Maths Champion of Askaboutmoney?

[DerKaiser]

Interesting occurence:


Is it really a 48 million to 1 shot? If so, how can it have happened more than once in the history of lotto?

Title of Maths Champion of Askaboutmoney will go to the first person who can explain how to calculate the true odds of six or more people sharing a single jackpot.

Assumptions:
1) Each line requires choosing six numbers from a possible 45
2) 2,000,000 lines filled out for the draw
3) Complete independence in how numbers are chosen

Hint:
Use of statistical distributions, excel functions, etc is strongly encouraged!

 

[Joe_90]

I always thought the odds of picking the numbers was 45x44x43x42x41x40 = 5,864,443,200 but that's the odds of picking the numbers in the correct order.

So you have 6 chances of getting the first number, 5 for the second, 4 for the third ect so 6x5x4x3x2x1 = 720.

Divide the big number by the little number and you get 8,150,160 the odds of picking 6 numbers in any order from 45.

Then I guess the odds of something happening 6 times, is the odds of something happening once x 6.

 

[Homer]

If you know the number of lines paid for in total, you can work out the probability of any specific number of winners using the Poisson distribution. This assumes that the numbers are completely random and that there is nothing causing a bias in the numbers chosen by those playing the game.

This latter point is not necessarily correct as, for example, it is more likely that there will be a number of people sharing the prize if all the numbers are 31 or less, because many people select numbers based on family birthdays, anniversaries, etc. It's also more likely that there will be several winners if the numbers are spaced out rather than clustered.

 

[cremeegg]

I always thought the odds of picking the numbers was 45x44x43x42x41x40 = 5,864,443,200 but that's the odds of picking the numbers in the correct order.

So you have 6 chances of getting the first number, 5 for the second, 4 for the third ect so 6x5x4x3x2x1 = 720.

Divide the big number by the little number and you get 8,150,160 the odds of picking 6 numbers in any order from 45.

I calculate the odds of choosing 6 from 45 as 1 in 8,145,060. That is 45!/(39!x6!) which is close to your number, though of course there should be no difference.

Then I guess the odds of something happening 6 times, is the odds of something happening once x 6.

I dont agree with you here. The odds of getting heads is 0.5 but the odds of getting heads twice is not 1. It is 0.5 by 0.5 = 0.25

SO the odds of 6 people getting the numbers is (1/8,145,060)^6. This is based on there being only 1 set of 6 correct numbers, but there are 720 sets of correct numbers, as you explain. So (1/8,145,060)^6 * 720

No I don't believe this either. maybe tomorrow I will be fresher. By the way probability is not real maths, it is maths for trainspotters.

 

[Homer]

Incidentally, if the odds of 48.5 million to 1 were calculated by multiplying the number of winners by the probability of winning from a single line, then that's about as valid as Stefan Klincewicz stating that if the numbers 1, 2 and 3 haven't come up for a couple of weeks, then there is a very good chance on the law of averages that at least one of them will come up in the next draw. I actually heard him say this on the radio a number of years ago and couldn't believe that the interviewer just accepted it without challenge.

Then again, I don't know why I was surprised as most people's grasp of probabilities is fairly abysmal. In the land of the blind, the one eyed monkey is king.

 

[Homer]

probability is not real maths, it is maths for trainspotters.

Classic! I bet you don't even see the irony in that statement.

 

[cremeegg]

Classic! I bet you don't even see the irony in that statement.

Is there irony there? Just let me get my Kagool and binoculars and see if I can find it.

 

[DerKaiser]

Most correct answer goes to......Homer.

A binomial distribution will give the likelihood of the number of times a particular set of numbers is likely to be chosen. The only parameters you need are:
the chances of a single line winning (1 / 8,145,060), and
the number of lines done (2,000,000 in this example).

Running the numbers there is:
78.2% chance no one will win
19.2% chance of one winner
2.4% chance of two winners
0.2% chance of three or more winners

The odds of six or more winners in this example are one in four million i.e. with two draws a week it should occur about once every 40,000 years.

Interestingly (I use that word loosely!) if double the number of people played in a particular week, the scenario would be 50 times more likely.

Of course, this example is completely academic as the process by which numbers are selected is not independent. 5 of Saturday's winning numbers form a column on the ticket.

As well as the (apparently) astronomical odds, the fact that there was no quick pick winners amongst the 6 was a very good indication that there was some human pattern at work.

 

[cremeegg]

Sorry Kaiser, but you have not addressed the challenge you originally set. You have not explained your answer, to say you looked it up in a table does not explain anything to someone who is not already familiar.

The first thing to consider is in how many ways can 6 numbers be chosen from 45. Those numbers are way to big to illustrate, so lets ask how many ways can we choose 2 letters from 4. For example a,b,c,d. Well the possibilities are ab,ac,ad,bc,bd,cd. That is 6 possibilities. You can try yourself to see if there are any others (I don't think you will find any).

This is calculated as 4 choices for the first letter, (a or b or c or d) then there are 3 choices for the second (you cannot choose the same one again). So that gives us 12 possibilities 4x3=12. However some of these are the same ab is the same choice as ba.

So we need to divide by the number of ways we have double counted the first choice 2 and the second choice 1.

The answer is 4x3 divided by 2x1 =6

Similarly the number of ways we could choose 3 out of 8 is 8x7x6 divided by 3x2x1 = 56

Or the number of ways to choose 6 from 2 million is

2,000,000 x 1,999,999 x 1,999,998 x 1,999,997 x 1,999,996 x 1,999,995 divided by 6 x 5 x 4 x 3 x 2 x 1 all of which is equal to 8.8888222e34 or approx 8 with 34 zeros after it.

That very large number is the number of ways the winning numbers could appear out of the drum.

The next issue is the chances of getting any one number correct. This is 1 divided by 45. Just like the probability of getting a coin toss correct is 1 divided by 2.

What are the chances of getting two coin tosses correct. Each one is a half, the probability if two in a row is a half multiplied by a half, thats a quarter. To illustrate, say you guessed two heads, well the possible results of tossing a coin twice are HH,HT,TH,TT. One out four of these results matches your guess.

So the probability of getting 6 correct out of 45, is 1 divided by 45 multiplied by itself 6 times. That is 1 in 8,145,060. The probability of 6 people getting this is this number multiplied by itself 6 times.

What about the probability of not getting 1 correct out of 45 that is 1-(1/8,145,060). In other words if there is 1 chance in 45 of a correct number then chances of an incorrect number is 1 minus 1 in 8,145,060. And this could happen not 6 ways but 2m minus 6 that is 1,999,994. So (1-(1/8,145,060)) multiplied by itself 1,999,994 times. That works out as 0.782276684

So the chance of there being exactly 6 winners is 8.88882222e34 x (1\8,145,060)^6 x 0.782276634 = 0.000000238 or 1 in 4,199,147


In summary

The number of possible results from 2,000,000 lines is 8.88882222 by 10 to the 34

The chances of getting any one number correct 1 in 45. Of getting each of the six correct is 1 in 45 multiplied by itself 6 times, or 1 in 8,145,060

The chances of 6 people getting each of the six correct is that multiplied by itself 6 times

We need to allow for the probability of not getting 1 correct from the 45 which is 0.782276634

So as above

8.88882222e34 x (1\8,145,060)^6 x 0.782276634 = 0.000000238 or 1 in 4,199,147

This is the probability of their being exactly 6 winners.

 

[cremeegg]

Of course, this example is completely academic as the process by which numbers are selected is not independent. 5 of Saturday's winning numbers form a column on the ticket.

Are you suggesting that the numbers players pick are not independent, which is a reasonable thing to say.

Or are you suggesting that the way the winning numbers are picked is not independent. That would be a serious accusation against the organisers.

To say that "5 of Saturday's winning numbers form a column on the ticket" points towards the idea that the picking of the winning numbers is not independent. Is that what you are saying?

 

[Duke of Marmalade]

Most correct answer goes to......Homer.

Fascinating!! I wish to object to the hasty decision, though. Homer has done well, indeed he got near explaining this phenomenon but he didn't finish the job.

DerKaiser your math suggests 4M to one against i.e. once in 40,000 years and this happened before

I agree your math but your model just has to be wrong. No way did a 1 in 40,000 year event happen twice since the Lotto began. If that were the case we may as well revisit Solvency II.

Homer nearly got it when he recognisd that well spaced out numbers are more likely to be selected. And indeed, apart from 4,5, these numbers more or less appear in each of 6 consecutive partitions of 45 numbers.

So consider this model (extreme of course but clearly not entirely invalid).

Every selection partitions the 45 numbers into 6 consecutive packets and then selects randomly from these packets. If then a result actually came up which was indeed so spread out the chances of having been successful are 1 in (45/6)^6 i.e. c. 1 in 180,000.

This would give a poisson parameter of 11 and near certainty that there would be more than 6 winners.

The true model of Human Lotto Selection Behavour will be somewhere between purely random and full Spreading with other cluster behaviours such as less than 31 (birthdays) and indeed 6 in a row.

The test of this theory would be to look back at the last "1/40,000 year" occurrence to see if the numbers that came up then were noticeably HLSB friendly.

So the moral is not to spread your selections nor indeed to do the opposite. Try to be truly random - the answer is Quickpick. As you have already pointed out there were no QP winners in the HLSB friendly result. That's what you want, a strategy that sets you apart from cluster behaviour.

BTW I am prepared as a gesture of good will to share the honour with Homer.

(I have just spotted that you essentially have made the same point with that observation about the columns)

 

[Duke of Marmalade]

Are you suggesting that the numbers players pick are not independent, which is a reasonable thing to say.

Or are you suggesting that the way the winning numbers are picked is not independent. That would be a serious accusation against the organisers.

To say that "5 of Saturday's winning numbers form a column on the ticket" points towards the idea that the picking of the winning numbers is not independent. Is that what you are saying?

Yes that is what he is saying and I have taken that proposition to its extreme and shown that if people did behave in this way it would be almost certain that there would be a plethora of winners when indeed the random outcome obliged in a similar fashion.

I note your own mathematical exposition. That looks like New Math to me but since you arive at an answer similar to the Old Math i.e. c.4m to 1 against this occurrence (assuming fully independent selections) then I guess the New Math is not that much different from the Old.

 

[Sunny]

Very interesting but at the end of the day 'it could be you'.......

 

[PMU]

A binomial distribution will give the likelihood of the number of times a particular set of numbers is likely to be chosen.

Derkaiser is right in that it is a binomial distribution, but it’s not the likelihood of the number of times a particular set of numbers is likely to be chosen.

The question is, in 2,000,000 trials (i.e. the number of lines entered in the Lotto according to DerKaiser in post #1), what is the probability of there being 5 other wins if you already have a win, given that the probability of a win is (1/8,145,060).

Now my maths aren’t good enough to work this out, but I suspect this is where the 1 on 48.5 million comes form.

By the way probability is not real maths, it is maths for trainspotters.

. Brilliant - I love it.

 

[DerKaiser]

I agree your math but your model just has to be wrong. No way did a 1 in 40,000 year event happen twice since the Lotto began. If that were the case we may as well revisit Solvency II.
.

Interestingly (again using the term loosely), the previous highest number of winners in a single draw since the selection was increased to 45 numbers was three. This type of thing should happen once every five years.

When the selection was lower, this would not have been as rare an occurence e.g. with 36 numbers having six winners would have been a one in fifteen year event.

So the moral is not to spread your selections nor indeed to do the opposite. Try to be truly random - the answer is Quickpick. As you have already pointed out there were no QP winners in the HLSB friendly result. That's what you want, a strategy that sets you apart from cluster behaviour.

Good moral - avoiding combinations that humans have a conscious or sub conscious predisposition to choose is a good strategy.

 

[DerKaiser]

Hi Cremeegg. I see that you've arrived at the right number and formula but some of the logic is a bit of a leap (I am more limited in English than maths!). It's extremely difficult to express in words how a probability function has been derived.

My best way of thinking about it is to first work out the probability a given set of numbers will be chosen if you play one line. There are over eight million combinations, all equally likely. Then look at the probability of a given set of numbers being chosen at least once if two million lines are played. This is still relatively simple as it is one minus the chance of not choosing the numbers two million times in a row. The tricky bit then is to split the probability the numbers get chosen into instances where thy are picked once, twice, etc, by two million players.

 

[jhegarty]

Each line has a 1 in 8.1 million chance of winning:
45 c 6 = 8,145,060

We play 2 million lines :
8,145,060 / 2,000,000 = 4.07253

So we will win 1 in 4 times.

But week need to Win 6 times all in one week:

4.07253 ^ 6 = 4562

So every 4562 draws it should happen. But this assumes an even distribution of all lines. Given no ones last week and 6 this week , we know this isn't the case.

 

[Duke of Marmalade]

But week need to Win 6 times all in one week:

4.07253 ^ 6 = 4562

This line is faulty. What you have calculated is (8m)^6/(2m^6)

The numerator is correct but the denominator is wrong. You have only 2mC6 combos of six numbers and this is the correct denominator. You need to divide 2m^6 by a factor of roughly 6! i.e. 720.

Let's see this in the simple example of two dice throws.

There are 36 possibilities. 11 of these have at least one six so the chances of at least one six is 11/36 but given that you have one six it is only 1 in a 11 that you have a second thus producing the right answer of 1 chance in 36 of two sixes. Intuitively one sees the flaw in your argument as follows. You start with 2m chances of a win but having got that win you have exhausted a lot of those chances and of course the more wins you demand the more exhausted the chances become.

You have correcty calculated the probabiity of there being a winner six draws in a row, which intuitively we feel is not too remote.

 

[Duke of Marmalade]

cremeegg I found your explanation so difficult to follow and yet you got the right answer so I decided to dissect your exposition to see where my confusion arose.

The first thing to consider is in how many ways can 6 numbers be chosen from 45. So we need to divide by the number of ways we have double counted the first choice 2 and the second choice 1.

The answer is 4x3 divided by 2x1 =6

Similarly the number of ways we could choose 3 out of 8 is 8x7x6 divided by 3x2x1 = 56

Or the number of ways to choose 6 from 2 million is

2,000,000 x 1,999,999 x 1,999,998 x 1,999,997 x 1,999,996 x 1,999,995 divided by 6 x 5 x 4 x 3 x 2 x 1 all of which is equal to 8.8888222e34 or approx 8 with 34 zeros after it.

I agree this result but it is not the answer to how many ways can you select 6 numbers from 45 as you started off purporting it to be. I think at times you get confused between the two 6s, the 6 winning lines and the 6 winning numbers.

The next issue is the chances of getting any one number correct. This is 1 divided by 45. Just like the probability of getting a coin toss correct is 1 divided by 2.

So the probability of getting 6 correct out of 45, is 1 divided by 45 multiplied by itself 6 times. That is 1 in 8,145,060. The probability of 6 people getting this is this number multiplied by itself 6 times

This is where I really lost you. This the correct answer for how many ways can 6 numbers be selected from 45 but your method seems all wrong. For a start 45^6 is 8,300million, 1,000 times greater than you stipulate. What you have (correctly) given is 45C6.

What about the probability of not getting 1 correct out of 45 that is 1-(1/8,145,060). In other words if there is 1 chance in 45 of a correct number then chances of an incorrect number is 1 minus 1 in 8,145,060. And this could happen not 6 ways but 2m minus 6 that is 1,999,994. So (1-(1/8,145,060)) multiplied by itself 1,999,994 times. That works out as 0.782276684

Again I agree the number but your reasoning seems all wrong. The chance of a correct line is 1/45C6 and so the chances of an incorrect line is 1-1/45C6.

So the chance of there being exactly 6 winners is 8.88882222e34 x (1\8,145,060)^6 x 0.782276634 = 0.000000238 or 1 in 4,199,147

Agreed and this is NCr/(LCs)^r*(1-1/LCs)^(N-r) where:
N is the number of lines submitted by punters (2M)
r is the number of correct entries (6)
L is the panel of Lotto numbers (45), and
s are the number of selections in a line (6)

 

[Duke of Marmalade]

[broken link removed]

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