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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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.
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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.
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