P(rolling a 6 twice) = P(rolling a 6 the first time) X P(rolling a 6 the second time) = 1/6 X 1/6 = 1/36 approx 2.8%

Similarly, note that the multiplication of probabilities is often associated with the use of the word “and” — whenever we say that some event E is equivalent to all of the events X, Y, and Z occurring, we use multiplication to combine their probabilities (if they are independent).

More info on this wikipedia entry: Probability, Mathematical treatment

Does this apply to the lotteries, like Powerball? There are two events, though separated by days. The consumer, player, picks a set, then later the operator picks there own set. And, they are independent, neither event is dependent on the other. So, the problem, to me, is interpreting the “experiment”. I contend that the whole game, which takes place over a few days is one thing, an experiment, and so the multiplication rule applies.

**Rolling a 195,249,054 sided Spherical polyhedron**

Another way of relating them is to use two die rolls. But now instead of a die with six faces we use N faces, where N is the total number of possible number sets we could pick in a lottery game. This “die” is really a form of Spherical polyhedron. Lets say N is 195,249,054 possible unique numbers, which correspond to each possible set. So when we roll two dice the total probability would be (1/195,249,054 X 1/195,249,054). Remember, these are “normal” die, just having a ginormous number of faces.

The above is not even mentioned in the Lottery math references, for example, this Wikipedia entry, Lottery mathematics. So some conceptual misunderstanding on my part is very likely.

**Example**

Lets take an actual example, the Powerball lottery states on their “Powerball – Prizes and Odds” page that to win the Grand Prize the odds are: 1 in 195,249,054. This is derived by application of math stuff to determine the combinations of the five white balls (1-59) and a red ball (1-39).

If we apply the multiplication rule the actual probability of winning is:

1 in 38,122,193,087,894,916

That’s 1 in 38 quadrillion. Big difference!

In scientific notation: 3.8122193087894916 x 10^{16}

What is the Expected Value now?

**Error?**

This analysis couldn’t be correct. First, the number is too large, there are too many winners. Second, I have never heard of anything like this. Surely if this were the case it would be news. So where is the mistake?

I think it has something to do with the “same set of numbers”. Then its not just a simple multiplication of probability? If I find out, I’ll update this post.

**So what?**

You should not be paying the “idiot tax”. True, but when the prize reaches 100 million I bet there are some math professors out there buying a ticket too. Further, it is an interesting math subject.

**Updates**

This article analyzes the occurrence of a lottery draw that duplicated the same numbers and argues that my kind of instinctive analysis above is incorrect. Adventures in Probability. So, perhaps, the way to look at this issue is to compute the probability of the same winning combination being picked twice in a row? What the article says that it is 3.8 X10^16 but this has to be multiplied by the amount of combinations, so:

(1/((3.8 x 10^16) * 195,249,054))*195249054 = 1/195249054. That same as what the lottery provider quotes!

I don’t get it yet. Then why is the two dice example where each die has ‘N’ faces not calculated in the same way?

**Further Reading**

**Off Topic**

**Appendix**

Groovy program to print the product:

x = new Long('195249054');
printf('%,d',(x * x));

This work is licensed under a

Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.

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