Eli5 Is there a rule in statistics where choosing the same number X times in a row decreases each time you choose that number?


Was watching a podcast recently where a girl called another girl dumb for choosing all 6 for a lottery ticket saying that after one 6 is chosen, the probability of each subsequent 6 being chosen decreases. I.e you’re better off choosing 10 random numbers than 10 6’s.

The other 2 in the podcast called the girl an idiot because each six is chosen separately. So the probability of arriving at all 6 is the same as any other combinations. This seems to make sense to me. Rolling 10 dice, the probability of one 6 doesn’t magically effect the other result of the other di.

However I seem to vaguely remember being taught something similar to the supposedly idiot girl when I was a kid.

So basically, who’s right and why?

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9 Answers

Anonymous 0 Comments

There’s no such thing. All 6s is the same as all different in terms of likelihood to be picked.

This is because lottery numbers are sampled “with replacement”. This means even if the lotto picked a 6 for the first number, the same 6 is eligible for the second number, etc. Further the numbers are picked independently.

In games where items are picked “without replacement”, the events are not independent – each selected item influences the possibilities of the next. So for example in those games you will never see duplicates. An example of this is if you have labeled numbers and you pull them out of a bag. Each number you pull is now no longer in the bag.

Anonymous 0 Comments

If each choice is made independently of the previous choices (meaning the previous choices do not affect the probability of the next choice), then the probability of choosing the same number each time is simply the probability of choosing that number once, raised to the power of the number of choices. For example, if you’re choosing a number between 1 and 10, and you choose the number 7 with probability 1/10, then the probability of choosing 7 three times in a row is (1/10)^3 = 1/1000.

Because your dealing with 2 events, the choice and the series, which contains the choice, each successive choice make the chance more remote of getting the same number. The probability on choice 500 would be (1/10)^500.

Anonymous 0 Comments

The key issue at hand is “independence” of the observations.

The first girl is making an assumption that the probability of getting a second six is affected by whether or not you got a six on the first draw.

The other two are making the opposite assumption, that each draw is unaffected by the results from the others.

If you’re drawing numbered golf balls out of a bowl without replacing them then the first girl would be right. There are only so many “6” golf balls in the bowl and removing one makes it less likely to be drawn in future rounds.

However, if the observations are independent, such as rolling a die a bunch of times, then the other two girls will be right.

Anonymous 0 Comments

There is a thing called the gambler’s fallacy, which is the mistaken belief that each selection influences the next.

If you rolled a dice each number has a 1/6 chance so you have a 1/6 chance to get it right. If you roll that dice a dozen times your chance to get it right each time is still 1/6 regardless of the number before or after.

Anonymous 0 Comments

It really depends on how the lottery numbers are selected. Lotteries and their televised ball machines make it possible to see the actual system, but it will depend on how they do it.

If there’s one giant collection of balls and you pull out a number from the collection each time… Like, if there are 10 of each number (100 numbers/balls total) and you pull out a `6`. Well now there are only 9 of the number `6` in the machine and 10 of each other number. So the number `6` is less likely to come up next time. It gets worse as each subsequent `6` is drawn, if they are. So the all-`6` ticket is statistically speaking least likely to win, and a ticket with all unique digits has a slightly higher chance at winning.

If there are several collections of numbers, isolated from each other, then each drawing of the number `6` means nothing to the others and so all `6` tickets are just as likely as any other combination.

So I’d have to see how the lottery numbers are selected before deciding who’s right.

Anonymous 0 Comments

If the choices are independent, then yes, it doesn’t matter what number you choose each time, they all have an equal chance of coming up.

If the choices are dependent, i.e. there are 5 balls with the same number and after one comes up there are only four left, then yes, you should change the number you choose as that one you picked is now less likely to come up out of the pool of number balls.

Anonymous 0 Comments

First, I’m not sure where these people are playing the lottery but either they all don’t know what they are doing or it’s a very weird lottery.

In most areas there is no repetition of numbers in the lottery. In other words you can only pick each number once. You could play multiple times in the same lottery (ie buy multiple combinations of numbers) and have one six in each one, but you can’t purchase just one draw and have two, let alone six sixes.

Ignoring that it would depend on how this weird lottery is being done.

With replacement, meaning after you draw a ball (or however the number is chosen) you put it back and draw again then there is no difference in the odds between drawing say six 6’s or the sequence 1, 2, 3, 4, 5, 6.

If you don’t replace the already chosen number, then yes the odds would be worse because there are now less 6’s available to draw. In the case of most lottery’s that less 6’s is actually zero 6’s because there is only one of each number to begin with.

Anonymous 0 Comments

I think if they allow the person to select that sort of number set, then the game is made to have that as a possibility.

If once the lottery draws a 6, then if they draw from the same dish and there’s no 6 in there, then it won’t come up again, which also means that when they create the ticket you can’t select 6 twice. So the fact that you can select 6 more than once sounds like they draw from different dishes for each digit. If that’s the case, then the chance of 6 coming up is the same for each independent draw. Therefore the chance of 66666 coming up is the same as 12345 coming up.

You can select 1-2-3-4-5 as your lottery number for years and people will tell you you’re crazy because the chances of it coming up are so low. It’s the same chance as 12-19-38-54-7

Anonymous 0 Comments

The whole crux of the situation is that the statistics are different for the following scenarios:

* *”Draw a card (and write it down as ‘Card-1’), then put it back (and shuffle) and draw a card again (and write it down as ‘Card-2’).”*
* *”Draw a card and call it ‘Card-1’, then (without putting Card1 back) draw another card and call it ‘Card-2’.”*

How? If I draw the Ace-of-Spades as my Card-1, then there is an exact 1-in-52 chance (~1.92%) that I ‘draw it again’ as Card-2 in the first case scenario… but there is also a precise 0-in-51 odds (0.00%) to ‘draw it again’ in the second case! (And the difference only gets more divergent as you cut down the number of possible card-suits or face-values in a deck.)

To get down to the raw statistics, each and every individual step in an algorithm should be boiled down whether it constitutes a “Draw another card. (Remembering what has already been drawn in previous steps ‘without replacement’.)” situation or a “Roll another die. (Because this is a ‘with replacement’ step equivalent to being reshuffled.)” situation.


Who among these podcast people is (most) correct will ultimately depend upon the rules of how the lotto-organizers algorithmically choose winning numbers. Do the rules of their number-choosing-process adhere to the first-case scenario, the second-case scenario, or is it a more complex algorithm switching between the two in a specific predefined pattern?

Let’s pretend that only numbers the ten numbers 0-9 can “hit” in this lotto’s choice of dice/decks and that players guess 7 specific numbers…

If the lotto-organizers have decided that each individual slot in the sequence is individually decided by a “roll-a-10-sided-die” step, and “winner-grading” is based upon a right/wrong answer for each individually-ordered number slot… then there are 10^7 possibilities. Furthermore, this ‘specific’ ordered sequence of “6666666” is just as likely/unlikely as the specific ‘ordered’ sequence like “3456789”, or a specific ‘random’ sequence like “1086385”, to win. Any specific individual combination has odds equal to one slice of the total (dice-values)^(number-of-dice) possibly-guessable slices of pie. In this case, Miss Sixty-six-six-six is correct that a specific ‘non-random’ guess of “6666666” is equally likely to win as any other specific ‘random’ guess like “7495724”.

However, if the lotto-organizer’s algorithm operates on a single “Draw-7 cards.” step, then “drawing 7-sixes” actually is more difficult than “Draw 6, seven times (with reshuffles).” because the odds decrease after every six drawn from a deck with a finite number of suits (the pool of possible hits decreases each time a hit is pulled out of the deck). In this case, Miss-six-six-six is more likely to lose than her podcasters (though the odds difference decreases as the number of “suits” approaches infinity).

Ultimately, another monkey-wrench might get thrown into the works if “order matters” becomes “order agnostic”. This is because a guess like “1666666” can overlap and “hit” on 10x more possibilities (like “6166666”, “6661666”, etc.) as a “one-1, five-6s” guess versus a “six-6s” guess which only hits exactly upon “6666666”. Unfortunately, things get more complicated if one starts digging into various “non-random” starter-guesses… since a “Bell-curve” of repeatability is likely better than the example “all-repeats” pattern or a “no-repeats” pattern. Ultimately, though, all of these order-agnostic cases can be accounted for by summing up all of the individual sequence odds of each possible ordering.