What's the chance of a six-number run in a lottery

[GwydionM]

I've asked this before, but not in this forum. It must be rare and be pretty remarkable, but maybe someone here can calculate it exactly.

I'm thinking mostly about the British lottery, but they are all much the same, I think.

------------------------------
An old man who lives in the UK

 

[strongm]

>It must be rare and be pretty remarkable

Nope; there is as much chance of a 6 number run as any 6 numbers coming up

 

[vongrunt]

Standard formula for number of combinations (K items from set of N items):

       N!
C = ----------
     K!*(N-K)!

! is factorial thingy

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[small]select stuff(stuff(replicate('<P> <B> ', 14), 109, 0, '<.'), 112, 0, '/')[/small]

 

[Tarwn]

I dunno, I would think it would depend on how the numbers were chosen. I think it would be differant depending on whethe they were just ranomly generated numbes or those little white balls they pull out of the reverse vacuum thingie...

 

[Tarwn]

Andfor the record, if I wasn't lazy I would crack open the book on the shelf behind me or recalculate both of those methods - I obviously wouldn't remember it, c'mon, how long have some of you guys known me?

 

[craigsboyd]

Nope; there is as much chance of a 6 number run as any 6 numbers coming up

My initial reaction was the same. However, given that we are talking about the chances for an entire combination (permutation depending on whether position matters), and there are fewer consecutive combinations possible than non-consecutive combinations... Wouldn't that mean that the chances are not the same?

Also, would any of the factors that contribute to the Birthday Paradox come into play here?

Just thinking out loud here.

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[Salem]

I take it you mean
1 2 3 4 5 6
up to
44 45 46 47 48 49

So 44 different sequences out of a total of 13.9M (I think)

1:317814

Or 6000 years at the rate of one draw a week ;-)

--

 

[vongrunt]

> Wouldn't that mean that the chances are not the same?

If I understand correctly, "consecutive combination" is nothing but single permutation. There are K! more perms than combos (720 for 6-from-whatever) - chances to get that are veryvery thin and likely not applicable for lottery.

> Also, would any of the factors that contribute to the Birthday Paradox come into play here?

Not exactly... slots in BP are not unique, while numbers in lottery combo are always unique. There are some vague similarities if you play systems (basically raising K and progressively paying extra $$$ for that).

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[small]select stuff(stuff(replicate('<P> <B> ', 14), 109, 0, '<.'), 112, 0, '/')[/small]

 

[Mike Lewis]

Are we talking about the chances of a particular six-number run coming up? In other words, if I bet on, say, 10, 11, 12, 13, 14, 15, are we asking if this has more or less chance of winning than any non-consecutive string?

If that's the question, the answer must intuitively be no. After all, if any class of six-number sets were more or less likely to win than others, that fact would be widely known, and punters would routinely favour or avoid the numbers in question. That's clearly not happening.

Or, are we asking for the chances that any six-number run will come up? In other words, what are the chances of today's winning hand being any six-number run?

If that's the question, the probability is easy to calculate, but must in any case must be low, because only a small proportion of all possible outcomes match that condition.

Hope this makes sense.

Mike



__________________________________
Mike Lewis (Edinburgh, Scotland)

My Visual FoxPro site: www.ml-consult.co.uk

 

[strongm]

And this is why people keep playing the lottery - because the odds are sometimes counterintuitive ...

 

[johnwm]

It's Budget day in the UK. We should all remember that the lottery was invented as a way of taxing the innumerate!

________________________________________________________________
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[vongrunt]

run = all consecutive digits?


(sorry, my babel fish took day off)

If order does not matter (combination) then Salem said it right. N-K+1 = 49-6+1 against total number of combos.

If exact order matters (permutation) then probability is K! times smaller.

Btw chances to get twins (2 consecutive digits in combo) are surprisingly high.

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[small]select stuff(stuff(replicate('<P> <B> ', 14), 109, 0, '<.'), 112, 0, '/')[/small]

 

[ChrisHunt]

See my answer in thread1091-1127959

the chance of a sequence of s balls appearing in a lottery where d draws are made from b balls is...

(1+b-s) * d! * (b-s)!
---------------------
    b! * (d-s)!

Slotting some other numbers into this (hurrah for Excel!), the chance of a...

one-number sequence occurring is 600% (you get 6 of them every time)
two-number sequence occurring is (approx) 61%
three-number sequence occurring is (approx) 5%
four-number sequence occurring is (approx) 0.3%
five-number sequence occurring is (approx) 0.01%
six-number sequence occurring is (approx) 0.0003%

Those percentages being for the UK National Lottery - 6 draws from 49 balls. As vongrunt observes, the chances of a 2-sequence are surprisingly high.

-- Chris Hunt
Webmaster & Tragedian
Extra Connections Ltd

 

[TAnselm]

It's not all that difficult...
assuming that any number can start the sequence the odds of that number coming up are 1 to 1 (no odds)

After that the next number has to be within 5 of the starting number... so the odds are 5/N

Then the next number would also have to be within 5 however 1 number is already taken so the odds would be 4/N and so on down to 1/N.

Multiply it all out and you have your answer

5 * 4 *3 * 2 *1 / N*N*N*N*N

using the example above with the nunmber of balls being 49 the chances are 120 in 282,475,249

 

[strongm]

Erm ... no. To demonstrate, let us assume we only have 6 balls, numbered 1 to 6.

According to your maths, the odds off getting a complete sequence are

5 * 4 * 3 * 2 * 1 / 6 * 6 * 6 * 6 * 6 = 120/7776

which is roughly 1.54%

Whereas we can clearly see that the odds are actually 100%

 

[TAnselm]

You are absolutely correct... I forgot one key ingredient. As each ball is chosen there is one less to choose from.
so the formula is actually
5*4*3*2*1 /N-1 * N-2 * N-3 * N-4 * N-5
120/120 = 100%

so for my other example it would be
5*4*3*2*1 / 48*47*46*45*44
or 120 / 205,476,480
or 1 chance in 1,712,304

 

[strongm]

Still incorrect...

If I have 11 balls, numbered 1 to 11 and first select say 6 ...

Well, I've got a choice of 10 balls that are within 5 of that (and only 10 balls left, so that's 100% chance of the second ball being within 5)

And, of course, the odds on the next ball being from a possible sequence of six are very much dependant on what the second ball was, not simply on whether it is within 5 of the first ball. e.g if second ball is 1 then remaining balls cannot be 7,8,9,10 or 11 even though they are within 5 of the first ball (our 6) if we want a sequence.

 

[TAnselm]

You are right... Hmmm, have to think harder.
The first ball is of course 1 to 1 or 100%.
Depending on where that ball lands in the sequence, the next ball could be anywhere in the range of 5 to 10 out of N-1.

After that of course my formula works because then no matter what the second ball was the sequence is set. So from then on, there are only 4 balls total that could remain in the sequence, making the odd 4 out of N-2, then 3 out of N-3, 2 out of N-4 and of course 1 out of N-5.

But seeing as the first ball could be anywhere in the sequence, how in the hell can you come up with the odds for the second ball? Unless of course you allow for the roll-over option. In other words the sequence could run like 48, 49, 1, 2, 3, 4. Then the odds of the second ball would be set as 10 out of N-1. Any thoughts????

 

[Olaf Doschke]

TAnselm,

there is no formula for the chance of the Nth ball being correct for an all-in-all 6-in-a-row result.

So it's much easier looking at the number of possible results and the permutations of that special result seperately.

The chances of 1,2,3,4,5,6 being drawn in that order are 1/49*1/48*...*1/44, as the chance for the first number to be one is 1/49, the chance for the second to be 2 is 1/48 and so on. The same chance goes with any special result.

If the numbers drawn are (1,2,3,4,5,6) then there are 6*5*4*3*2*1 possible ways to do so (permutations). 1 being the first drawn, 2 being second etc. is only one of those permutations.

As there are 44 straight-in-a-row sequences from 1-6 to 44-49, the chances are 44 of about 14 million. Or 1:317814.

With your roll-over option, there could be 49 sequences. That's all to it. 1:285384.

Your formula then would still not hold. (49*6*5*4*3*2*1)/(49*48*47*46*45*44) reduces to (6*5*4*3*2)/(48*47*46*45*44), not to (5*4*3*2/48*47*46*45*44). Where is that factor 6 lost?

The odds for the second number are only 5/48, if you want the first drawn ball to have the first number of the sequence. Taken this into account your result must be multiplied by 6, because the first drawn ball could be any of the six numbers of the result. Otherwise your thinking would then be okay, the roll-over option considered.

If the first ball drawn must be the start of the sequence, chances for the following balls drawn are 5/48, 4/47... down to 1/44.

Bye, Olaf.

 

[Zathras]

As Salem has pointed out, there are a total of 44 possible 6-ball sequences. Now, each of those sequences can occur in any of 720 ways (6!) for a total of 31,680 ways to draw the 6 balls.

Given that there are 49 balls to choose from, there are 49 x 48 x 47 x 46 x 45 x 44 ways to draw any 6 balls (49!/6!) for a total of 10,068,347,520 possible outcomes.

Dividing 31,680 by 10,068,347,520 gives a probability factor of 3.1464944904881471552543311496662e-6 which rounds to .000003146 which matches the result of .0003% posted by ChrisHunt. This also agrees with the result of 1 chance in 317,814 as posted by Salem.

Can we give it a rest now?