# How can I linearly decrease chance?

Suppose there are 10 different people, each holding 1 lottery ticket, and I give each of them 10% chance of becoming the 'winner'.

Now let's say one of those 10 people (I dont know which one) decided to buy a second ticket.

I could choose at that point to lower every ticket's chances to 5% (effectively increasing the chance that nobody will win). Since that person now holds two tickets, which both give 5% chance, whoever it was didn't increase their chance of winning, since he still has 10% total chance, like before. Now suppose another ticket gets sold, I can lower every ticket's chances to 3.33%, and still the person who may hold 3 tickets, gained absolutely nothing from doing so.

The problem with this solution (although it works) is that very soon I have to lower the chances to such a tiny percentage, that its almost garantueed that nobody will win.

Is there a simple scheme possible to prevent this, or is this unavoidable?

• Out of curiosity -- are you working on a defense against a "sybil" attack in computer science? Jan 13 '14 at 1:57
• @WillNelson Yes, can't believe you guessed that :)
– Muis
Jan 13 '14 at 8:00

What is it that you are trying to do? Why not using simple probabilities?

At first everyone has 1/10. If someone buys a 2nd ticket, his probability is 2/11. Everyone else has probability of 1/11.

Do you need, for some reason, to decrease the cumulative probability so that noone wins?

• Im trying to discourage people from buying a second ticket. And i cannot forbid them, because they are anonymous. So there may be multiple ways to discourage them. By lowering the chance, the price pot, the ticket price, etc. I just dont know how to combine that.
– Muis
Jan 9 '14 at 13:17
• Please answer the following questions, so that i can create a function? 1) How many tickets do you want to be sold? i can understand that up to 10 it is ok. and then you want to discourage them. How many tickets are "too many"? 20?50?100? 2)I bought the first ticket. I have probability 10%. supposing "too many" tickets are bought (see quest.1), what should be my probability (i didn't buy any more tickets)? Is 5% ok? Jan 10 '14 at 9:05
• I don't want to limit the number of tickets sold, I just want to make sure that every individual will only get 1 ticket, and that their chance doesnt increase by getting a second ticket. So there is not a problem when a million tickets get sold, thats fine as long they were not all bought by a small group of persons.
– Muis
Jan 10 '14 at 9:23
• but, if tickets are bought anonymously, you can never tell if someone buys a second ticket Jan 10 '14 at 9:27
• Maybe there is a solution where you don't have to tell.. Because you could give people an incentive for buying just one ticket. If they decrease the price-pot by doing so, or increase the ticket-costs, there must be a point where they stop buying.
– Muis
Jan 10 '14 at 20:43

Consider offering two large prizes, say £1,000,000 and £500,000. Do the drawing for the first one first. Here's the catch: noone is allowed to win both prizes. If you won both, you get nothing. If you one the larger prize and noone claims the second, you get nothing. Psychologically, this would encourage people to get one ticket.

This works best in small groups and terribly in large groups.

• This is exactly in the right direction, but I need a formula that would work for larger groups too.
– Muis
Jan 12 '14 at 1:05
• I'll get working on it! Jan 12 '14 at 1:10
• "If you won both, you get nothing. If you one the larger prize and none claims the second, you get nothing." The second condition is a) extremely unfair and b) not too helpful because most people have friends and relatives (not mentioning the option of anonymous donation, etc.). Jan 13 '14 at 3:37
• @Muis may i ask how Brian's solution is applicable in computer science? also you may want to explain the rationale behind the question further, because I admit I was a bit like 'what...' when I first saw it. Jan 13 '14 at 10:38

After the OP's clarifications, I will answer the question for the following specific set up: There are $n$ persons and $n-1+k$ tickets, $k=1,...$ (up to some finite number). Initially $k=1$, and all persons buy one ticket. The price of each ticket is fixed at $v$, and it doesn't change. The winning ticket is decided by a "random draw", namely, all tickets always have the same probability of winning. All tickets take part in the draw, irrespective of whether they have been bought or not. Therefore the probability of a ticket winning is $p_k= \frac 1{n-1+k}$. The winner collects an amount of $W$ which can be variable.
Question: how can we decide on the value of $W_k$, as a function of the number of tickets, so that there is no incentive for somebody to buy an additional ticket, if one is issued?

For $k=1$ the expected profit of each one of the participants (we are ignoring issues of risk aversion and so we deal in terms of expected profits and not of expected utility) is the expected value of collecting the prize, minus the sure cost of buying one ticket: $$EP_{11} = p_1\cdot W_1-1\cdot v \qquad $$

where the double index on $EP$ refers the first to the value of $k$ and the second to the number of tickets somebody holds.

Assume now that $k=2$. If somebody buys this ticket, his expected profit will now be $$EP_{22} = 2p_2\cdot W_2-2\cdot v \qquad $$ while the expected profit of all others will be $$EP_{21} = p_2\cdot W_2-1\cdot v \qquad $$

We want the prospective buyer of the 2nd ticket to have the same expected profit by buying the extra ticket as the expected profit he had when he had bought the first ticket, so we want to determine $W_2$ by equalizing equations $$ and $$:

$$EP_{11} = EP_{22}\Rightarrow p_1\cdot W_1-1\cdot v = 2p_2\cdot W_2-2\cdot v \Rightarrow W_2 = \frac {p_1W_1+v}{2p_2} \qquad $$

So assume that we issue this extra ticket and we announce that now the prize is $W_2$ as determined by $$. The buyers will do the calculation and they will find that they won't gain anything in terms of expected profit if they buy the extra ticket... does this mean that they won't buy it? No, it does not imply that. Because, the buyers have already invested in the lottery by buying the first ticket. So there is no point in comparing the expected profit if they buy the 2nd ticket with the expected profit they had before the extra ticket was issued, because it is no more relevant. What they will do is to compare the profit the will now have, in this new set of rules, if they buy the extra ticket and if they don't. Namely, being rational, they will compare $EP_{22}$ with $EP_{21}$. So if we as organizers calculate $W_2$ as above, then we have that

$$EP_{21} = p_2\cdot W_2-1\cdot v = p_2\frac {p_1W_1+v}{2p_2} - v = \frac {p_1W_1-v}{2} - =\frac 12EP_{11} = \frac 12EP_{22} < EP_{22}$$

So we see that, if we calculate $W_2: EP_{22} = EP_{11}$ we do not make the buyers indifferent in buying the extra ticket or not - on the contrary, buying the ticket is the only way to maintain the same expected profit as before - because if they don't, they will have expected profit $EP_{21}$ which is lower. So expect a lively bidding for this extra ticket! (and think how high you can set its price so that the buyers would still want to buy it).

If we want to make the players truly indifferent in buying the extra ticket we must equalize the expected profits

$$EP_{21} = EP_{22} \Rightarrow 2p_2\cdot W_2-2\cdot v = p_2\cdot W_2-1\cdot v$$ which is obviously possible only if

$$W_2 = v$$

...but if you do set $W_2$ like that, expect a lively beating instead of a lively bidding...

Following conversation with the OP in the comments, if we don't play around with the value of the prize and keep it fixed, then the "no-incentive-to-buy twice" result can materialize if a priori the announced rules of the lottery are that (denoting now $n$ the number of persons and $k$ the number of additional tickets above $n$):
a) The draw is "random" (equal probability per ticket)
b) Only bought tickets take part in the lottery
c) That the probability of winning depending on the number of tickets is such that

$$p_n=2p_{n+1} =3p_{n+2}...=(k+1)p_{n+k} \Rightarrow p_{n+k} = \frac {p_n}{(k+1)}$$ What is the consequence of such a rule? Since $p_n = 1/n$ we have that

$$p_{n+k} = \frac {1}{n(k+1)}$$

But the numbers of tickets taking part is $n+k$ so the total probability allocated to ticket buyers is $$\frac {n+k}{n(k+1)} < 1$$ In other words, except in the initial state with $n$ tickets, a "house probability arises", i.e. a probability that no-ticket wins. So for $k>0$, each ticket has the same probability of winning compared to the other tickets, but the "house" acquires necessarily a probability of winning too, and this probability is increasing in $k$ and tends to $(n-1)/n$ ($n$ is fixed here). This is a formalization of the initial idea described in the question.

The following is the initial post that asked for clarifications This question needs to be formulated in a clearer way. The OP states in the beginning that there are $10$ lottery tickets, in a one-to-one mapping with $10$ persons, and that "he gives everybody a $10$% chance to win". How does he achieve that? One way would be to have one "random draw" from these 10 tickets. But in this case, the probability that "no-one will win" is $0$. But is there any way for each one of the $10$ ticket-holder to have a $10$% chance to win and for the probability of no-one winning to be non-zero? No, it is not, because the events $\{$"Person $i$ wins; ${i=1,...,10}$"$\}$ are mutually exclusive and so the probability of their union is equal to the sum of their individual probabilities -hence it is equal to unity a.s. Therefore, these events partition the event space, and so an event "no-one wins" although it may exist, it can only have zero probability measure. So in the 2nd phase, the chance of "no-one wins" does not "increase" -although it becomes strictly positive since the event space will be changed.

Now let's move to phase 2: an $11$th ticket appears, bought by one of the $10$ persons. What does it mean here to "lower each person's chances to win, to $5$%"? First the OP initially says that, and then, in a parenthesis, says that the two tickets of the 2-ticket holder have each $5$% chance). These are different statements - either the OP means that the winning probability of each ticket-holder is 5%, or that the probability of each ticket is now $5$%. What is it?

I am posting this as an answer, because I plan to answer the question, once the OP clarifies the set up.

• I ment the winning probability of each ticket, not the holder (since we cannot identify those). However, maybe its not necessary to adjust the probability of the tickets winning, like I did in my example. Maybe its better solved by lowering the total pricepot with each new ticket sold? I just was thinking out loud.
– Muis
Jan 13 '14 at 7:59
• Thanks for your answer, I'm trying to wrap my head around it. Your conclusion is "expect a lively beating instead of a lively bidding..", could you clarify why that is? Do you mean that with 'W2 = v', every player has a negative expected profit?
– Muis
Jan 14 '14 at 18:56
• If the lottery prize is equal to the price of the lottery ticket, would you buy into it? It is not just "negative expected profit" -it is an angering laugh (hence the... beating of the organizer by the buyers). Another matter, pay also attention to the specific set-up I describe in the beginning -just to make sure that it is what you actually had in mind. Jan 14 '14 at 19:21