Too Many Lotteries

So recently I'd gotten really into this TV show - it's called 'Life', and it's a 'murder of the week' crime drama starring Damian Lewis. Lewis plays Detective Crews, who was previously in prison for a muder he didn't commit, and he's into zen and such. It's a bit silly. But in an endearing sort of way.

Unfortunately, the only way I could watch it (short of buying the DVDs) is to stay up til 4am. Which isn't really a problem; I don't have anything to get up for. Though it is tricky getting to sleep as the sun's coming up.

Anyway, the point is, by 4am, there are only, like, four adverts on rotation (on FX, at least). These are: upcoming shows on FX, whatever product JML is flogging, that tacky accident helpline ad with Esther Rantzen, and this 'Win Trillions' thing.

Win Trillions is a lottery sindicate thing - from what I understand, it works like this: a bunch of people from around the world get together, each buying tickets for their respective countries' lotteries. If one of the sindicate members wins, they split the winnings with their fellow sindicate members (and presumedly, Win Trillions takes a cut).

Really, though, the details aren't that important. What is important is their slogan - "play more lotteries, get more chances".

Which isn't a false claim. It's just kind of misleading; playinng more lotteries does mean more chances of winning, but only in a similar way as buying more tickets for a single lottery means more chances of winning - your odds of winning are so small to begin with (~1 in 13,983,816) that you'd have to play thousands of lotteries (or buy thousands of tickets) to significantly improve your chances. And there are only so many lotteries in the world.


Strategies

You'll notice I used the word 'similar' back there; the fact is, buying more tickets, and playing more lotteries don't 'improve' odds in the same way.

So the question is, which is the better strategy - buying many tickets for a single lottery, or buying one ticket for each of several lotteries?

First, imagine I have a dice; I'm going to roll the dice, and you have to bet on the outcome. Your choice is this - would you rather place three bets on one dice roll, or one bet on each of three dice rolls?

For the first case, say you place your bets on it being 1, or 2, or 3. Your odds of guessing the correct number (and winning) is 3 in 6 (=0.5).

For the second case, say you bet that each of the rolls will be 6. Then your odds of winning 'something' is odds of winning just one of the dice rolls, plus the odds of winning just two of the dice rolls, plus the odds of winning all three.

Which is trickier to work out. For example, the odds of winning the first roll only, is odds of getting the first roll right (1 in 6), and not getting the second roll right (5 in 6) and not getting the third roll right (5 in 6) - which makes the probability of getting just the first roll right = 1/6 * 5/6 * 5/6 = 25 in 216

And you have to work out the probabilities for all possible winning outcomes (in this case, there are seven winning outcomes, one losing.).

Alternatively, we can use the trick that the odds of winning something is one  minus the odds of winning nothing. The odds of winning nothing is the odds of getting all the bets wrong: 5/6 * 5/6 * 5/6 = 125 in 216 (0.58)

So the odds of winning something is 91 in 216 (=0.42)

So for this dice game, it's better to place your three bets on a single dice roll.

But what about in general?

We say each game has a probability p of winning, and that we're going to place n bets (with n>1).

For the single game bet, the probability of winning is n*p

For the multi-game bet, the probability of winning is

Which is bigger?

Well for n = 1, the games are idential. For n = 2, the single game is 2p, the multi-game is 2p - p^2; the single game has higher odds. For n >= 1/p, the single game gives a probability greater than or equal to 1 - i.e. guaranteed win - where the multi-game always gives a values less than one; again, the single game gives better odds.

So it's looking like the single game is always better. But how do we prove it?

For this, we look at the binomial expansion of (1-p)^n

A Binomial Expansion works like this

where k! is the factorial of k. For example: 2! = 2*1; 3! = 3*2*1; 4! = 4*3*2*1; etc.

In this case, x = 1 and y = -p, so the expansion becomes

Which makes the probability of winning the multi-game

So whether or not the mutli-game has higher odds than the single game depends on the sum of the terms after np.


First, we compare the k-th and (k+1)-th (adjacent) terms in the expansion

Which simplifies to

The righthand side has it's maximum value when n is maximum. We previously defined the maximum value of n as 1/p, so

since p and k are always greater than 0.

Which means

In other words, each term in the expansion is smaller than the ones before it.

This means that, if we pair up terms,

i.e. the sum of each pair will be negative. Therefore, the sum of all the pairs (all the terms in the expansion after np) is negative. So necessarily

Which means the odds of winning the single game is ALWAYS better than the odds of winning the multi-game.


Pay-Outs

You might have noticed that the single game has a single possible payout, where playing the multi-game means you could win double, triple, or even more of the prize. So the odds of winning the multi-game are lower, but you stand to win more.

Does that mean it's worth the extra risk?

For this, we look at the expected winnings for each game. The expected winnings is calculated as the probability of winning multiplied by the prize amount.

So if you were guessing the outcome of a dice roll, and the prize was £1, then the expected winnings from playing that game would be ~17p.

Or think of it like this - if you had to guess the outcome of a dice roll six times, you would expect to be right once out of six. So your expected prize for 6 rolls is £1, or 17p per roll.

For this single game, the payout is easy: w*n*p (where w is the prize for winning one game)

For the multi-game it's more complicate. We can't use the same trick as last time; instead, we have to work out odds of winning one game times prize from one game + odds of winning two games times prize from two game + etc.


The general form is the series sum

Which can be manipulated and simplified to give

Now, if we expand it out again

Then there's a factor of np in each term, so we can take that outside the bracket

And since n is some arbitrary interger, we can substitute m = n - 1 and put it back in summation form

Now, you might notice the summation part is actually a binomial expansion; in this case we have x = (1 - p) and y = p, so

For all values of p and m = n - 1.

Therefore, the expected payout of the multi-game is always w*n*p


You'll notice that this is the exact same expected payout as for the single game. As in, regardless of which strategy you use, your expected payout is the same.

So the question is, would you rather a low risk low prize game, or a high risk (potentially) high prize game?


Complexities

You'll notice for this I assumed all lotteries have the same odds of winning and the same prize amount. This is not necessarily true.

Even so, from this, the best strategy would be to find the lottery with the best single game single bet expected payout (prize * probability of winning), and buy several tickets for that game, rather than spread your money around.

The other thing in lotteries is you can usually win smaller prizes for matching fewer numbers. But I don't imagine that changes which strategy is best.

So now you know. For what it's worth. The odds of winning playing, say, 70 tickets on one lottery is only about two 10,000ths of a percent better than playing one ticket on 70 lotteries.


Incidentally, those derivations up there count as mathematical proofs. So these conclusions I've drawn are irrefutable*. And the results apply to any game where you're placing bets on a single probability random outcome - not just lotteries.


Oatzy.


[*- Assuming I haven't cocked it up...]

With Enough Tries..?

Probability is tricky. It isn't always intuitive. Coincidences aren't necessarily as rare or as unusual as they might seem.

I can't remember how I got to it, but the other day I came across the wiki article on the Law of Truly Large Numbers. An interesting idea to say the least.

Then a couple of days later I was looking through one of my books for blog ideas, and came across an essay with an example strikingly similar to that in the wiki article (in never gave it a name).

Coincidence?


So what is the Law of Truly Large Numbers?

The Wiki page gives this description:

[The law] states that with a sample size large enough, any outrageous thing is likely to happen.

The example given on the page is a little inelegant, so I'll go with the (abridged) similar example from the book,

Suppose that a really memorable, once in a lifetime coincidence is one which has a one in a million chance of happening today, and that during any particular day there are 100 opportunities... [T]he chance that one of these coincidences will happen to you tomorrow is 1 in 10,000. Still very unlikely...

[But] the chance that every one of the next twenty years will have no one-in-a-million coincidences for you is.. 0.48, or a 48 per cent chance.

According to this extremely rough and ready calculation, there is actually more than a fifty-fifty chance that in the next twenty years you will experience a memorable one-in-a-million coincidence. This also means that for every twenty people you know, there is a greater than 50% [chance] that one of them will have an amazing story to tell during the course of a year.

Now this is an interesting thought.

And it raises an interesting question - If you play the lottery enough times, does winning eventually become significantly more likely? Inevitable?

It's an often quoted 'fact' that you're more likely to be stuck by lightening on your way to buy your ticket, than you are to win. But what does 'the law' have to say on the subject?


Preamble

For this we're assuming a good old fashion, six balls from a pool of 49 lottery.

Probability of winning the jackpot (matching all six balls) with one ticket is 1/13983816 or about 7 in 100million

If you play two lotteries, then your odds of winning are (Odd of winning the first) + (odds of winning the second) + (odds of winning both).

OR, and this is easier to work out,

Let 'Odds of not winning', q = 1-p(winning)

'Odds of winning at least once in two games' = 1 - (odds of winning neither) = 1 - (q*q)

This can be generalised to 'Odds of winning jackpot playing n games', p = 1 - (q^n)


Round One: Will I hit the Jackpot in My Lifetime?

First of all, odds of winning the jackpot by playing every week for a year

p = 1 - [1-p(winning)]^52 = 3.7 in 1million

So not great. How about if you play ever week, starting on your 16th birthday and giving up (dying) on your 86th. Or basically, playing for 70 years. Probability of hitting that jackpot?

About 1 in 4,000 chance. So still not great.

Of course, if you buy 40 tickets a week, then that gives you a 1 in 100 chance of winning the jackpot at some point in your life. But by that point you're spending £2,080 a year on lottery tickets. The average jackpot would have to be more than £14.6 million for the expected return (prize*chance of winning) to make it worth playing.


Round Two: What About Immortality?

So we've got the equation p = 1 - (q^n)

The question is, can we find n - i.e. the number of games you'd have to play - such that the probability of winning (p) is 50:50

The trick is logarithms, and the formula is

n = log(1-p)/log(q)

So for p = 0.5, n = 9,692,842 games, or about 186,400 years.

For a 1 in 4 chance of winning? 77,363 years

1 in 100 hundred chance?! 2,703 years

Alternatively, to have a 50:50 chance of winning in your lifetime (70 years) you'd need to buy 2,663 tickets a week. Yeah.

Basically, even by the Law of Truly Large Numbers, and immortality, you'd be waiting a ridiculously long time and you'd still be lucky to win.


Round Three: I'll Take Anything!

Now wait a minute, I hear you say, I could still win something by matching 5 numbers, or even 3. Okay, that's a fair point.

So you need to match 3 or more numbers to win something. Probability of winning anything in any given game? ~6 in 100,000

So once again, chance of winning something if you play every week for 70 years? 195 in 1,000

Now that's interesting. That's just short of a 1 in 5 chance. But to be worth playing, the average prize value would have to be ~£18,666. Worth it? I'll let you decide*.

And finally, how long would you have to play to have a 50:50 chance of winning something? ~223 years. Or 45 years if you buy 5 tickets a week.

Which is going to be a real kick in the balls if that something turns out to be £5.


Or To Put it Another Way

* Imagine a game you only get to play once. You pay me £3,640 to play, then you pick a number between 1 and 5. I then generate a random number between 1 and 5.

If the number that's generated is the number you chose then you will win some randomly chosen prize between £5 and £5million; you're more likely to win a smaller prize than a larger one, and you can't know in advance what the prize will be.

Want to play?

If you play the lottery, but answered no to the above, you should probably reconsider.


tl;dr As has been said many times before, your odds of winning the lottery jackpot are catastrophically minute. Even if you were to play every week of your life.


Oatzy.