# Lottery probability question:  consecutive numbers

**URL:** <https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470>\
**Category:** Factual Questions\
**Created:** [January 24, 2005, 9:28pm UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470 "2005-01-24T21:28:01Z")\
**Posts on this page:** 20\
**Page:** 1

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**Author:** ![Earl\_Snake-Hips\_Tucker](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/earl_snake-hips_tucker/32/3269_2.png) [@Earl\_Snake-Hips\_Tucker](https://boards.straightdope.com/u/Earl_Snake-Hips_Tucker)\
**Post date:** [January 24, 2005, 9:28pm UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/1 "2005-01-24T21:28:01Z")

</div>

I heard some people who regularly buy lottery tickets saying that it’s interesting how often two (or possibly more) of the numbers selected are consecutive. What are the odds that in selecting numbers from 1-53, with no duplications, that at least two of them will be consecutive. (Not necessarily _selected_ consecutively, though.)

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [January 25, 2005, 12:54am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/2 "2005-01-25T00:54:39Z")

</div>

That’s a tough question.

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<div class="post-metadata">

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [January 25, 2005, 1:22am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/3 "2005-01-25T01:22:25Z")

</div>

I’m not sure how to approach the problem from an analytical standpoint, but [sub]53[/sub]C[sub]5[/sub] is a fairly small number, so brute force is fine and dandy here. There are 2,869,685 possible outcomes, and 536,130 of them have at least two consecutive numbers (not considering the order in which the numbers were picked). That works out to a probability of a little over .18.

For those who might want to check for errors, here’s my code:

```auto

#include "stdafx.h"
#include "math.h"
#include "windows.h"

// Code taken from http://doodleproject.sourceforge.net/algorithms/

bool enumerateCombinations(int n, int k, int j[])
/*-------------------------------------------------------------------
 Description:
   Enumerates all possible combinations of choosing k objects from n
   distint objects. Initialize the enumeration by setting j[0] to a
   negative value. Then, each call to enumerateCombinations will
   generate the next combinations and place it in j[0..k-1]. A
   return value of false indicates there are no more combinations to
   generate. j needs to be allocated with a size for at least k
   elements.

 Language:
   C++

 Usage:
   int comb[10] = {-1};
   while(enumerateCombinations(10, 5, comb)){
       // do something with comb[0..4]
   }

 Reference:
   Tucker, Allen. Applied Combinatorics. 3rd Ed. 1994.
-------------------------------------------------------------------*/
{
    int i;
    if(j[0] < 0){
        for(i = 0; i < k; ++i){
            j* = i;
        }
        return true;
    } else {
        for(i = k - 1; i >= 0 && j* >= n - k + i; --i){}
        if(i >= 0){
            j*++;
            int m;
            for(m = i + 1; m < k; ++m){
                j[m] = j[m-1] + 1;
            }
            return true;
        } else {
            return false;
        }
    }
}

void insertionSort(int array[], int lo, int hi)
/*-------------------------------------------------------------------
 Description:
   Sorts the elements of array[lo..hi] into ascending numerical order
   using an insertion sort.

 Language:
   C

 Usage:
   insertionSort(array, 3, 6); Sort elements 3 through 6 of array.
-------------------------------------------------------------------*/
{
    int i;
    int j;
    int m;

    for(i = lo + 1; i <= hi; ++i){
        m = array*;
        j = i - 1;
        while(j >= lo && array[j] > m){
            array[j + 1] = array[j];
            --j;
        }
        array[j + 1] = m;
    }
}

// My code

void check( int j[], int n, unsigned long& x )
{
	bool b = false;
	for ( int k = 0; k < n - 1; ++k )
		if ( (j[k + 1] - j[k]) == 1 )
			b = true;
	if ( b )
		x++;
}

int main()
{
	unsigned long x = 0;
	int j[5];
	while ( enumerateCombinations( 53, 5, j ) )
	{
		insertionSort( j, 0, 4 );
		check( j, 4, x );
	}

	printf( " X: %d
", x );

	return 0;
}

```

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<div class="post-metadata">

**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\
**Post date:** [January 25, 2005, 1:34am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/4 "2005-01-25T01:34:01Z")

</div>

First of all, events with a low probability that “look cool” (like the clock saying 12:34 or 11:11) have a much higher perceived probability, because you notice them and remember them. No one says “Wow! The average difference between any two pairs of winning lottery numbers for this week was 6.75, just like it was two weeks ago!” because your mind doesn’t automatically notice things like that.

Now, without using a computer (because I don’t know how), let’s say the first number you pick is n. Now, of the 52 possible numbers you could choose for your second number, (n-1) or (n+1) will form a consecutive pair. (Let’s say that 1-1=53 and 53+1=1) Already we have a probability of 2/52 = 0.0385. But assume you chose one of the other 50 numbers (m) instead. There are now three or four numbers that will form a pair, (n+1), (n-1), (m+1), and (m-1). In the case that |n-m| = 2 (which happens with probability 2/51 = 0.0392), there are only three. See where I’m going with this? If someone hasn’t already, I’ll finish up this line of thinking in a day or two.

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<div class="post-metadata">

**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\
**Post date:** [January 25, 2005, 1:35am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/5 "2005-01-25T01:35:39Z")

</div>

Yes indeedy. The thing is that you need to take into account case where some numbers are distributed _close_ (ie, only one number away) from other numbers. I think I can give an upper and lower bound, though (and **ultrafilter** can check me if I’m wrong).

Assuming all numbers wind up distributed with exactly one number in between (1-3-5-7-9, for example), I get the chance of NOT getting consecutive numbers as (51_49_47_45_43)/(52_51_50_49_48), and the chance of getting consecutive numbers as 27.11%. This should be a lower bound.

Assuming all the numbers get distributed with more than one number in between, I get the chance of NOT getting consecutive numbers as (50_47_44_41_39)/(52_51_50_49_48), and the chance of getting consecutive numbers as 48.35%. This should be an upper bound.

I suspect the real answer should lie closer to the upper bound than the lower bound, as numbers typically tend to not _all_ be clustered. I’ll take SWAG at 44%.

I’m curious what the actual answer is, though.

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**Author:** ![aahala](https://avatars.discourse-cdn.com/v4/letter/a/a88e4f/32.png) [@aahala](https://boards.straightdope.com/u/aahala)\
**Post date:** [January 25, 2005, 1:50am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/6 "2005-01-25T01:50:10Z")

</div>

I believe the answer is about 33%.

I thought this exact question was asked and correctly answered sometime last year, but I can’t find it with search.

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**Author:** ![Gfactor](https://avatars.discourse-cdn.com/v4/letter/g/9de053/32.png) [@Gfactor](https://boards.straightdope.com/u/Gfactor)\
**Post date:** [January 25, 2005, 2:00am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/7 "2005-01-25T02:00:11Z")

</div>

This person says that the answer for a 45 number lottery is 40/45C6

[http://www.aims.ac.za/~discus/discus/messages/8/178.html?1064862906](http://www.aims.ac.za/~discus/discus/messages/8/178.html?1064862906)

> [@](#):
>
> For South Dakota’s “Dakota Cash” in which one chooses 5 numbers without replacement from 1 to 35 (with no powerball) the chance of at least one pair of consecutive numbers is 1 - C(31,5)/C(35,5) or about .477. Likewise, the chance of at least one consecutive pair of numbers among the winning white Powerball Lottery numbers (i.e. excluding the powerball) is 1 - C(45,5)/C(49,5) or about .359.)

[http://silver.sdsmt.edu/~rwjohnso/module5.htm](http://silver.sdsmt.edu/~rwjohnso/module5.htm)

Apparently, betting consecutive numbers is a good way to avoid sharing one’s winnings.

[http://news.bbc.co.uk/1/hi/sci/tech/240734.stm](http://news.bbc.co.uk/1/hi/sci/tech/240734.stm)

But according to this source, it has no winning ticket has ever conained six consecutive numbers:

> [@](#):
>
> For example, six consecutive numbers (such as 1-2-3-4-5-6) have never been drawn in any state or international Lotto game.

[http://www.gailhoward.com/gailtips.htm](http://www.gailhoward.com/gailtips.htm)

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**Author:** ![Pasta](https://avatars.discourse-cdn.com/v4/letter/p/ecccb3/32.png) [@Pasta](https://boards.straightdope.com/u/Pasta)\
**Post date:** [January 25, 2005, 2:01am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/8 "2005-01-25T02:01:05Z")

</div>

Well, if you’d like _another_ answer…

I like to do these things statistically since it requires **no** clever thinking, which means it’s harder to mess up. Anyway, I just ran 10 million lotteries and asked how often I got at least 2 consecutive numbers. My answer:

prob = 0.46560 +/- 0.00016

where the error is due to finite statistics.

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**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\
**Post date:** [January 25, 2005, 2:01am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/9 "2005-01-25T02:01:20Z")

</div>

It is just as likely that two selected numbers will be consecutive as it is that they will be any other two numbers that are _ **selected in advance.** _

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<div class="post-metadata">

**Author:** ![Pasta](https://avatars.discourse-cdn.com/v4/letter/p/ecccb3/32.png) [@Pasta](https://boards.straightdope.com/u/Pasta)\
**Post date:** [January 25, 2005, 2:17am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/10 "2005-01-25T02:17:15Z")

</div>

By the way, I assumed you meant a 6-ball lottery…

Anyway, here’s what I get for the probabilities of getting at least _N_ consecutive numbers, for all six values of _N_. (100 million lotteries run. Again, errors shown are due to finite statistics.)

```auto

P(1) = 1.0 +/- 0.0
P(2) = 0.465583 +/- 0.000068
P(3) = 0.040926 +/- 0.000020
P(4) = 0.0024483 +/- 0.0000049
P(5) = 0.00009873 +/- 0.00000099
P(6) = 0.00000202 +/- 0.00000013

```

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**Author:** ![Freddy\_the\_Pig](https://avatars.discourse-cdn.com/v4/letter/f/a587f6/32.png) [@Freddy\_the\_Pig](https://boards.straightdope.com/u/Freddy_the_Pig)\
**Post date:** [January 25, 2005, 3:51am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/11 "2005-01-25T03:51:10Z")

</div>

It’s easy enough to verify the lower numbers in **Pasta** ’s table analytically. (I also assumed a six-ball game.)

There are 48 possible outcomes with six consecutive numbers, running from 1-6 through 48-53. So:

P(6) = 48 / 22,957,480 = 0.00000209

A string of 5 can begin with 1 through 49, with the sixth ball being any of 48 choices. But this doulbe-counts the strings of 6. So:

P(5) = 49\*48 / 22,957,480 - P(6) = 0.00010036. This is about one and one-half standard deviations greater than **Pasta** ’s number.

A string of 4 can begin with 1 through 50, with the fifth and sixth balls being anything. This double-counts the strings of 5 and triple-counts the strings of 6. So:

P(4) = (50_49_48/2) / 22,957,480 - P(5) - P(6) = 0.002459

By extension,

P(3) = (51_50_49\*48/6) / 22,957,480 - P(4) - P(5) - P(6) = 0.04098

However, this is a slight error, even though it agrees with the simulation. It double-counts the small epsilon of outcomes consisting of two discontinuous strings of 3. There are only about 1,000 of these, so it doesn’t much skew the answer.

But for P(2), there are loads of two-pair outcomes, and my methodology breaks down. Perhaps someone else can finish it.

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**Author:** ![ccwaterback](https://avatars.discourse-cdn.com/v4/letter/c/df705f/32.png) [@ccwaterback](https://boards.straightdope.com/u/ccwaterback)\
**Post date:** [January 25, 2005, 4:18am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/12 "2005-01-25T04:18:26Z")

</div>

I pondered this question a few years ago myself. I was thinking about the Illinois lottery, I believe at the time there were 50 balls to choose from and you had to match all 6 balls chosen. I’m not sure MY inquiry was exactly the same as the OP, my question was:

After the 6 lottery balls are chosen, what are the odds there will be AT LEAST two numbers that are consecutive?

So, this includes, all possible 6 consecutive number draws, all possible 5 consecutive number draws … all possible 2 consecutive number draws.

Note that a draw may have TWO 2 consecutive numbers, i.e. 10 11 17 23 24 48, so you need to take into account ALL of the permutations of this type of draw. Likewise, a draw may contain 3 consecutive numbers and 2 consecutive numbers, i.e. 12 13 19 20 21 43. So, you need to think about all the possible 2, 3, 4, 5, and 6 consecutive number permutations.

I got tired of trying to figure it out theoretically, so I wrote a program that generated all the possible combinations for the drawing. I belive my program calculated somewhere around 52 percent of the time the lottery drawing will contain AT LEAST 2 consecutive numbers. Perhaps the converse is easier to think about, that is, what is the probability NO two numbers will be consecutive in the drawing.

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**Author:** ![ccwaterback](https://avatars.discourse-cdn.com/v4/letter/c/df705f/32.png) [@ccwaterback](https://boards.straightdope.com/u/ccwaterback)\
**Post date:** [January 25, 2005, 6:16am UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/13 "2005-01-25T06:16:46Z")

</div>

Here’s a Basic4GL program that figures the percentage we are looking for. I know, it’s not a very elegant solution, but it works. 🙂

BTW, the answer for a “draw 6” lottery is:

46.5% for a 53 number pool  
47.3% for a 52 number pool  
48.0% for a 51 number pool  
48.7% for a 50 number pool (\<-- my recollection was wrong about this in my previous post)

'-----------------------------------------------------------------------------------------------------

const MAX\_NUM = 50

dim a  
dim b  
dim c  
dim d  
dim e  
dim f

dim con\_count  
dim total\_count  
dim percent#

a = 1  
b = 2  
c = 3  
d = 4  
e = 5  
f = 6

con\_count = 0  
total\_count = 0  
percent# = 0

while (a \<= MAX\_NUM - 5)

```
total_count = total_count + 1    

'printr
'print "N" + total_count + "a" + a + "b" + b + "c" + c + "d" + d + "e" + e + "f" + f 

if (a = (b-1)) or (b = (c-1)) or (c = (d-1)) or (d = (e-1)) or (e = (f-1)) then 
  con_count = con_count + 1
endif  

f = f + 1  

if (f &gt; MAX_NUM) then
e = e + 1
f = e + 1  
endif

if (e &gt; MAX_NUM - 1) then
d = d + 1
e = d + 1
f = e + 1
endif

if (d &gt; MAX_NUM - 2) then
c = c + 1
d = c + 1
e = d + 1
f = e + 1
endif  

if (c &gt; MAX_NUM - 3) then
b = b + 1
c = b + 1
d = c + 1
e = d + 1
f = e + 1
endif

if (b &gt; MAX_NUM - 4) then
a = a + 1
b = a + 1
c = b + 1
d = c + 1
e = d + 1
f = e + 1  

printr
print "ROLL"
endif  

```

wend

percent# = con\_count  
percent# = percent# / total\_count

printr

print "cc " + con\_count + " tc " + total\_count + " P " + percent#

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<div class="post-metadata">

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [January 25, 2005, 4:18pm UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/14 "2005-01-25T16:18:09Z")

</div>

I just realized something–the OP doesn’t specify how many numbers are drawn. I was working off 5 earlier, but with 6, here’s what I get:

P(6) = 10685968/22957480  
P(5) = 939672/22957480  
P(4) = 56448/22957480  
P(3) = 2304/22957480  
P(2) = 48/22957480

P(n) is the probability of getting at least n consecutive numbers in a sorted combination.

Here are the relevant changes in my code:

```auto

	unsigned long x = 0;
	int j[6] = { -1 } ;
	while ( enumerateCombinations( 53, 6, j ) )
	{
		insertionSort( j, 0, 5 );
		check( j, 6, x );
	}

	printf( " X: %d
", x );

```

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<div class="post-metadata">

**Author:** ![aahala](https://avatars.discourse-cdn.com/v4/letter/a/a88e4f/32.png) [@aahala](https://boards.straightdope.com/u/aahala)\
**Post date:** [January 25, 2005, 4:24pm UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/15 "2005-01-25T16:24:35Z")

</div>

My guess was based upon 5 balls also. I had roughed out a mathematical WAG. Then since 53 balls were in powerball, I looked at last years results, that’s how I got my guess of .33, which was the approximate frequency for last year.

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<div class="post-metadata">

**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\
**Post date:** [January 25, 2005, 4:53pm UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/16 "2005-01-25T16:53:40Z")

</div>

> [@ultrafilter](#):
>
> I just realized something–the OP doesn’t specify how many numbers are drawn. I was working off 5 earlier, but with 6, here’s what I get:
> 
> P(6) = 10685968/22957480  
> P(5) = 939672/22957480  
> P(4) = 56448/22957480  
> P(3) = 2304/22957480  
> P(2) = 48/22957480
> 
> P(n) is the probability of getting at least n consecutive numbers in a sorted combination.

I think you got a transposition there, pardner. How about:

P( **2** ) = 10685968/22957480  
P( **3** ) = 939672/22957480  
P( **4** ) = 56448/22957480  
P( **5** ) = 2304/22957480  
P( **6** ) = 48/22957480

---

<div class="post-metadata">

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [January 25, 2005, 5:51pm UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/17 "2005-01-25T17:51:47Z")

</div>

Thanks.

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<div class="post-metadata">

**Author:** ![nivlac](https://avatars.discourse-cdn.com/v4/letter/n/3bc359/32.png) [@nivlac](https://boards.straightdope.com/u/nivlac)\
**Post date:** [January 25, 2005, 6:50pm UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/18 "2005-01-25T18:50:17Z")

</div>

> [@Earl Snake-Hips Tucker](#):
>
> What are the odds that in selecting numbers from 1-53, with no duplications, that at least two of them will be consecutive. (Not necessarily _selected_ consecutively, though.)

The quick solution (without resorting to writing code) is  
1-C(48,6)/C(53,6) = .46546781 if you’re drawing 6 numbers from 53, or  
1-C(49,5)/C(53,5) = .33550756 if you’re drawing 5 numbers from 53.  
It’s easier to work with the complementary event.

---

<div class="post-metadata">

**Author:** ![pretend\_my\_name\_is\_witty](https://avatars.discourse-cdn.com/v4/letter/p/b9e5f3/32.png) [@pretend\_my\_name\_is\_witty](https://boards.straightdope.com/u/pretend_my_name_is_witty)\
**Post date:** [January 25, 2005, 7:01pm UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/19 "2005-01-25T19:01:29Z")

</div>

My Notation “P2[sup]nd[/sup][sub]cons[/sub]1[sup]st[/sup]” means the probability of the 2nd being consecutive to the 1st

If we let 49 & 1 be ‘consecutive’, then we can say that the probability of the 2nd ball being consecutive to the first is 2/48, as there are going to be two of the remaining 48 balls that are consecutive to the first one picked.  
P2[sup]nd[/sup][sub]cons[/sub]1[sup]st[/sup]" = 2/48  
= 0.041666666  
The probability of the 3rd being consecutive to the 2nd is going to be 2/47 MINUS the probability that one of these two balls was removed on the first ball (2/47 x 1/49)  
P3[sup]rd[/sup][sub]cons[/sub]2[sup]nd[/sup] = (2/47) - (2/47.49)  
= 96/2303  
= 0.041684759

The probability of the 4th being consecutive to the 3rd is going to be 2/46 MINUS the probability that either of these two balls was the 1st OR 2nd ball:  
P4[sup]th[/sup][sub]cons[/sub]3[sup]rd[/sup] = (2/46) - [(2/46 x 1/49) + (2/46 x 1/48)]  
= 2255/54096  
= 0.041685152

_and so on…_  
P5[sup]th[/sup][sub]cons[/sub]4[sup]th[/sup] = (2/45) - [(2/45 x 1/49) + (2/45 x 1/48) + (2/45 x 1/47)]  
= 0.041665862

P6[sup]th[/sup][sub]cons[/sub]5[sup]th[/sup] = (2/44) - [(2/44 x 1/49) + (2/44 x 1/48) + (2/44 x 1/47) + (2/44 x 1/46)]  
= 0.041624671

Probabilities of being consecutive to the preceding number:  
2nd = 0.041666666  
3rd = 0.041684759  
4th = 0.041685152  
5th = 0.041665862  
6th = 0.041624671

total = 0.20832711

So roughly every 1 in 5 draws of the UK National Lottery will have at least one set of consecutive numbers (including 1 & 49).

Now that number seems about right to me, but I’m sure I can’t have got the calculations right. Someone will be along to correct me soon!

I only got an ‘E’ on my first attempt at Statistics 1 (A-Level) and a ‘C’ on my retake. I’m good enough at Pure Maths, but Stats is tedious (as proven above)!

Oh, and usually I wouldn’t work with decimals, hate the things! But my calculator wouldn’t work in fractions after the first couple.

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**Author:** ![pretend\_my\_name\_is\_witty](https://avatars.discourse-cdn.com/v4/letter/p/b9e5f3/32.png) [@pretend\_my\_name\_is\_witty](https://boards.straightdope.com/u/pretend_my_name_is_witty)\
**Post date:** [January 25, 2005, 7:04pm UTC](https://boards.straightdope.com/t/lottery-probability-question-consecutive-numbers/286470/20 "2005-01-25T19:04:51Z")

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Ok, now I realise that you wanted a lottery system with 53 balls. But the principle is the same, just adjust some of the numbers. Not doing it again. If it’s right, that is!

Cheers,

Harry

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