Winning on Prize Bonds

In ireland the goverment issue a special "Savings Bond" called Prize bonds. You dont get any interest on them, but you can get your money back any time you want, and you are put into a weekly draw with a chance to win various money prizes as follows:

         1 x €20,000 x 52 - 12 (12 prizes of €1M) = 40 prizes per year of €20K
5 x € 1,000 x 52 = 260 prizes per year of €1K
10 x €   250 x 52 = 520 prizes per year of €250
Over 7,400 x €    75 x 52 = 384,800 prizes per year of €75


Currently there are 7,400 prizes in each draw. So given the figures, currently there is 7400 - (10 + 5 + 1) prizes each week. (origionally this was 7000) Lets ignore this minor calculation as additional prizes are added as the fund grows.

Also there is 1 monthly jackpot prize of €1,000,000, which replaces the €20,000 prize in that week. As the prize bond fund grows they add additional prizes to the €75 category. Even if you win, no matter the prize, you can keep winning. So your tickets are not removed once you win.

On their www.prizebonds.ie site they say a customer who has €1,000 invested has a 3.6 to 1 chance of winning. But I dont get how they work this out. Or even what it means. 3.6 to 1 (3.6 of what to 1 of what). If its 3.6 chances to win for each bond then every bond would be winning 3.6 times in a year. This doesnt seem real. But I am open to being wrong. I just dont really get this.

To purchase a prize bond, you get them in lots of 4 at a time, each €6.25 each or €25 in total. This entitles to you 4 entries in each draw.

In 2010 the amount of money in sales of prize bonds was €1,328,000,000. From this figure we can obtain the number of investors. €1,328,000,000 divided by €6.25 each equals 212,480,000 prize bonds have been issued.

My question is two fold. How do I work out my chances of winning if I had invested €1000 and leave it there for the period of the full year either 31-dec-09 or 1st-jan-10, to 31-dec-09 or 1st-jan-11 (I know maths guys are picky so was not sure exactly what dates to use to represent the full year, lets not get into leap years :-> ).

Simple as this may seem, when I calculate this, I dont get 3.6 to 1 as they do in their FAQ. So this is the second part of the question. How are they getting this figure. And what exactly does 3.6 to 1 mean. The part of the FAQ that I refer to is following:

FAQ here

I would prefer my chances of winning expressed as a percentage. For example, I get X number of bonds for my €1,000 euro, each prize will appear X number of times over the 12 months, and there are X number of prize bonds compteting with mine for those prizes. This makes sense to me. But how do I convert my answers to something like 3.6 to 1.

Edits for consideration
The edits have been highlighted in bold. If anyone is woundering where the previous answers got their figures I want new people to know that above the origional figures were €398,000,000. These were the number of sales in 2010. As at least one answer pointed out, I should have used the number of actual bonds outstanding as opposed to the number sold. So I changed that number above. The figures are got from the following link:

Sales figures

When there scroll down to the second heading under "PRIZE BONDS SALES GROW BY", there is a short paragraph which shows figures. The one I am interested in is "Fund Value at Year End", which has a figure of €1,328,000,000 (1.328 billion).

This can be used to tell us that at the end of 2010 exactly €1,328,000,000 divided by 6.25 equals 212,480,000 (origionally I had 63,680,000).

So I am updating those figures above.

Also, I am a bit of a mathematical idiot. So I dont understand most of the first 3 answers I have received. What is that squiggly sign that looks like an equals having cramps. Is there an easy way to explain this stuff. I hope this is not offensive. But I dont have strong mathematical background. Thanks again.

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I've updated my answer with your new values, and get quite a bit closer (3.2 to 1 - still not exact agreement, but in your favour); I've also added an explanation of the notation I use. It's worth noting that the probability of winning isn't necessarily as useful to know as the average effective interest rate, which my Googling suggests to be 3% (although given the large imbalance between the largest prize and the typical prize, nearly everyone will get much less than that). –  Peter Taylor Aug 14 '11 at 7:50

Regarding odds, Wikipedia has a clear explanation stating that usually odds-against rather than odds-on are stated, and that m to n odds corresponds to success probability of n/(m+n). So, odds stated as "3.6 to 1 chance of winning a cash prize in a 12 month period" corresponds to likelihood of $1/4.6 \approx 21.7\%$.

In computing the odds yourself, two problems arise. First, the 2010 sales you mentioned (€398,000,000) may be much less than the actual total capitalization, against which your €1,000 investment is competing. (I didn't find any sales or cap. figures for prizebonds, so will just continue with the numbers you gave.) Second, last month's payout at prizebonds was ca. €3.9M, about 65% more than the ca. €2.4M payout one would expect from the table of draws you mentioned (from prizebonds faq #11).

The first problem will make the odds worse; the second, better. All that said, I think getting within a factor of two or so on the odds may be as good as one can expect. Here's one such calculation:

Probability of success on one draw = investment shares / cap. shares = investment / cap. = 1000/398000000 $\approx$ .0000025.

Probability of 1 success in a year = Binomial dist.(1; n, p) where n = draws per year, so nominally n $=52\cdot7016=364832$, and p = .0000025 from above.

Now $n\cdot p \approx 0.96 < 10$ so a Poisson approximation with $\lambda = 0.96$ is ok. From a Poisson article, Prob($k;\lambda)$ = $\lambda^k\cdot e^{-\lambda}\over{k!}$ whence Prob(1; 0.96) = $s \approx 0.37$, or 37% chance of a payment within 12 months time. Hence, the odds based on this calculation are $(1-s)/s = 0.63/0.37 =$ 1.7 : 1, vs. official odds of 3.6 to 1.

Update 1 Although I was aware Prob($1;\lambda)$ isn't the desired answer, I assumed (incorrectly) that Prob($k>1;\lambda)$ would be of negligible magnitude. However, $\text{Prob}(k>1;\lambda)\approx 0.237$. $\text{Prob}(k\ge1;\lambda) = 1-\text{Prob}(0;\lambda)\approx 0.60015$, giving odds of 2 to 3.

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Hi dude. Thanks for the detailed answer. I can see you looked over the site and really did your homework. All I can say is well done. After your answer I edited my post. I dont have a clue about most of what you said. Binomial dist, Poisson approximation, the squiggly equals and the funny upside down y. But because you have done good research on my question, I will make an effort to find out more about the parts of your answer I dont understand. Thank you kindly for all your effort. I think I need to dig out my spreadsheet for this. –  Francis Rodgers Aug 13 '11 at 22:48
@francis, the "squiggly equals" stands for approximately equal. The "upside down y" is a greek lambda, as commonly used for the parameter of a Poisson distribution. Re Binomial distribution and approximations to it, the Wikipedia page I linked to explains in detail; briefly, a Binomial dist. expresses the probability of getting k out of n flips to come up heads, with a given probability p for a head. Re "probability of success on one draw", I approximated a binomial by a geometric-probability ratio; the other answers don't make that approximation. –  jwpat7 Aug 14 '11 at 21:45

If you have $n$ bonds out of $N$ and $r$ distinct bonds are drawn then the probability that none of your bonds are drawn is $\frac{\binom{N-n}{r}}{\binom{N}{r}} = \frac{(N-r)^\underline{n}}{N^\underline{n}}$.

If $d$ independent draws are held in a year then the probability that you don't win any prizes (assuming that $n$, $N$, and $r$ remain unchanged) is $\left(\frac{(N-r)^\underline{n}}{N^\underline{n}}\right)^d$, so the probability that you win at least one prize is $1 - \left(\frac{(N-r)^\underline{n}}{N^\underline{n}}\right)^d$

The values you give are $n = 160$, $N \approx 212480000$, $r \approx 7000$ (they'll have adjusted it up from 7000, if I understand correctly), $d = 52$. Taking those as exact values, the probability that you win a prize is $1 - \left(\frac{212473000^\underline{160}}{212480000^\underline{160}}\right)^{52}$

If my calculation with Sage Notebook is correct that's about 0.24 or 3.2 to 1.

Postscript on notation:

• $\binom{a}{b}$ is the number of ways of choosing $b$ different items from $a$ items.
• $a^\underline{b}$ is the "falling factorial" $a \times (a-1) \times (a-2) \times \ldots \times (a-b+1)$
• $x \approx y$ means "$x$ is approximately equal to $y$"
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Interesting and topical question with excellent replies to date as to the principles of working out the odds. One of the specific challenges to arrive at the published odds is knowing the number of bonds in actual circulation. Prior to advent of the Euro about 10 years back,bonds were sold in units of 5 Ir pounds and way earlier again may have been 1 Pound units at launch. Plus we don't know the redemptions over the years as mentioned.

Currently bonds with Serial No's QA 000001 to QA 999999 are being issued having worked all the way up from AA 000001 to AA 999999 then AB AC.....(each letter increment means another million bonds sold) etc and prior to that single alpha A 000001 to Z 999999. Going on those issuances, some 400+ million bonds may be out there, minus the unknown redemptions of course.

Whereas estimating 212 million bonds @ 6.25 Euro from the current "kitty value" makes us all feel more optimistic, sadly it may be distorting the odds greatly. FexCo / An Post would need to tell us the numerical quantity N of valid bonds to clear this up.

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From what you've written, an individual ticket is €6.25, which means that €1000 is 160 tickets. If there are $k$ tickets out there (you indicate that $k= 63,680,000$ the probability that a ticket wins nothing is approximately $(1-\frac{7000}{k})$, and the probability that none of your tickets win anything after a year is approximately $(1-\frac{7000}{k})^{52\cdot 160}$ (This is not quite correct, as one ticket can't win multiple times in a week, but if $k>>7000$, this is a reasonable approximation). For small $x$, we have an approximation $e^{-x}\approx 1-x$, and hence $e^{-nx}\approx (1-x)^n$ (assuming $nx$ is still small). Applying the approximation here, we have

$$\left(1-\frac{7000}{k}\right)^{52\cdot 160}\approx e^{-\frac{7000\cdot 52 \cdot 160}{k}}.$$

With your value of $k$, $\frac{7000\cdot 52 \cdot 160}{k}$ is approximately $1$, so the approximation is a little suspect, but we get that the odds of never winner are about $1/e$, and hence the win/loss ratio is about $(1-1/e)/(1/e)=e-1\approx 1.71$ to $1$. Without using the approximation, we can calculate the chance of never winning as being much closer to $2/5$, and so the win to loss ratio would be $1.5$ to $1$ (which is not far from the approximation we had).

Of course, if they did their calculations with a different value of $k$ (in particular, some of the bonds may be traded in, or they might have calculated their numbers when many fewer bonds had been bought), that could explain the discrepancy. In particular, if they change the value of winning at the lowest level but not the number of winners, your chance of winning goes down as the lottery grows.

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I've been invested in Prize Bonds for quite a number of years with a sufficiently large amount of money to get some statistically meaningful results. Before I invested I did a similar calculation to what you've done however I did take a slightly different approach.

I emailed the Prize Bond company to confirm what the annualised interest rate that was being used to determine the prize fund. This was confirmed as 3%. As the weekly prizes are funded from the interest of the money invested in Prize Bonds, you can use the weekly total prize to calculate the total size of the fund. You can then calculate your odds. So let's work it through.

If the weekly number of €75 prizes is 7400 then that means that in that week, the total prize fund was:

€20,000 + (5 x €1000) + (10 x €250) + (7400 x €75) = €582,500 (mid month)

or

€1,000,000 + (5 x €1000) + (10 x €250) + (7400 x €75) = €1,562,500 (end of month)

We'll annualise this to determine the number of Prize Bonds that are outstanding to have earned enough interest to pay for the prizes with an annual interest rate of 3%.

So in a year the total prize fund will be:

40.178571 x €582,500 + 12 x €1,562,500 = €42,154,017.86

(Why 40.178571? There are slightly more than 52 weeks in a year, leap year or otherwise. 365.25 / 7 = 52.178571.. This therefore also accounts for leap years.)

With an annual interest of €42,154,017.86 at 3% means that:

€42,154,017.86 / 0.03 = €1,405,133,928.57 Prize Bond fund outstanding

So now let's calculate the odds.

The outstanding number of Prize Bonds is: €1,405,133,928.57 / €6.25 = 224,821,428.57

If we purchase €1000 worth of Prize Bonds we will receive 160 Prize Bonds.

So the odds of you winning one of the weekly draw with 160 Prize Bonds is:

160/224,821,429 = 0.000000711675931

As there are 7400 + 10 + 5 + 1 draws a week in which all Prize Bonds are eligible the annual probability is:

0.000000711675931 x 7416 x 52.178571 = 0.2753874 (approx 27.5%)

Convert this in to odds: 1/0.2753874 = 3.631

So you would have to wait a little over three years and half on average to win a prize which in all likelihood will be the €75 prize.

As the number of €75 prizes increases as people invest in Prize Bonds the odds of winning a prize doesn't decrease. However the odds of winning a prize other than the €75 prize decreases as the number of draws of these other prizes remain static.

From my own investment in Prize Bonds, these odds are proving to be correct.

If you are considering investing in Prize Bonds you really need to put in at least €4000. Why €4000? Well if investing €1000 means you have to wait some 3.6 years before you win a prize which is most likely to be €75. But as averages go you could wait 5 years for the first prize and then 2 years for the second one and then 8 years to win two more in quick succession, etc...

With €4000 the odds are you will have to wait on average a little less than a year (actually 47 weeks) to get a prize. Mathematically the odds don't change, it's just as humans waiting around for something to happen can be quite boring and also we have a limited life span.

If you invest less than €1000 the odds of you winning something, anything, in your lifetime starts to get really small and it becomes more of wishful thinking than actual investing.

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