Suppose the winning combination consists of $7$ digits, each digit randomly ranging from $0$ to $9$. So the probability of $1111111$, $3141592$ and $8174249$ are the same. But $1111111$ seems (to me) far less likely to be the lucky number than $8174249$. Is my intuition simply wrong or is it correct in some sense?

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    $\begingroup$ There are 107=10,000,000\,10^7=10,000,000\; of possible combinations. I'd say your probability to win choosing one number is the same: 1107=0.0000001\,\frac1{10^7}=0.0000001\; , no matter of what number you choose... $\endgroup$ – DonAntonio Aug 14 '13 at 16:46
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    $\begingroup$ It assumed the \$ signs were LaTeX markers. You can get around that by escaping them: \\\$. $\endgroup$ – Christian Mann Aug 14 '13 at 17:06
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    $\begingroup$ @DonAntonio Careful: The hopeless gambler will continue the advice this way: "better to get a job so that you can afford more tickets." :) $\endgroup$ – rschwieb Aug 14 '13 at 18:27
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    $\begingroup$ Just a thought, but if the number did come up "1111111", which to flawed human brains seems extra improbable, there's a decent chance that the result would be thrown out. Surely such an "impossible" number is self-evidently the result of some kind of hacking! $\endgroup$ – Jon of All Trades Aug 14 '13 at 19:42
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    $\begingroup$ If you think about how ridiculously implausible it is that 1111111 would come up... that's exactly how you should be thinking about any number -- ideally before you spend good money on a ticket. $\endgroup$ – sh1 Aug 14 '13 at 22:28

19 Answers 19


Your intuition is wrong. Compare the two statements

A. The event "the lucky number has all its digits repeated" is much less probable than the event "the lucky number has a few repeated digits"

B. The number 1111111 (which has all its repeated digits) is much less probable than the number 8174249 (which has a few repeated digits).

A is true, B is false.

BTW, this can be related to the "entropy" concept, and the distinction of microstates-vs-macrostates.

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    $\begingroup$ I think the first statement explains why I got the wrong idea $\endgroup$ – arax Aug 14 '13 at 16:50
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    $\begingroup$ Leonbloy, I'd suggest you edit 2 by adding the word 'specific' before 'number like 8174249' to avoid confusion. There are 10 numbers of 10,000,000 like the 1111111, but millions that "[have] only a few repeated." (and +1, by the way) $\endgroup$ – JoeTaxpayer Aug 14 '13 at 19:57
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    $\begingroup$ I don't understand the second argument. You're finding a pattern $A = \{1111111, 2222222, \dots\}$ and saying it's much less probable, since $A << \tt{all}$, but can't you think up a pattern for any numbers? I might say that the chance of hitting my pattern of numbers $B = \{8471925, 6581824, 8571824, ...\}$ is just as unlikely as hitting your pattern $A$. Why does it matter if it's part of some pattern? Especially pattern $A$ is only present with base 10. Numbers are numbers; they don't care about representation. $\endgroup$ – kba Aug 15 '13 at 6:56
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    $\begingroup$ @deedo2392 - Let me put it this way. Suppose we replace the lottery ticket with just picking 1 person at random from the US population to win a million dollars. Now, it's very unlikely that this person is going to be, say, an albino midget. That's because there aren't many albino midgets out there in the population. But if you're an albino midget, you have the same odds as anyone else (and equal protection under the law, and equal human dignity, to boot). Likewise, there aren't a lot of 7-digit numbers with all the same digit. They're rare, but no less likely than any other number. $\endgroup$ – fennec Aug 15 '13 at 12:17
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    $\begingroup$ @deed02392: It's because there's only one way for digits to be equal, but there are 9 ways for digits to be unequal. Simple to see with numbers 00..99: 10 of those have equal digits, 90 don't. If those were lottery numbers, the chance of a price falling in the last group is 9 times bigger, but only because there are 9 times more tickets in that goup. $\endgroup$ – MSalters Aug 15 '13 at 14:07

You should never bet on that kind of sequence. Now, every poster will agree that the odds of any sequence from 000000000 through 999999999 has an equal probability. And if the prize is the same for all winners, it's fine. But, for shared prizes, you will find that you just beat 10 million to 1 odds only to split the pot with dozens of people. To be clear, the odds are the same, no argument. But people's bets will not be 100% random. They will bet your number as well as a pattern of 2's or other single digits. They will bet 1234567. I can't comment whether pi's digits are a common pattern, but the bottom line is to avoid obvious patterns for shared prizes.

When numbers run 1-50 or so, the chance of shared prizes increases when all numbers are below 31, as many people bet dates and stick to 1-31. Not every bettor does this of course, but enough so shared prizes show a skew due to this effect.

Again - odds are the same, but human nature skews the chance of split payout. I hope this answer is clear.

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    $\begingroup$ I had always thought there's nothing to be studied about lottery... $\endgroup$ – arax Aug 14 '13 at 16:57
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    $\begingroup$ So now I'll wind up sharing my PowerBall winnings with a bunch of MathSE readers? $\endgroup$ – DJohnM Aug 14 '13 at 17:24
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    $\begingroup$ Odd; I would have thought that superstition would cause fewer people to bet "1111111111" than average. But, I suppose, I haven't done any actual studies of what people pick. $\endgroup$ – Hurkyl Aug 14 '13 at 17:51
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    $\begingroup$ This happened in Florida a couple of years ago. The pick 5 was 14-15-16-17-18, and 47 people had to share the prize. $\endgroup$ – Ben Miller Aug 14 '13 at 21:29
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    $\begingroup$ "I can't comment whether pi's digits are a common pattern" Sure you can. They are. You just need to go far enough through the decimal representation to find them. $\endgroup$ – RoadieRich Aug 15 '13 at 12:26

Your feeling is incorrect, but there is more to it.

It is in the interest of the lottery organizer for the lottery to be fair (because they have much more to lose in a scandal than they can gain by cheating). Thus it is fairly safe to assume that the lottery combinations are indeed drawn with a uniform distribution, which is to say that all combinations are equally likely. So you are wrong to think that 1111111 is less likely to be drawn than 8174249. Both are equally likely.

Many people are like you, they think some combinations are special, and that these are either more likely or less likely to appear. Your example is 1111111, you find it less likely. Some people find last week's combination to be less likely. Some people think more likely the combination made from those numbers that have occurred most in the past.

My non-scientific explanation of this goes as follows: people's brains automatically look for patterns everywhere. When a pattern is recognized in a thing, the thing gets categorized as "special" and "worthy of attention". This happens with lottery combinations too: any combination that has an obvious pattern, or follows some rule that is easy to describe, will be categorized by our brains as "special". Such special combinations will then be deemed less likely to appear.

In other words, humans are quite bad at dealing with randomness because they cannot help themselves from seeing patterns where there are none.

So, should you play 1111111, or should you avoid 1111111? To answer the question we have to take into account the fact that the prize is shared among all who guessed it. Now, since people are unable to generate random combinations well they tend to play combinations with recognizable patterns: visually or arithmetically pleasing combinations, birth dates, telephone numbers, etc. This seriously skews the combinations that are actually played. For instance, numbers above 31 are less likely to appear, while numbers below 13 are more likely to appear, because people play birth dates.

The upshot of this is that if you play a combination that your brain recognizes as special, and you happen to win, then you will have to share the prize with lots of other people whose brains thought of the same combination. In this sense, even though 1111111 is equally likely as all other combinations, the expected profit is smaller because we know that many other people will play the same combination.

The best strategy to play the lottery is to not play it, because the game is rigged so that your expected profit is negative. However, you may not care about this. For instance, you find pleasure in dreaming about what you would do with the prize, and so you are willing to pay something for it. (This is a perfectly legitimate reason for playing the lottery, I pity those who play because they actually think they can come ahead.)

Anyhow, if you do play the lottery, you should not play obvious combinations, or anything that can be described in one sentence, such as "the birthdays of my pets, increased by 5" (yes, there is going to be someone who has pets born on the same days as you, and who also thinks 5 is his lucky number). By the same reasoning, you should not avoid special combinations because that can be described by "Do not play a combination that has a nice pattern" (many people will use this strategy). The safest procedure is to choose a random combination, and use it no matter what your brain is telling you about its likelyhood. So even if you throw dice and get 1111111, you should use it.

Many lottery organizers will give you the option of choosing a random combination for you. It is in their interest to convince people that they should play randomly chosen combinations, because of the possible fiasco when a pretty combination gets drawn and there are several dozen winners. You should use the organizers random number generator if you believe their programmers are competent enough to get them right. History shows that this is often not the case. For instance, there have been a number of security problems on the web because various components (servers, browsers) used bad random number generators. Just throw dice.

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    $\begingroup$ And if you use dice, just make sure they are n-sided dice according to the range of digits allowed for numbers in your chosen lottery. No good to use a standard 6-sided die if the numbers can range 1..59 :) $\endgroup$ – Jeffrey Kemp Aug 15 '13 at 7:30
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    $\begingroup$ My state currently uses the range 1-54. But it's so hard to find the 54-sided dice in the stores. $\endgroup$ – Dan Aug 15 '13 at 14:05
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    $\begingroup$ I disagree with "you should not avoid special combinations". Even if the vast majority of players were afraid of having to share their prize with too many competitors, I think we can safely assume a small fraction p* of players preferring special numbers. These players will crowd on the sparse set of special numbers, whereas only the tiniest fraction of "special number avoiders" would need to throw their dice again because they landed on a special number. I assume that p* is no smaller than 1/100, at least enough to populate the special numbers by the dozens (see @Ben-Miller's comment). $\endgroup$ – quazgar Aug 15 '13 at 17:34
  • $\begingroup$ Let me rephrase: You should not do anything "special" because other gullible people who play lottery will think of the same special thing, and so you will have an increased probability of sharing the prize with them. Apart from not playing at all, the safest thing is to choose randomly. $\endgroup$ – Andrej Bauer Aug 19 '13 at 9:19
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    $\begingroup$ @Dan In that case, using most common RPG dice, I'd use a d10 (and throw out a result of 10) to get a digit from 1 to 9, and a d6 to get a digit from 1 to 6, and play 6(d10-1)+d6 as my number. $\endgroup$ – David Millar Aug 27 '13 at 21:30

These two statements are directly contradictory:

"So the probability of 1111111, 3141592 and 8174249 are the same."

"But 1111111 seems far less likely to be the lucky number than 8174249."

You cannot simultaneously believe that A is "less likely" than B, and that A and B have the same probability. This is regardless of whether or not the lottery is fair, or whether it is stacked in favor of some numbers.

If you translate this into mathematical terms, A is "less likely" than B is written $P(A) < P(B)$, and same probability is written $P(A) = P(B)$. That is to say, likelihood and probability are exactly the same thing.

We cannot have $X < Y \cap X = Y$; you must choose which side you believe.

Believing two contradictory statements is worse than believing in a falsehood. Believing in a falsehood could be the result of a mistake or deception, but holding contradictory statements to be simultaneously true is a flaw of reasoning.

However, consider security instead of a lottery: another area in which we reason about combinations, and where probability finds application. Suppose that you have some system of seven digit passwords, such as a numeric keyless entry, or some kind of mechanical padlock with a seven digit combination. Should you configure 1111111 as a combination, on the basis that they are all equally likely to be randomly guessed? Of course not; attackers will try such patterned combinations before doing a brute force search. If a brute force search is sequential, then a low number like 0012345 will be found earlier.

Do not mix up your intuition about what might be a good lock combination with probability in random events like lotteries. The way password spaces are attacked does not obey a uniform, random distribution, because the choice of combinations in the attack follows some cunning strategy driven by an intellect.

The lottery balls neither not prefer nor avoid "nice" numbers whose digits follow patterns. To believe that they do is to anthropomorphize the machines: endow them with human qualities, or to endow probability itself with intelligent qualities (like that it is driven by supernatural forces or beings which make choices that guide human fate).

There is one more angle to this and it is the mistake of interpreting the probability of a pattern with that of a single instance of a pattern. Suppose we are dealing with seven digit numbers whose digits are 0-9. A number with all digits which are the same might be called "seven of a kind". There are ten seven-of-a-kind numbers, which makes them seem rare. On the other hand, say, numbers in which all digits are different are far more numerous: $\frac{10!}{3!} = 604800$. So in fact it is far less likely that a randomly drawn number will be one of the ten sevens-of-a-kind, than that it will be one of the 604800 all-digit-uniques. This can lead to the wrong intuition: because 1111111 is part of a set (seven of a kind) which is rare, you might think that it is less likely. Our intuitive reasoning is that we impart a category's property on the individual member: if a number is part of a rare group, the number itself is regarded as rare. However, a number's membership in a rare set has no bearing on the likelihood of a specific number; such subset membership is just a categorical view that we impose on the structure of the numbers. Each number is equally "rare", simply because it is distinct from all others, and the random choice is not biased by categories like seven-of-a-kind.

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    $\begingroup$ I dislike that you're calling out on OP's contradictory statements when they have just posted a question on MSE to try and get this contradiction resolved. $\endgroup$ – Lord_Farin Aug 15 '13 at 9:59
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    $\begingroup$ What the Lord said. Also, he said it "seems" less likely. That means he recognizes that it shouldn't be so, but it seems like it is, and so he wants to understand why that is $\endgroup$ – Ray Aug 15 '13 at 13:28
  • $\begingroup$ @Ray Hence the answer "seems" like it is not useful and deserves a downvote, I see. $\endgroup$ – Kaz Aug 15 '13 at 14:00
  • $\begingroup$ @Lord_Farin Note that in the last section of the answer, I give a psychological hypothesis about why 1111111 seems unlikely. It could be because it is an element of a set ("seven of a kind" numbers) and that set is unlikely compared to some other sets, like numbers with all seven digits distinct. There is a mode of reasoning which imparts the properties of a set that something belongs to, to that something. Often, this reasoning is correct: after all, instances take attributes from classes. And often, the reverse is wrong: generalizing to classes from individuals. $\endgroup$ – Kaz Aug 15 '13 at 14:16
  • $\begingroup$ I like your issue of n-of-a-kind argument. But, I did not get why all-1 is equiprobable with random digit number, after you have aptly proven that the probabilities are different. $\endgroup$ – Val Aug 15 '13 at 18:03

Your intuition is indeed wrong. It is correct in the sense that it's true that getting seven 1s in row is indeed very unlikely. But it's incorrect to think that it's more unlikely than any other 7-digit number.

Another way to think about this is that base 10 is completely arbitrary. Imagine you were an alien with 8 fingers. In base 8, your number 1111111 is 4172107 (according to the handy calculator here). Now do you think that the same number in base 8 is more or less likely?

  • $\begingroup$ Can the downvoter comment? Good to know what I got wrong / could do better. $\endgroup$ – TooTone Aug 14 '13 at 17:43
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    $\begingroup$ (Not my vote) To that alien, the distribution of each digit would NOT be random at all. Well, the least significant digit would be almost random, but the chance of the first digit being 0 would be far greater than the chance of it being 7. $\endgroup$ – MSalters Aug 15 '13 at 14:14
  • $\begingroup$ @TooTone Commenting and downvoting runs the risk of serial revenge downvoting, which is one reason some users do just one or the other. $\endgroup$ – Kaz Aug 15 '13 at 14:19
  • $\begingroup$ @MSalters thanks, you mean like Benford's law. My answer is strongest w.r.t the distribution of the number as a whole, and weakest w.r.t the distribution of the individual digits. The latter viewpoint is natural in lotteries such as that in the UK where balls are drawn one after the other. Some of the other answers make some v good points about the relative quantity of numbers with and without repeated digits. $\endgroup$ – TooTone Aug 15 '13 at 14:21
  • $\begingroup$ @TooTone: No, I don't mean Benford's law. I'm simply referring to the fact that 7777777777 octal is not a possible outcome, but 0000000000 octal is. $\endgroup$ – MSalters Aug 15 '13 at 14:30

A couple of people have commented on how to increase the odds that you won't have to share your lottery winnings with other people, so it's worth mentioning a book on precisely this: How to Win More, by Norbert Henze and Hans Riedwyl. Here's a brief review, written by David Aldous:

Despite the title, this is a well written and serious book on the modern "pick 6 numbers out of 49" type of lottery. Of course you can't affect your chance of winning but you can try to choose unpopular number combinations to maximize your share if you do win. Uses empirical data from around the world to describe "foolish ways to play" (based on previous winning or non-winning numbers, patterns, etc) -- what makes these foolish is simply that too many other people use them. Concludes with a non-obvious recommendation: choose randomly subject to several constraints (one of which requires a bit of math to understand: a quantified measure of non-arithmetical-progression). An appendix has some upper-division college level math probability analysis, but non-math folks can just ignore it.

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    $\begingroup$ So, the better the book sells, the less helpful it will be? $\endgroup$ – DJohnM Aug 14 '13 at 22:21
  • $\begingroup$ @User58220 LOL, well, the harder it will be to find the constraints that will maintain your low sharing probability $\endgroup$ – Oliver Aug 15 '13 at 14:31

I think your intuition is right in some cases. For example, it may be likely that other people have chosen $1111111$ and you would be forced to split the prize. And what if the lottery is rigged against such numbers being chosen in order to defend against allegations of corruption? I suppose that itself would be a form of corruption but if $1111111$ is chosen someone might say the lottery isn't really random and the lottery officials would be in hot water.

But if it is just a simple situation of trying to guess a uniformly random number then of course it doesn't matter.


The reason $1111111$ seems less likely is because it is part of an easily identifiable pattern, and the pattern itself is less common than the patterns that you see in $8174249$.

For example $8174249$ belongs to the set of numbers between $0000000$ and $9999999$ which have no repeated digit. That set is quite large, it has $10 * 9^6 = 5314410$ numbers in it.

$5314410/10^7 = .531441$

So you have a greater than $50$% chance of the number having no repeated digits.

Whereas there are only $10$ numbers which are a single digit repeated $7$ times, so you have a very small ($10/10^7 = 0.000001$) chance of getting a number in that set.

So at this point, it seems like picking a number with no repeated digits is a much better choice, but the size of the set is so huge that you will end up with the exact same probability of winning. This is no coincidence.

If you pick a number with no repeated digits, you have a $.531441$ chance that the result will be in the same set, but there are 5314410 numbers in that set, so the odds of winning the lottery, given that the chosen number will have no repeated digits, are still $1$ in $5314410$. The probability of both events happening (You picking the right number out of the 5314410 choice, and the lottery system picking a number with no repeated digits) is exactly what you would expect your odds of winning to be: $.531441 * 1/5314410 = 0.0000001$

The odds of winning given that the lottery system picks a number with all repeating digits are quite high for a lottery, $1$ in $10$, you only have 10 choices to choose from! But the probability of that pattern being picked are so low, that the probability of both happening are exactly the same as the probability of $8174249$ being picked: $0.000001 * 1/10 = 0.0000001$

That is true for any pattern. The larger the set of numbers that fit the description of a pattern, the more likely it is that the pattern will be picked, but as it becomes more likely for that pattern to be picked it becomes less likely for you to pick the correct number in the pattern. It balances out perfectly like you might expect, and the odds of winning are the same no matter which number you pick.

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    $\begingroup$ As many have pointed out, the O.P.'s assertion "1111111 seems far less likely to be the lucky number than 8174249" is erroneous. However, had it been worded as: "But a number like 1111111 seems far less likely than a number like 8174249," then that could be true, depending on how we define "like". As you say, though, it's one thing to pick a lottery number in the right set; it's another thing to pick the winning number. $\endgroup$ – J.R. Aug 15 '13 at 10:46

Most of the answers are making two assumptions about the nature of the lottery being played. Firstly that order matters (that a drawing of "17, 23, 31" is not the same as a drawing of "23, 31, 17"), and secondly that balls are replaced (that you put back the "17" ball after it's been drawn, before you pick the next number).

Depending on the lottery being played, one or both of these assumptions may be wrong. Suppose you have balls numbered 1 through 49:

  • If order matters and balls are replaced, then "1, 1, 1" is as likely as "1, 17, 23".

  • If order doesn't matter and balls are replaced, then it depends on how many of each number there are. If there's one of each number, "1, 1, 1" is six times less likely than "1, 17, 23", as there' six different ways to make the latter (since "1, 17, 23" and "23, 17, 1" are the same), but only one to make the former. If there's three balls of each number, they're equally likely.

  • If order matters and balls are not replaced, then it depends on how many balls of each number there are. If there's one of each number, "1, 1, 1" is impossible. With three of each number, "1, 1, 1" is still less likely than "1, 17, 23", as, e.g. for the second number, there's two "1"s but three "17"s. (More specifically, there's six ways to draw "1, 1, 1" (3*2*1), but 27 ways to draw "1, 17, 23" (3*3*3)).

  • If order doesn't matter and balls are not replaced, then it depends on how many balls of each number there are. If there's one of each number, "1, 1, 1" is impossible. If there's three of each number, "1, 1, 1" is way less likely than "1, 17, 23" - there's six ways to draw "1, 1, 1", but 162 ways to draw "1, 17, 23" (9*6*3).

For example, in the UK national lottery, there is one of each number (from 0 to 49), balls are chosen without replacement, and order doesn't matter (giving a total number of possiblities of 49Choose6 = about 14 million). So "1, 1, 1, 1, 1" is impossible, and "1, 2, 3, 4, 5, 6" is as likely as "1, 3, 15, 27, 41"

  • $\begingroup$ Most lotteries in the US have either a large number of balls with no replacements, or else an N-digit number where the order matters, but, instead of doing replacements, it's actually N indepedently drawn digits, often with some kind of mechanical device like this one or these machines. (I understand "replacements" is a common term in probability, and that you weren't necessarily meaning that the balls were literally replaced.) $\endgroup$ – J.R. Aug 15 '13 at 17:12

There are $10^7$ ways of writing out a sequence of 7 values between 0 and 9. Imagine you have an infinite supply of each digit, each of them selected uniformly at random for each position in the sequence. Hence every sequence is equally likely unless sampling is somehow affected by the previous outcomes.


I think the source of the confusion is that human intuition lends itself to some very fallacious reasoning when it comes to probability (and large numbers). The fallacy is this, your intuition groups numbers into two categories: "nice" numbers, and not-so-nice numbers. Suppose we call a "nice number" any number that is just a full sequence of repeated digits. By all means, the probability of getting a nice number is far smaller than the probability of getting a not-so-nice number. Extend this to any number that "stands out" to our perception, and they're still outnumbered by numbers that don't.

The problem with that is, we're not choosing between $2$ "categories", but in fact $10^7$ outcomes.


Another argument in favor of choosing a random-looking number is the following: suppose 1111111 is drawn. The scientifically uneducated audience will likely complain that it "can't be random" and something went wrong (or maybe that you cheated), and in the end they'll have the draw cancelled or repeated.

Sadly, there's no arguing against that --- I bet you can convince the average judge and jury that "1111111 is not random".

(That said, in real life 1111111 is indeed more likely to appear than 8174249, since a mechanical or programming error in the drawing machine could make it more likely to have repeated numbers drawn than completely random ones).

In short, real life is not like mathematics. :)

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    $\begingroup$ "...in the end they'll have the draw cancelled or repeated." That is ludicrous. The idea that there would be widespread protest after a repeating digit lottery result, with the lottery commission then bowing to pressure and redrawing the number, is preposturous. That scenario seems even less likely than 1111111 being the winning number. I'm not sure I agree with your assertion that 1111111 is more probable than 8174249, either, since lotteries don't use computer programs to select winning numbers, for a number of obvious reasons. $\endgroup$ – J.R. Aug 15 '13 at 10:55
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    $\begingroup$ @J.R. "lotteries don't use computer programs" en.wikipedia.org/wiki/Category:Computer-drawn_lottery_games $\endgroup$ – Federico Poloni Aug 15 '13 at 12:07
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    $\begingroup$ Federico: Hmm, that's an interesting link. This crow tastes delicious. (To be honest, I'm a bit surprised; thanks for enlightening me.) I suppose that brings another element into the question, one that's outside the purely mathematical view. I'm still not sure that 1111111 would be "redrawn", though (not unless it started showing up weekly, or was traced to a verified bug). $\endgroup$ – J.R. Aug 15 '13 at 12:52
  • $\begingroup$ I have no idea what would happen, either; we are in the realm of guessing. I was surprised to see that many computer-drawn lotteries, too --- and that list seems to be US-only. Another interesting bit of Wikipedia gold is the following: en.wikipedia.org/wiki/1980_Pennsylvania_Lottery_scandal (not very relevant to our discussion, though, because there was some real rigging involved in this case). $\endgroup$ – Federico Poloni Aug 15 '13 at 13:11

This is a case of the devil being in the details.*

So the probability of 1111111, 3141592 and 8174249 are the same. But 1111111 seems(to me) far less likely to be the lucky number than 8174249.

With a small stretch of English grammar, these two statements are true, but it is the difference between the statements that is the key to understanding your confusion.

You are conflating two quite different things as if they were one.

  1. The possibility of X being the winning number.
  2. The possibility of the winning number being X.

On the one hand, all numbers in the range have an equal possibility of being the winner. 1111111 is one number out of a million and has once chance out of a milion—the same one chance out of million that 111112 has, as does 923652.

On the ther hand, the chance of the winning number being a specific number (or specific pattern) is bounded by the number of patterns in your set. Assuming zeros are allowed, there are 10 sets of repeated digits. In other words the winning number has a 1 in 100k chance of being a set of repeated digits.

The chance of the winning number being any specific number pattern does not in any way change the chance of a specific number pattern being the winning number.

* For the sake of simplicity, I've glossed over complexities in the math to show the general idea at stake.


Putting in a number like $111111$ in a lotto series, is as likely to win as any other series (last week's super 66 here was $411511$ for example). What is likely to happen is that people are more likely to select this number, or some bleedingly obvious number, like $142857$ or $262144$, than some fairly anomonous number.

Since when there are several winners in the pool, the prise is split equally between them, you are not likely to get 100 pence in the pound, say 50 or 11.111, One sees that even with a pretty random set of lotto numbers, if the first division is big, say \$10,000,000 it might be split up among nine winners who get $1,111,111 each.

So while the chance of getting a win on $111111$ is no bigger than any other set of digits, the chance of having to share it with a dozen others is.


As my Math teacher used to say, "I don't know the way to win more often in lotery, but I know the way to win MORE":

In fact, in a situation where the gain is shared between the winners, you have your interest in taking numbers that, if you win, should have less winners possible. Actually, people often choose their birth date, so if you avoid combinations like 19xx and 01-31 when you choose your numbers, you're more likely to not to share your gain with somebody wif you win :D


The intuition might be wrong for the following cause: You compute the probability of the event "any special number is extracted" and is X.

Knowing first probability you might be tempted to choose from the "special set", thinking that the probability to win will be higher. What you forget is that you will have 100% probability to lose in all cases that the number is not in special set, combined with some probability to win from the special set.

Combining these probabilities will still give you 0.0000001. The same will happen if you choose from the "not special set".


Another point to consider is how is the winning number selected. Here the lottery numbers is decided through a powerball mechanism (http://en.wikipedia.org/wiki/Powerball). So given that there is only a fixed amount of numbered balls the winning number is drawn from, this skews the probability of the winning number being 111111 versus another random number.

For example normally, if the numbers chosen are truly random, the odds of 1111111 being the winning number is 1 / 10 000 000.

But lets say in the powerball bin, there is 7 '1' balls, 7 '2' balls and so forth, then the odds that first drawn number will be '1' is 7 / 63. That ball is then removed from the bin. The odds that second number will be '1' is now 6 / 62 etc.

That brings the odds of 1111111 to 7*6*5*4*3*2*1 / 63*62*61*60*59*58*57 = 2 / 1 000 000 000.

Quite a bit less... The effect will vary depending on how many balls of each number there is in the bin, and how balls are replaced once selected. Some powerball lotteries has a preselection draw which determines which balls are in the final draw, but the same argument is still valid.


In canada, lottery tickets are manually drawn and each number cannot be drawn more than once.

For instance, Loto 6/49 (6 numbers, 1 to 49), an "human instution" inprobable number would be 1-2-3-4-5-6 (6 number in a row) and I beleive this is less probable then a more general number like 1-11-20-30-31-45.

Why? Because of the way it's drawn. If the number were drawn all at once by a computer for instance, I wouldnt beleive this. However, it's not the case in this example. Each time a number is drawn, it gets removed from the pot and the pot is scrambled again. Thus, there less numbers in the first 10 digits after each draw, making them less likely to be drawned.

Call me crazy, but I would never bet my money on a ticket that has 6 numbers in a row.


As mentioned by all other people here there is no disadvantage in picking 11111111 as your lottery number.

However I would just like to add that there is a pattern in these kinds of thinking mistakes.


Let's say that :

  • A = a very unlikely event
  • B = a very unlikely event

Our brain is tempted to use transitivity whenever it can. Immediatly we want to know what happens if A and B would happen at the same time. Usually, the chance of A and B happening at the same time would then be even smaller. We would have to multiply the chances.

Cards, An example of transitivity

When taking a random card from a deck of cards (52 cards):

  • The chance to take an Ace = 1/13
  • The chance to take a Spade = 1/4
  • The chance to take the ace of spades = 1/13 * 1/4 = 1/52

Here it makes perfect sense.

The lottery problem , transitivity ?

It is tempting to apply this reasoning to the lottery problem.

  • A = the chance to guess the lottery number is very small
  • B = a number with all equal digits seems like something rare

And here again, our brain is tempted to use transitivity The reasoning would then be that ...

  • the combined chance of (A and B) is smaller than the chance of A.

Unfortunately this is incorrect. You cannot apply transitivity here. But why not ? I will illustrate with some more examples.

Dice problem , another transitivity failure

Here is an obvious example that makes the same mistake:

  • the chance to throw a 1 with a dice = 1/6
  • the chance to throw a 2 with a dice = 1/6
  • What is the chance to throw a 1 and a 2 at the same time with the same dice ? Is it 1/36 ?

It is of course impossible to throw a 1 and a 2 at the same time.

When can you apply transitivity ?

So, what is the difference between the dice and the cards? The difference is that (as with the cards) when applying transitivity we have to make sure that the individual conditions that we are combining do not interfere with each other. If there are no such dependencies then you can blindly apply transitivity and multiply the individual chances.

Applying it !

There is 50% chance that I am a man, there is 50% chance that I am a woman. However, the chance that I am both is not just 25%. This is a perfect example of interfering conditions.

So, yes a digit with all equal numbers is something special (0.00001%), and yes guessing the lottery number is difficult (0.000001%). The problem is that you are trying to combine these 2 conditions, but the conditions are interfering. So in conclusion, even while it seems that you should multiply the chances (0.0000001 * 0.00000001), this is in fact a mistake.

I hope this explanation gives you a better insight.

  • $\begingroup$ Somebody voted to delete this post, but in all honesty, I have no idea why. Please explain. $\endgroup$ – bvdb Nov 11 '15 at 15:24

protected by Willie Wong Aug 15 '13 at 13:19

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