Question about blockers.
Does a 2 in our hand,reduce the amount of times our opponent is holding A?

Unless you have some more information, no. It increases it.

Having an Ace in our hand reduces the amount of times our opponent can have an ace.
Likewise, having a spade in our hand reduces the amount of times our opponent can have a spade

It's straightforward to say it increases the chance of an A being held by villain due to reducing number of cards villain can get.
I don't follow how the chances of the villain holding one of a suit increases if hero is holding another of that suit. Could you explain?

Nah, knowing one card which means all the other are more likely. I.e. Rusty didn't say anything about suits.

Can someone tell me why I'm not getting this?

In NLHE, Hero holds 1 ace and so villain has a 3 in 50 chance having an ace, whereas hero holding 0 aces will increase villains odds to 4 in 50.

Hero holds 1 spade and there is a 8 in 50 chance of villain holding at least 1 spade, whereas the chances of villain holding at least 1 spade increases if hero holds 0 spades.

You're correct. But OP is not asking about the chance of having 'a spade' he's asking about the chance of having the ace of spades. Having a spade blocks spades, but it can't block specific spades.

Okay, thanks, I thought I was losing my mind. Perhaps OP was using "block" in the wrong sense?

i.e. If hero holds a spade, villain always has the exact same odds of holding the A no matter what spade hero holds, but the chances are reduced compared to if hero held no spades.

Not sure if this is correct.

Think about it this way: if you know any one card (spade or not) the odds of villain having any other card (like A or any other) will be more likely as there are now only 51 cards left in the deck.

I think that "2 blocks spades" -idea trips you. That is true but as Rusty said while it reduces one spade (i.e. 2) from the deck, it doesn't reduce the change of getting A at all!

You've basically just repeated what I said.

It doesn't reduce the chance of getting exactly A but reduces the overall chance of getting any other spade, and this includes the A thus it reduces the chance of getting that card.

I think we're going in circles here because of our wording and I think we're both thinking the same thing but it's coming across wrongly.

No, I think you're making a mistake. It does not reduce the chance of you getting any particular spade - it reduces the chance of you getting one of the remaining spades.

If you ask the question: "is it more likely for him to have a spade" the answer is "no, it's less likely"

If you ask the question: "is it more likely you him to have the As" the answer is "yes, it's more likely"

Let's consider a much simpler example. Your opponent has just 1 card.

If you don't have a card, then:
he has the As: 1/52
he has a spade: 13/52

If you have the 2s, then:
he has the As: 1/51 (1/51 > 1/52 so he'll have it more often)
he has a spade: 12/51 (12/51 < 13/52 so he'll have a spade less often)

(and in fact, the chance for him having any particular spade except for the 2s is 1/51)

DecisionMade I am on the same page as you now.

Should have phrased the question better, but you guys figured it out and answered it right.

I still think we're on the same page and I've just been interpreted wrongly.

So is my quote below not correct? Edit: Actually this looks like I'm referring to just hero's spade as the main focal point and not the villains chances of holding a specific spade. I can see why this may have confused you.

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For the above quote, is this quote not also correct? My wording might be misleading but what I'm saying is that the odds of holding a spade is reduced ("he'll have a spade less often") whereas the odds of it being a specific spade is the same as any other specific spade (but still increased due to hero holding a spade).

Anyway, I fully understand your post and it was basically what I was trying to say, you just did it a bit more clearly!

This is what I object to.

The chance of getting "that card" (the As) is NOT reduced when you hold the 2s. It's increased.

Actually I can't really figure out what you're even trying to say here because you statement boils down to "it doesn't reduce the chance of getting the As, something something, it reduces the chance of getting the As"

(I'm not disagreeing with you - I think we may just have a semantic difference, I just can't quite parse the sentence I quoted above into anything I can agree with)

DecisionMade, your wording is definitely not quite right, not sure if it's semantics or misunderstanding. Assume 1 card dealt:
If hero has a spade, villain has a spade 12/51
If hero has a non-spade, villain has a spade 13/51

If hero has the As, villain has the As 0/51
If hero has any card other than the As, villain has the As 1/51

The only card we can have that has any effect on villain's frequency of holding the As is the As.

Aside: I feel like this talk of our card increasing his probability of having the As because it's gone from 1/52 to 1/51 is misleading at best. We have a card. After we look at it villain either has the As 1/51 or 0/51, I think it's kind of silly to say having the Qs "increases the chances" villain has the As.

Just to clarify, it does looks like an oxymoron from the outside because we may think that hands will distributed suitwise equally.
One thing I forgot when asking this question is that we are always dealt 1 out of 1326 hands.
say we have 3h 2s, we are only removing 1/1326 combinations, and now our V has 86/1325 which basically increases his chances of having As X combos, but decreases amount of spades in the combos by 87.
Hence we only eat one out if he is chasing a FD, but we don't block him from having As in his range, we actually increase it.

Reason I asked this question, was I wanted to experiment with merging, to add more mathematical bluffs into my range, but looks like this idea floped

We remove a lot more than 1 combination! Having the 3h stops him from having every hand that could have the 3h: Ah3h, As3h, ... , 2d3h, 2c3h and having the 2s stops him from having every combination that has the 2s in it. So there are no longer 1326 combinations for him to have, but 50*49/2=1225 combinations.

So before we look at our cards he has 51/1326= 3.85% chance of having the As
After we look at our cards, if we have the As he has 0% chance of having the As
If we don't have As, he has 49/1225 = 4.00% chance of having the As

that's right, I threw all the numbers on the run, and never bothered putting them all together.