# Assigning Probabilities to Lottery items

So I am trying to set up a imaginary lottery with multiple items consisting in it. For realistic purposes house should have a edge over the player. Hence to assign probabilities I am trying to solve these equations:

p is the price of item x is the probability for the item.

$p_1 x_1 + p_2 x_2 + p_3 x_3 + ... + p_n x_n = Price Of Lottery(1-HouseEdge)$

and since x's are probabilities they should sum up to 1:

$x_1 + x_2 + ... + x_n = 1$

With these conditions my assumption is that there are infinitely many solutions. How can I find solutions for these? Also should I put more conditions like probabilities of items are inversely propotional to price so lottery outcomes are more stable? Or this system is stable enough?

In probability exercises, one is so used to seeing $p$ represent a probability that it is a little confusing to see $p$ used for something else, but I'll continue using your notation.

Since the $x_i$ are probabilities you also have the conditions $0 \leq x_i \leq 1$ for each $i.$

There are a number of other conditions that lotteries usually seem to obey, but they seem to be matters of style rather than mathematically required. For example, the player is usually not guaranteed to win anything of value, which you can enforce by setting $p_n = 0$ and $x_n > 1.$ But note that if $p_{n-1} = p_n = 0$ and $x_{n-1} + x_n > 0,$ you can make infinitely many variations of $(x_{n-1}, x_n)$ without having any effect at all on how the lottery is conducted, so I'll assume from this point forward that you have already decided how many different prizes there are and have set $n$ to one greater than that number.

Conventionally the "first" prize is more valuable than the "second" prize, which is more valuable than the "third" prize, and so forth, which gives us $$p_1 > p_2 > p_3 > \cdots > p_n = 0.$$ Note that this implies there are no prizes whose value is negative. These assumptions do not apply in a Shirley Jackson story.

It also makes sense that it should be possible to win the first prize, that is $x_1 > 0.$

The more valuable prizes usually are given out less frequently, so if there are exactly $k$ prizes of non-zero value that the player can win, $$x_1 < x_2 < x_3 < \ldots < x_{n-1}.$$ You may or may not want to require that $x_{n-1} < x_n$ (the most likely outcome is not winning anything) or $x_n > \frac12$ (the player usually does not win anything).

All these conditions still leave infinitely many solutions. You may want to impose additional conditions such as setting bounds on the ratios between the prize amounts, but there are many variations of that type in real life, so which additional constraints you place is a matter of psychology, personal preference, or other considerations.

• Thank you for a rigorous answer. However it still does not solve my problem. My main objective is to find an algorithm that gives x vector given p vector and the houseedge also the price of lottery. As you said, there are infinitely many solutions for that. For my purposes I want the solution to be making player feel the game is fair or natural. As you again said this is psychology preference. After reading your answer main solution that pops to me considering maybe normal distribution for probabilities, I don't know how can I do it or even it is possible. Jul 27 '17 at 11:49
• Also because this is an algorithm I need general solutions that I can apply to a programming language Jul 27 '17 at 11:50
• I don't see any lottery in real life that has anything looking like a normal distribution of probabilities. This leads me to think that a normal distribution would not be understood by the players as fair and natural. The point of my last paragraph is that the part of your question that I did not answer is (I think) outside the scope of this stackexchange site. Jul 27 '17 at 18:12
• By the way, in real life there are several government-run lotteries with "jackpots," meaning that there is one prize (the jackpot) whose value changes from one drawing of the lottery to the next--each time it is not won, some of the funds from ticket sales are added to the jackpot. Because of this, the house edge (in your formula) varies from one drawing to the next. Jul 27 '17 at 18:17