Lottery - Probability Discrepancy

I hope that you are familiar with binomial coefficients, because they are often used in these type of problems.

1. In the 6/49 lottery a player chooses a set of 6 numbers from the values 1 ... 49. In the draw 6 random numbers are obtained plus an extra bonus number. In the case "3 of 6", this means that there are exactly 3 numbers of the player's set among those first 6 of the draw. Whether the bonus number is part of the player's set or not is actually not important, because in either case it is still "3 of 6". The reason is that there is no special pay-out for "3 of 6 + bonus". In the case "2 of 6 + bonus", there are exactly two numbers of the player's set among the first 6 of the draw, and in addition the bonus number needs to be a different number in that set as well. Note that this resembles a bit the situation "3 of 7", but this is not exactly the same because of the particular requirement for the bonus as the 7th drawn number rather than simply 3 of any of the 7.

2. Yes, you might expect that. Here this just happens to be the same due to the particular combination of the set-up.

I'll show here the formula's that are required to compute the various probabilities. Note that 6/49 is for instance also discussed here. I will do this for the general situation, where

n = the range of available numbers

k = the size of the set selected by the player

l = how many numbers of the player's set are among the first k numbers of the draw are correct

For 6/49 this gives n=49,k=6, l=0,...,6. And like-wise for 7/50 n=50,k=7, l=0,...,7.

The probability for l correct in a draw of k numbers in the range 1 ... n is given by (ignoring the bonus number): $$ P(n,k,l) = frac{binom{k}{l}binom{n-k}{k-l}}{binom{n}{k}} $$ This is the ratio of the total number of possible winning combinations over the number of all possible combinations. There are $binom{k}{l}$ ways to choose the l correct ones among the set of the player. The other $k-l$ numbers need to be selected from the total set excluding those from the player, i.e., $binom{n-k}{k-l}$. Multiplying these two values with each other gives the total number of possible winning combinations for the player with exactly l correct. The total number of possible draws of k out of n is simply $binom{n}{k}$. Note that the order in which they are drawn is not relevant here.

Now we add the bonus number as the $(k+1)$th number of the draw. There are two possibilities, either the bonus number is in the set of the player or it is not. Suppose we calculate the probability $P_text{bonus}(n,k,l)$ that $l$ numbers are correct in the draw and that we also want the bonus number to be correct. (This of course implies that $0 leq l leq k-1$). After the initial draw only $(n-k)$ numbers remain available to be chosen as the bonus number, which include $(k-l)$ unselected numbers in the set of the player. Hence the probability to have a good bonus number is $frac{k-l}{n-k}$ and we find

$$ P_text{bonus}(n,k,l) = frac{binom{k}{l}binom{n-k}{k-l}}{binom{n}{k}} frac{k-l}{n-k} $$

Like-wise we can have the situation of l correct in the draw of k numbers, but with an incorrect bonus number, which of course than gives

$$ P_text{no bonus}(n,k,l) = frac{binom{k}{l}binom{n-k}{k-l}}{binom{n}{k}} frac{n+l-2k}{n-k} $$

Examples:

"3 of 6" in 6/49 : $P(49,6,3)=frac{8815}{499422}approx0.0176504$ and 1:56.6559

"2 of 6 + bonus" in 6/49 : $P_text{bonus}(49,6,2)=frac{1025}{83237}approx0.0123142$ and 1:81.2068

"3 of 7 + bonus" in 7/50 : $P_text{bonus}(50,7,3)=frac{287}{71346}approx0.00402265$ and 1:248.592

But:

"4 of 7" in 7/50 : $P(50,7,4) - P_text{bonus}(50,7,4) = P_text{no bonus}(50,7,4) =frac{287}{71346}approx0.00402265$ and 1:248.592

The reason is that "4 of 7" can be with or without a correct bonus number. With a correct bonus number, however, there is already a special pay-out and hence it would not result anymore in this lesser pay-out. In the case "3 of 6" n 6/49 this is not the case, since there is no special pay-out for "3 of 6 + bonus".

Also note that you get here 1:248.592 in stead of 1:82.8641. This is because in this particular lottery the player has three sets on a single ticket, hence the chance of winning per ticket is three times as high and the odds are a factor three smaller.

As far as the pay-outs are concerned, this is mainly psychology. For the lottery is doesn't matter how they distribute the winnings over the different combinations, as long as they make a profit, which is warranted by the fact that only a fixed percentage of the ticket sales is payed-out. They simply want to sell the maximum number of ticket. Bottom line, most people will loose more money than they will ever win and the lottery is the true winner.

The key point is that the bonus drawing must be the last ball, while the non-bonus balls can be any of the first six balls that are drawn.

To illustrate why order matters, suppose you are watching the Lotto 6/49 drawing. You have a ticket that has the numbers 01, 02, 03, 04, 05, 06, and you're assigned a bonus number 07. The first two draws are 01 and 02.

At this point, are you more likely to match one of the balls 03, 04, 05, or 06 on one of the next four draws? Or are you more likely to match 07 on one of the next four draws?

(This isn't a perfect analogy, since it deals with the conditional probability that you win if your first two draws are correct. But it does illustrate why requiring the bonus ball to be a particular drawing makes a difference.)

I can't really explain why the payoffs are set as they are, though. That may not be a mathematics question so much as a economics and psychology question (i.e., maybe it's a method of encouraging people to play more by giving them $10 payoffs more often.)