Here is a rather practical question, I and my son have different opinions about, please bear with me, it's a probability question:
With NS&I bonds, is it better to buy many smaller value bonds or fewer (eventually one) larger value bonds?
For example, say you have $£1000$ available, is it better to buy
- $10$ bonds with $£100$ value each or
- one bond of $£1000$ value?
Options:
- I say - one larger value is better
- My son says - it doesn't matter, applying geometric distribution.
By better I mean the chance of winning anything, no difference between large and small wins.
A short intro.
NS&I bonds are like a perpetual lottery in UK. Each $£1$ buys one lottery number. A bond (or bond value) consists of one or more such numbers, technically the smallest value is $£25$ or $25$ numbers. The number structure is like ($||$ means concatenation) $$\text{up to3 digits } || \text{ 2 characters } || \text{ 6 digits } \tag{1}$$ for example $295\text{PC}150165$. The numbers are consecutive, e.g. $£100$ will buy a bond value of $£100$ with a possible sequence like (again, this is an example) $$300\text{PC}000001 - 300\text{PC}000100$$ Unfortunately, you can't choose the numbers, the system does it for you. The maximum number of tickets one can ever hold is $50000$, in one large value bond of $£50000$ or in many smaller - doesn't matter, $50000$ is the upper limit. Winning numbers are selected randomly and uniformly.
Why I think "one larger value is better".
If I group the winning numbers, for the last almost one year, by the $2$ characters (let's call it a basket) in the middle (see $(1)$), I get a pretty uniform distribution (more or less) per basket, here is an example:
There are $529$ such baskets. Uniformity of the random selection of the winning numbers makes the spread uniform per basket. We can "approximate" each such basket with an independent lottery. Buying a bond (any value, $50000$ upper limit fits into one basket) means buying tickets in one basket/lottery (I haven't seen cross baskets purchesments, let's assume it's not possible). Then the problem is similar to
Is it better to buy tickets in one lottery or many lotteries, given all lotteries are similar?
Similar in this case means each lottery has maximum $M$ tickets possible and $n$ random draws are performed. This is hypergeometric distribution. The answer is buy all in one lottery. Let's consider a simple example of $2$ lotteries (and then by induction ...) and the intuition of even splitting. I can buy $2k$ tickets, either all in one lottery or $k$ tickets per lottery. The chances of winning are
- All in one lottery case $$1-\frac{\binom{M-2k}{n}}{\binom{M}{n}} \tag{2}$$
- $k$ tickets per lottery case considering independence $$1-\frac{\binom{M-k}{n}}{\binom{M}{n}} + 1-\frac{\binom{M-k}{n}}{\binom{M}{n}} -\left(1-\frac{\binom{M-k}{n}}{\binom{M}{n}}\right)^2=\\ 1-\left(\frac{\binom{M-k}{n}}{\binom{M}{n}}\right)^2 \tag{3}$$
Then $$(2) \geq (3) \iff \frac{\binom{M-2k}{n}}{\binom{M}{n}} \leq \left(\frac{\binom{M-k}{n}}{\binom{M}{n}}\right)^2 \iff \binom{M-2k}{n}\cdot \binom{M}{n} \leq \binom{M-k}{n}^2 \iff\\ (M-2k)(M-2k-1)...(M-2k-n+1)\cdot M(M-1)...(M-n+1) \leq (M-k)^2(M-k-1)^2...(M-k-n+1)^2 \iff\\ \left(1-\frac{k}{M-k}\right)\left(1-\frac{k}{M-k-1}\right)...\left(1-\frac{k}{M-k-n+1}\right) \leq \left(1-\frac{k}{M}\right)\left(1-\frac{k}{M-1}\right)...\left(1-\frac{k}{M-n+1}\right)$$ which is true, compared term by term. All in one lottery in this case is similar to all in one basket or one larger value bond is better.
This is also backed by the observation that large value bonds are leading the winning board. Here is a histogram from the data this March:
Why my son thinks "it doesn't matter".
The winning odds per basket aren't really independent, which is fair, since I only assumed it. Also, if we "round" the winning odds per ticket to a value $p$ (NS&I suggests the odds of winning anything are $1$ to $34500$), then:
- All in one lottery case $$1-(1-p)^{2n} \tag{4}$$
- $k$ tickets per lottery case considering independence $$1-(1-p)^{n} + 1-(1-p)^{n} -\left(1-(1-p)^{n}\right)^2=\\ 1-(1-p)^{2n} \tag{5}$$
final result is the same.
Who is right, thoughts?
P.S. This should be treated as a probability problem, not an investment advice!
P.P.S. A bit of brainstorming and the Python code to produce the spreadsheet (as per the screenshot) are available here. Historical winning numbers can be downloaded from here.