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?
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.
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.
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.
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.
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.