Expanding on the logic riddle:

[braalian]

How is there ever four permutations? There’s two black cards, two red, or one of each.

 

[frigidweirdo]

But that's specious.

That Thunderball stat site is just rude, IMO. Just trying to suck people into losing more money. They are generating noisy data from too-small a sample size to try to fool people into believing one number may have an advantage over the other, causing them to falsely inflate their expectation.

There are 39 Thunderballs, 5 balls drawn = 69 million+ combinations. How many historical drawings? 5,000? 10,000 maaaaybe?

Those noisy charts will trend to even frequencies for each number, over time.

1,2...7 has no less chance than any other 7-digit number.

Again, I think it depends on how you look at it.

Statistics is just that. It's math. It can be manipulated in many different ways. Simple statistics is 50/50. There's a 50/50 chance that anything will happen. Either it will, or it won't.

So you could say 1,2,3,4,5,6,7 has a 50/50 chance of happening. Or you could say that complete nonsense because this other thing says something different.

 

[frigidweirdo]

"Before I lay the two cards down, I shuffle them behind my back."

Sorry for the confusion, if any.

It's okay, though I still think it's 50/50

You have two cards. You shuffled them, you pick one up. 50/50.

 

[frigidweirdo]

Good point. I’ll agree that a string of consecutive numbers obviously has less chance than a string of non-consecutive ones.

But I maintain (or I think I do) that the numbers I said have just as much chance as any other *single* combination of numbers.

Times like this I wish I had taken some math courses beyond high school. Hated math with a passion, but I think I’d dig stuff like this.

Yes, but it's like with anything, it's how you approach it.

Obviously we can look at all the lotteries around the world and I doubt there's ever been a 1,2,3,4,5,6 or anything like that.

I'd say patterns are much less likely, not because of math, but perhaps because we see patterns. If you put the same balls into the pot WITHOUT the numbers on them, would the chances increase? They might do. Purely because the balls might be put into the machine in a random order. (I don't do the lottery, but I assume they have the numbers in some kind of order when they go in).

 

[Fort Fun Indiana]

BUT, you do have a semantic point. That's the semantic debate: whether or not the cards are then "re-ordered", left to right.

Whatever reason it was put there, it made the odds 50/50.

Meaning, your point is correct, if you consider them re-ordered, because it is a rule that the left card is always flipped.

Probably better to leave out that the card on the left is flipped, and just say I flip over a card.

 

[Fort Fun Indiana]

Again, I think it depends on how you look at it.

A randomly generated 7-digit number is just that, though. Random.

 

[Toddsterpatriot]

BUT, you do have a semantic point. That's the semantic debate: whether or not the cards are then "re-ordered", left to right.

Meaning, your point is correct, if you consider them re-ordered, because it is a rule that the left card is always flipped.

Probably better to leave out that the card on the left is flipped, and just say I flip over a card.

You flip over a card, either card, it's red.

One card left, 50/50 chance it's also red.

 

[Fort Fun Indiana]

How is there ever four permutations? There’s two black cards, two red, or one of each.

But you agree that, if you collected pairs of draws from an infinite deck, what you would eventually have is:

25% both black
50% one black, one red
25% both red


Right?

 

[frigidweirdo]

A randomly generated 7-digit number is just that, though. Random.

Assuming it is random. Assuming that everything is pure. Is it pure?

 

[Fort Fun Indiana]

Assuming it is random.

Yes, true.

 

[braalian]

But you agree that, if you collected pairs of draws from an infinite deck, what you would eventually have is:

25% both black
50% one black, one red
25% both red


Right?

Ok, I’m too tired to think about it anymore tonight, but at least it got my brain working

 

[braalian]

Assuming it is random. Assuming that everything is pure. Is it pure?

I guess there’s infinitesimal differences in real life lotto drawings like some being slightly different shapes or weights, or being put into the machines in a certain order that might slightly give certain numbered balls a greater chance.

But, in a pure environment, I’d have to assume any combination of numbers has just as much chance as any other combination of numbers.

In fact, there’s just as much chance this weeks numbers will be the same as last weeks numbers as there is it will be any other single combination. Right?

 

[Toddsterpatriot]

This is your sample space of combinations: {BB, RB, RB, RR} (note the repeated combination "RB", to denote it being 50% likely)


There are only three combinations of two red or black cards possible.

Both black, both red or one of each.

RB, RB is not two separate combinations.

 

[frigidweirdo]

I guess there’s infinitesimal differences in real life lotto drawings like some being slightly different shapes or weights, or being put into the machines in a certain order that might slightly give certain numbered balls a greater chance.

But, in a pure environment, I’d have to assume any combination of numbers has just as much chance as any other combination of numbers.

In fact, there’s just as much chance this weeks numbers will be the same as last weeks numbers as there is it will be any other single combination. Right?

Well, again, depends on how you view statistics.

I could also go for the predestination argument as well, but....

 

[Orangecat]

They are the same combination.
They are distinct permutations.

One might contend that is a distinction without a difference.

 

[Orangecat]

Because, in and of itself, each individual card draw will always have a 50/50 chance, but the chance that *two* consecutive draws will be the same color lowers the probability. Am I right?

I see what you're saying, but that really only works in a finite deck of cards. Removing one red card still leaves an equal amount of red and black cards in the infinite deck, meaning the odds are still 50/50.

 

[Orangecat]

Hmmm, but if you flip a coin twice there’s a 1/4 chance both flips will be heads.

But there's a 50/50 chance that they will match. If you already know what the first coin flip reveals, the odds are 50/50 that the second will match.

 

[Toro]

I would’ve thought it’d be 50% for all four scenarios with the infinite deck.

More importantly, I’m just really impressed that you can shuffle cards behind your back!

 

[mamooth]

It is 50% for all cases. It really is that simple.

So, in scenarios 3 and 4, if a flip a card, and it is red, you can only eliminate 1 permutation. Three remain:

BR
RB
RR

No. The fundamental error there is assuming BR and RB have the same probability as RR. They don't.

One way to look at it is that either "BR" or "RB" was a possibility, but not both. One has to be discarded. That red card could be the right card, it could be the left card, but it can't be both. That leaves 2 cases, so 50%.

The other way of calculating is that BR and RB each have half of the possibility of RR. That makes the possibility of a second red card ...

1 / (1 +0.5 +0.5) = 1/2

 

[mamooth]

In this case, the claim is that if I deal two cards, and flip one over, the odds of the second card being the same color as the first are 50% ... unless you shuffle those two cards randomly before flipping them over, which will somehow reduce the odds of same-colorism to 33%.

That's absurd, so we know it's wrong. It's just a matter of showing why. If one could adjust the odds that way, gamblers would have figured out how to use it to their advantage.