Econstudentlog

A repost: Some stuff on lotteries

I’ve reconsidered my decision not to update the blog over the next days. But as I don’t have time to actually do blogging-worthy stuff I’ve decided to repost some material from the archives that I’m not too ashamed of having posted here (the posts I’m reposting should incidentally in no way be construed as an argument for going through my archives – there’s a lot of crappy stuff hidden away there and it’s just not worth it…). The original post is from May 2011 and the only change I’ve made to the post is to remove the last part requesting feedback from readers (I probably would not consider writing such an addendum if I were to write a similar post today). Anyway, here goes:

Let’s say you have a population of n (ex ante) identical individuals each making an income of w. Say you now decide to set up a simple voluntary tax-transfer type scheme, where all individuals who choose to participate are required to pay an amount (/tax), t. The contribution/tax t is used to finance a transfer T, which is equal to n*t (the sum of all contributions, i.e. there’s no administration or anything like that to start with). Each individual has a 1/n probability of receiving the transfer T, so that the expected payoff of this scheme is equal to the probability of receiving the transfer times the size of the transfer minus the contribution, or 1/n*(n*t)-t. Which is equal to 0 and independent of both t and n. The expected income of an agent participating in the scheme is w + EP = w.

A risk averse individual will always choose not to participate. A risk neutral is indifferent between participating and not participating given that the reservation utility is 0. Note that even if the expected payoff of the scheme is ‘mathematically’ zero, the way most people think about a scheme like this (..out of context at least, when talking pure math) is that you’re most likely to lose if you participate, especially if n is sufficiently high. If a million people participate and there’s one transfer each month, then the likelihood that you’ll have gotten your money from the contributions back after a year is not very big.

It’s probably even lower than you realize, if you’re not familiar with statistics. To illustrate why this is, let’s get a little more technical. There’s one transfer T each timeperiod. There are n people who participate in the scheme. Now assume that your likelihood of getting a transfer next period does not depend on who got it last period. You can think of it as an assumption stating that an individual can receive several transfers if he or she is very lucky. This assumption is important, but I also think it’s justified in the empirical framework I’m attempting to apply this to – it would be completely justified if the scheme was mandatory, but regarding lotteries we know that a) at least some lottery winners play on after they’ve won anyway and, far more important, b) that the number of participants in real world lotteries is pretty much independent of the behaviour of the winners ex post (1 marginal lottery winner does not translate to one less lottery participant in general) [where ‘behaviour’ here relates only to the decision as to whether to participate in future lotteries or not]. If you don’t like to think of it as an assumption about past winners playing along after they’ve won, you can think of it as new people entering the scheme after past winners decide to exit, keeping the probability of winning constant over time.

Now perhaps a not uncommon way to misunderstand how this works is for people who don’t know statistics to think/assume that if you have 52 participants and 52 weeks of contributions/transfers, then the probability that you receive a transfer is equal to 1 after one year. It’s not, it’s lower than that, because some lucky guy might win 2 times and get the transfer instead of you. The only case where you can be certain to have won after a year is in the case where nobody can win more than once. In that case, the conditional probability of winning is increasing over time – the chance of winning the first lottery is 1/52, if you don’t win the first lottery you have a 1 in 51 chance of winning the next lottery, ect. I’d like to instead look only at the case where the conditional probability of winning is constant over time.

The probability that an individual i will receive a transfer before period k, where k is equal to 1,2,3…, follows in that case what is called a geometric distribution, which is itself a negative binomial distribution (I know I’ve linked to that one not long ago here on the blog) with r = 1. The cumulative distribution function, which in this specific case can be thought of as a function telling us how likely we are to have gotten a transfer by the time we reach period k, is equal to 1 – (1 – p)^k. To make this a bit easier, think of throws with a die. After one attempt, the likelihood of rolling a 6 is 1 – (1 – 1/6)^1 = 1 – 5/6 = 1/6 (we knew that!). The likelihood of rolling a 6 after exactly two attempts is equal to: 1 – (1 – 1/6)^2 = 1 – (5/6)^2 = 1 – 25/36 = 11/36. Note that this is smaller than 1/3 (or 12/36) for reasons already mentioned; when outcomes are independent, you can’t just add the probabilities to get your estimate. Also note that the probability of getting that damn 6 is of course increasing in the number of attempts. Now what’s the probability that you will not have rolled a 6 after 10 throws? Probably higher than most people think: 1 – [1 – (1 – 1/6)^10] = 0,1615, which is a tiny bit lower than the probability of rolling a 6 in the first attempt. Note that here I take advantage of the fact that there are only two outcomes [roll 6 or don’t roll 6] and that the probability of not rolling 6 in a sequence is equal to 1 minus the probability of doing it (mutually exclusive & collective exhaustive and all that..).

Now if we have a lottery with 1 million people participating (p = 1/1.000.000) and one transfer handed out each week, what’s the probability that you’ve not gotten a transfer after 10 years of participation (k=520, 52 weeks in one year…)? Putting in the numbers we get 1 – {1 – (1 – 1/1.000.000)^520} = 0,99948 = 99,948%. The funny thing here is also that the transfer is uncertain but the contributions are not, so if you assume weekly contributions of value \$5 over the 10 year period, the certain costs are \$5 * 520 weeks = \$2.600. So if you play along in this lottery, you pay \$2,600 and get nothing with 99,9% certainty. The expected payout from the lottery is of course the same as the amount you pay, as the transfer is \$5.000.000 and and the probability of getting the transfer each period is one in a million, so that expected payout is 520/1.000.000*5.000.000= 520*5 = 2600 and the expected total payoff is 0.