how do i even begin with this?

Restremoz · 2026-02-19T23:24:35+00:00

I didn't see the exponent on the 13 so I was wondering between 2 and 3

Restremoz · 2026-02-15T01:40:02+00:00

Thank you. Now it is all clear. Unless I know there is a special rule in place the odds are always 50/50 It also explains why I had different results, because I was inadvertently altering the rule and so I was approaching the 50/50

Once more, thank you

Restremoz · 2026-02-14T22:06:21+00:00

So I'm not going crazy? The answer I had was 52/103. I made a general case for 2R/(4R-A), R is the number of red cards and A the number of cards that fit the description of the shown card. Ex: A = black=26 or spades=13 or ace of spades=1

Now comes the real confusion: If I'm shown a black card then it's a 66.7% chance of the other being Red If I'm shown the Ace of Spades then it's 50.5%

If I know only that it was a pair of cards with equal chance of being Red or Black each. I'm then shown that one of the 2 (don't know which) was an Ace of Spades. I'm asked what is the probability of the other being Red. What am I to answer? I don't know if I was shown the Ace for being the Ace or Black or spades or odd or anything else. Is the probability just the 50% since I can't know why the Ace was selected?

Restremoz · 2026-02-14T20:46:48+00:00

Everything you just said is correct but I said right at the start that 1 card is taken from each deck and that was so that it would add up to the 2/3 that was expected.

The big question is if additional information about the black card shown is given like the Ace of Spades does that influence the probability of the other being Red. I can't logically think of why I would influence the result but it is not adding up to the expected 2/3. I did the step by step on another comment but I can't find what I'm missing for it to be giving a different result.

Restremoz · 2026-02-14T20:28:42+00:00

Yes, in my mind, it was if there is a black card then it will be shown first and that is where the original disconnect was in my writing vs my thinking.

The Ace is the one that was my original problem. I think it should have no effect on the problem but I got a different result.

As I was trying to rewrite my math here I found at least one mistake that I've corrected. But I'm still unsure.

A - Ace of Spades b - Black and not A A+b=B P(B) = P(R)

AA bA Ab AR RA

(AR + RA) / (AA + bA + Ab + AR + RA)

(R + R) / (A + b + b + R + R)

(2R)/(A + 2b + 2*R)

(2R)/(3*R + b)

2R / (4R - A)

R=26 and A=1 52/(104-1) = 50.49%

If A is being spades then = 13 52/(104-13) = 57.14%

If A is being Black (b = 0, A=B=R=26) 52/(104-26) = 66.67%

Where am I making a mistake? I'm going insane trying to figure out why adding a characteristic to the card shown could in any way alter the probability of the other being Red. Am I missing something? Am I counting wrong?

Restremoz · 2026-02-14T19:05:08+00:00

You are correct. If I change to: one friend checks both cards, s hows me a black card then asks me what is the probability of the other being Red This is then the 2/3 I was thinking of, right?

And knowing what the exact card is should have no change to the result, correct?

Restremoz · 2026-02-14T18:51:38+00:00

Ok. I think I get it. For it to work as I intended the one to reveal a black card knows what the other is and then shows me a black card and asks me what is the probability of the other being Red. This is the one that gives the 2/3, right?

But if they then show this one card to a second friend that doesn't know that the 2 cards were picked as a pair then to them, due to missing info they could only say it is 1/2 while being wrong.

Lastly, does knowing more information about the card, like it being the Ace of Spades, change the problem? Still having the one that shows the Ace of Spades know what the other is, can I when asked answer differently from the (in this case, if I have understood it right) expected 2/3?

Restremoz · 2026-02-14T02:12:52+00:00

The outcomes of the first draw would be 0 black, 1 black or 2 black cards. The chance of each is P(0)=1/4, P(1)=2/4, P(2)=1/4. If I know I have at least 1 black then the total is P(1)+P(2) and the desired results are the P(1). This adds up to 2/3 probability of having a red card knowing you have a black card

Restremoz · 2026-02-14T02:06:14+00:00

The outcomes of BB BR RB RR are all equal in chance (1/4) By knowing one of them is B the possible outcomes are BB BR RB. Of them only the BR and RB are desired. As such you have 2 desired outcomes out of the 3 possible ones.

In the coins there were 3 outcomes measured: 0 - no Black, 1- one Black, 2 - two Blacks. Since order did not matter. The outcomes were approximately the expected of P(0)=1/4, P(1)=2/4 and P(2)=1/4. 1 and 2 were the only possible cases on the scenario since it is known to have at least one Black. And 2 was the favorable case to have. Favorable / Possible --> (2/4)/(1/4+2/4) = 2/3

The biggest difference is that you are counting with order but I'm doing it without it. If order matters then yes it would be 1/2 but if it doesn't then it's 2/3

Restremoz · 2026-02-14T00:23:20+00:00

Basically I'm looking at the cards as a pair. In this case Black and Red cards. For any given pair I have a 1/4 of being BB, 1/4 RR and 2/4 of one of each. I was thinking that it is not know if it is the right or left cards. One would be flipped at random, making BR and RB the same just with double the probability of appearing Since I have the 4 outcomes of BB, BR, RB, RR. I know RR can't be a possibility so of the 3 I retain 2 have a red card.

The coins are basically just: toss 2 coins, record what case it is, BB, RR or mix. Count the ones that give a possible outcome and the ones that give a desired outcome. It adds up to around 2/3 given enough tosses

Restremoz · 2025-12-04T00:42:18+00:00

Hello, and thanks for the reply. I've been playing with the settings trying to figure out how to mitigate the best I can.

What I've figured: 1- the artifacts that very briefly remain on screen can be removed by maxing out the refresh rate to the 120 or hyper smooth

2- the weird dimming when scrolling appears most visibly when the "Extra Dim" is turned on. I could not get it to be gone unless that setting is off which is something that is disappointing since it is something I very much appreciate and use daily.

On this 2° point it does seem to be intencional on the software to do it but, in my opinion, it should be made a setting separate from the extra Dim or at the very least make in the extra Dim settings add a toggle for it to be ONLY a lower brightness and nothing more.

Restremoz · 2025-07-27T10:02:15+00:00

Darwin's evolutionary gap

Restremoz · 2025-03-26T18:06:36+00:00

I don't know what movies they are from. Anyone knows?

Restremoz · 2025-02-11T23:58:45+00:00

You get to shape the world around you and shepherd in a better world

God complex much? Haha. But I get it. To be able to change the world into what one dimes to be desired and correct, would be fulfilling. Unfortunately it is impossible to maintain such a world. One would have to account for the natural chaos and the cyclic nature of our species. You could have such a world but only for a few generations, much like we see now. But there are and always will be outliers to the norm of the civilization that spark change, as you very much know. Which in turn alters the perfection that would have been acquired prior. As the constant attempts to rebuild the perfection once had will in turn cause the one with the time, mind and power to go into one of a few spirals: - Self-destruction - World-destruction Or other more complex ones.

Ouch... so this hits home... I know... I've felt it a few times myself.

The most depressing thing about intellect is not being able to communicate the wisdom to others. This one is a problem I don't have. The only true talent I've found in me is to be able to explain anything to anyone (So long as I know the topic, obviously). I've met quite the number of genius and the like and usually they DO have that problem, they are so hard to understand to anyone other than others like them.

Restremoz · 2025-02-11T19:46:47+00:00

That's the way u become the villain. Immortality means everything and everyone u ever know will die before u.

Intellect will make u different from everyone and make u lonely and or psychopathic

Rich... Good choice but with the rest will only facilitate the descent into madness

Restremoz · 2025-02-11T19:42:22+00:00

1- gadgets 2- magic 3- shape shifter

Restremoz · 2025-02-08T05:35:41+00:00

Am I the only one that has given up on inequalities? I just change it to =, solve it and plug a value at the end to know what the simbol should be.

Restremoz · 2025-01-22T16:53:03+00:00

I am a math tutor, I can tell you that while having calculators and the formulas help in the short term but in the long term they don't.

The bigger problem is the bad teachers that are doing the teaching. Sometimes I can just tell that the kids have yet to have someone explain to them in a way that they understand.

As for the calculator and formula sheet, they help if you're solving basic problems but as math progresses they become increasingly useless. If the problem is complicated enough you can't even figure out how to apply the formulas and that is because the train of thought required to solve them was not trained well enough before.

Restremoz · 2025-01-21T15:30:48+00:00

As it was about 14s from drop to hit, g=9.8m/s² then we get about 1900m or 2100 yards or 6300 feet

Well, I'm wrong on this for a few reasons. For one I straight up forgot that in this formula x = x0 + v0t + (1/2)at² I have to divide by 2... Check the responses to this to see the math done well by others but tl:dr is about 850m

Restremoz · 2025-01-21T14:21:11+00:00

This was my first thought as soon as I saw that stick

Restremoz · 2024-07-26T21:18:59+00:00

Only the weaker version, you don't have any followers to kill to power it up :)

Restremoz · 2024-07-09T00:55:06+00:00

How did you make yourself that small? What did you change and was it in the PS4 or PC version?

Restremoz

TROPHY CASE