[lazide]

The second pick will have better odds of a blue ball, assuming the first one picked was red and the ball was not put back however - correct?

Even more so if you’ve gone 50 picks like that, and are now down to 49 red balls and one blue ball.

That is what card counting helps you with - knowing the odds based on your current state, as compared to the initial state.

[zamadatix]

Depends if the balls are placed back or not. Similar to cards, any game which never replenishes the deck and lets you draw all the cards down to the last one lets the last pick (but not the prior ones) be non-random.

[lazide]

I literally stated ‘if the balls don’t get put back’.

And the last card was still random, btw. But can be guessed now with perfect precision due to the process of elimination.

Random at shuffle doesn’t mean unguessable or unpredictable as the game goes on.

[zamadatix]

Sorry, meant to say the more generic "replenished" but typed a bit of your comment while reading it at the same time :p. That's what I was saying by the next part, it doesn't matter _how_ it's replenished just that the pool doesn't dwindle to 1 option for any reason. E.g. you could not put them back pull and mix from multiple decks/ball sources as the game progresses and even though 51 cards/99 balls are discarded you don't know what n+1 is for certain yet.

I disagree the last card is "still" random, at least in the sense of statistics. It "was" random up until the measure of uncertainty of the next event reaches 0 (i.e. entropy reaches 0). At that point it's no longer a "guess", there is no uncertainty and the remaining pattern is always 100% predictable in that regardless which proceeding events occured to get there it can always be known what the next value is without uncertainty. Since there ceases to be any uncertainty in what the remaining pattern will be there ceases to be randomness in the next value generated. That the card's value was not known at n=0 does not affect whether the n+1 card still is/isn't random when n=51. In another form, that you didn't previously know the value of card n=52 with past information holds no influence whether the value is random or not with new information. Statistical randomness is all about what you know of the future predictability, not about how something came about.

This is also true of events which fall into predictable patterns at any point along the path. E.g. if I had a (relatively useless) hardware random number generator that generated random numbers 0-127 once per second until it generated a 0 at n=17, at which point it ceased being able to pull randomness from it's dead circuits and always produced 0 afterward, the first 17 values were all statistically random at the time of their draw but n=[18,inf) are all now predictable and no longer random from that point on.

[lazide]

I think you’re getting confused? Or we’re talking past each other?

Knowing the outcome of a die roll after the roll doesn’t make the roll of the die non-random.

Any more than the die coming up 6 a bunch of times in a row does. Though maybe it would be a good idea to check the weight/balance of it, hah.

Knowing what the last card has to be after all the other cards have been played doesn’t make the process which made that card the last card any less random either.

Picking cards off a deck isn’t a randomization/non-randomization event, shuffling the deck is what does that.

Probability however, which directly impacts gameplay in some kinds of games, is directly impacted by knowledge exposed during gameplay.

So for instance, the second to last card (and also the last card) can be known with about 50/50 odds (probability wise) if someone is good at counting, which is damn good! Way better than 1/2704 odds of guessing a sequence of two cards at the start of the deck.

In the example of the broken hardware random number generator, knowledge of the defect can be used to attack a system if it assumes a continuous probability distribution out it’s outputs, if the attacker knows this.

The same as an attacker at a casino could probably manipulate a game to make money if they knew something was broken in the deck shuffling and the same two cards were always at some position in a ‘new’ deck.

[zamadatix]

> Knowing what the last card has to be after all the other cards have been played doesn’t make the process which made that card the last card any less random either.

Exactly, same page then. The card not being random at the end of game frame of reference places no limitation that it was random from the beginning of the game frame of reference. In the end of game reference the last card is never random, it's only ever the remaining card. In the beginning game reference which card will end up being the last card is still random at that point despite it ceasing to be from the later frame.