## Lottery Choice – Pick a Winner

Posted by Karl Sharman on October 15, 2011 – 5:48 am

The lottery – wouldn’t we all like to win it?

Well there’s a new one in the UK – The Health Lottery. The National Lottery is the other, more ensconced one. Which one of these two should you play?

The statistics…

The Health Lottery

Pick 5 numbers out of a pool of 50

Match 3 – £50

Match 4 – £500

Match 5 – £100,000

The UK National Lottery

Pick 6 numbers and a bonus ball out of a pool of 49

Average potential winnings

Match 3 – £10

Match 4 – £75

Match 5 – £1750

Match 5 + Bonus Ball – £225,000

Match 6 – £2,500,000

October 15th, 2011 at 9:52 am

when playing the Health lottery you win on average £0.3339…

at the UK National Lottery your average winning is £0.3779…

so it’s wiser to play the National Lottery.

but as the price for one lottery ticket is £1 each, the best strategy would be not to play lotto at all.

October 15th, 2011 at 9:55 am

So, does each cost the same, or different? How much?

October 15th, 2011 at 10:35 am

Hi Ragknot, each cost £1

October 16th, 2011 at 5:37 am

I’ll start with an example. Consider the UK Lotto (formerly known as the UK National Lottery) with 49 balls, where 6 are drawn and we want the odds of picking exactly 6 of them. Firstly, the number of possible ways of drawing 6 balls from 49 is 49*48*47*46*45*44 = 49!/43!. But that counts e.g. 1,2,3,4,5,6 as different from 2,1,4,6,5,3. As the order doesn’t matter, we must divide that by 6! as that’s the number of ways that we can arrange 6 numbers. So the number of possible draws are 49!/(43! 6!) = C(49,6) = 13 983 816.

If a ticket has exactly 4 matching numbers, and so 2 that don’t match, we have 49-4-2 = 43 possible numbers to not match the 6-4 = 2 numbers on the ticket. That gives us 43*42/2! = 43!/(41! 2!) = C(43,2) possible winning tickets (with a particular 4 matching numbers). The 2! is because we don’t care about the order the non-matching numbers are drawn in. Because there are 6 numbers on the ticket, that gives C(6,4) sets of 4 numbers. So the probability of you winning (with exactly 4 matching numbers is) C(43,2)*C(6,4) / C(49,6) ≈ 1/1032.397.

By examining the equation, we can deduce that if we had b balls, picked n and matched m that we’d get the probability of winning is:

P(b,n,m) = C(b-n,n-m) * C(n,n-m) / C(b,n).

The Lotto has an additional special case. A seventh ball (the bonus ball) is drawn. If you have matched five numbers on your ticket, then if the bonus ball matches your sixth number, you win. As there are 49-6=43 balls remaining, you have 1 chance in 43 of matching the bonus ball. The probability of getting 5 balls and the bonus ball is P(49,6,5)/43. Conversely, the probability of the bonus ball not matching your sixth ticket number is 42/43 and so the probability of matching 5 balls but not the bonus ball is P(49,6,5)*42/43

Putting that altogether we get, with b = 49, n = 6: m = 3 => 1/56.655927396483264, m = 4 => 1/1032.3968992248062, m = 5 no bonus => 1/55491.33333333333, 5 with bonus => 1/2 330 636, m = 6 => 1/13 983 816.

For the Health Lottery, b = 50, n = 5: m =3 => 1/214.0161616161, m = 4 => 1/9416.7111111, m = 5 => 1/2 118 760. In each of those we simply had to calculate P(50,5,m) with m = 3,4,5.

What we want is the average winnings. For that we need to calculate Sum(expected winnings * probability of winning). Using Karl’s data, I get 0.5560052420598212 for the Lotto and 0.3339217278030546 for the Health Lottery.

The Lotto pay out is far better that the Health lottery pay out. Of course on average you’ll lose and so it’s a mug’s game. However, those people who’ve won with 5+bonus or 6, won’t agree with me. Winning on 6 would change your whole life. You could buy a very nice house and never

haveto work again.October 16th, 2011 at 6:36 am

I am going to try and understand Chris;s procedure. I don’t understand how the Bonus ball works. But I did do the health one. I didn’t try to figure the probability. I just bought a million $1 tickets and added the winnings for each. Winnings/cost was just less than 25%.

October 16th, 2011 at 6:50 am

Hi Ragknot. The Lotto actually draws 7 balls. But the seventh is called the bonus ball. It only affects people who have got 5 matches on the main draw. If the bonus ball is the same as their 6th number (that happens 1 in 43 times), they then win £225 000 rather than £1750.

You’ll find simulating this will not work very well. You’re probably better off trying all

13 983 816 possible draws. I might do that because I’m unhappy that Carc’s Lotto expectation is so different from mine. But I think Carc’s goofed.

October 16th, 2011 at 11:04 am

Hi Ragknot. I’ve just done a test. VB Rnd() function cycles length is 16777216 = 2

^{24}. So it’s a longer cycle than I thought, but not a prime number length.October 16th, 2011 at 12:49 pm

I’ve written a proggy that finds the Lotto average winnings. I’ve optimized it to skip many of the no possiblility of a win cases, and used probability to deal with the awkward bonus ball case. I get 0.556005242059821, as before. Here’s the code:

http://trickofmind.com/wp-content/uploads/2011/10/lottery.txt

October 16th, 2011 at 4:58 pm

PS The number of possible draws with Health Lottery is 2 118 760. I tweaked the above proggy and confirm the 0.333921727803055

October 16th, 2011 at 6:32 pm

Chris?

I don’t see what the 0.333… means in how much is won.

I only worked on the “health” deal.. I still didn’t get

how the “bonus” works.

This is the health deal…

Getting 3 to match the five I got 0.431322% (less than 1/2 a percent of the time.

Matching 4 of 5, I get 0.004900%

Matching 5 of 5. I get 0.000013%

Since …

Match 3 = £50

Match 4 = £500

Match 5 = £100,000

This means for buying £100 worth tickets, the average winning would be £24.25 (or win 24.25% of the cost of the tickets.

The 5 matching was very small. Out of 120,000,000 I got 15 that were 5 out of 5 matches.

Chris, does this match your 0.333..? I thought the answer would be the cost vs. winnings to decide which to play.

But I never got a percent for the one with the bonus.

October 16th, 2011 at 8:04 pm

Hi Ragknot. The 0.3339… and 0.556… are the average number of £s you win on a £1 ticket. i.e. Your £1.00 wins you £0.3339… or £0.556…, on average. Or if you prefer, £100 wins £33.39… on the Health Lottery, or £55.60… on the Lotto. When you win, you don’t get your ante back. So on the Health, on average, for each £100 you end up being £66.61 worse off. Of course that’s on average. What you hope and pray for is to get very lucky, and win the £100k. Or on the Lotto, to win the £2.5 million.

I’ll try to explain the bonus ball once more. The Lotto ticket has 6 numbers. The Lotto draw involves selecting 7 balls from the lottery machine. The first 6 balls are the main game balls, and they are the ones that you normally match against your ticket, to get 3,4,5 or 6 matches. The seventh lottery ball is only applicable to tickets which have exactly 5 matching number on the main game. If the sixth number on the (5 matching) ticket also matches the bonus ball, then you get the larger prize. Example. Your ticket is 1,2,3,4,5,6. The drawn balls are 1,2,3,4,5,7;6. So your ticket only has 5 of the main game numbers (i.e. 1,2,3,4,5), but the bonus ball matches your remaining number (6) – so you get the £225 000 rather than the £1750 that you’d have got if the bonus ball had been any of 8 thru 49.

On the Health Lottery you should be getting 3 matches 0.46725…% of the time, 4 matches 0.01061941…% of the time, and 5 matches 0.000047197417…% of the time. As 3 matches gives £50, 4 gives £500 and 5 gives £100 000, the for £100 investment you get back: 50*0.46725 + 500*0.01061941 + 100000*0.000047197417 = £33.392 (approx, especially as I didn’t use the full precison in the calculations). Unless you’ve got a 3rd party random number generator, there’s no point in going beyond 16 777 216 draws, as the VB Rnd() will simply be repeating previous values (in the same order) after that. You’re definitely not getting anything like enough 5 matches, after 120 000 000 draws you should have 56.636… matches.

October 17th, 2011 at 4:43 am

As I haven’t actually written it down, the expected winnings for the Health Lottery is: 50P(50,5,3) + 500P(50,5,4)+100000P(50,5,5), and for the Lotto it’s:

10P(49,6,3)+75P(49,6,4)+(1750*42+225000)P(49,6,5)/43+2500000P(49,6,6)

where P(b,n,m) = C(b-n,n-m) * C(n,n-m) / C(b,n), where

C(n,r) = n!/((n-r)! r!) and is the well known combinations function.

Aside. Way back in the Stacked Deck problem, I introduced an extension for C(n,r), to wit: C(n,a,b,c,…) = n! / ((n-a-b-c-…)! a! b! c! …). I wondered if Mathematica had that function. It doesn’t seem to have; it has a prettier function. Mathematica actually calls the combinations function “Binomial”, so

C[n,r] = Binomial[n,r] = n!/((n-r)! r!). It also has a Multinomial function:

Multinomial[a,b,c,d...] = (a+b+c+d+…)!/(a! b! c! d! …).

Therefore, Binomial[n,r] = Multinomial[n-r,r].

My extension C[n,a,b,c,...] = Multinomial[n-a-b-c...,a,b,c,...]

If I ever use it, I’ll just write M[...]. Also, I often use [,] rather than (,) for functions, because Mathematica only uses [,].

October 26th, 2011 at 8:09 am

The health lottery. It has a fixed winnings and if 3 people match 5 numbers all five get £100,000 each

November 28th, 2011 at 10:56 am

This trick sounds interesting. I would love to use it when I play lottery. I hope to get favorable results as the post explained the method in a comprehensive manner.