>>16920587
4 choose 2

Total quantity of cards = 52
Divided into two colors 50% = 52 divided by half = 26.
26/52 = 13/26 = 6.5/13 = 3.25/6.5 = 1.625/3.25 = 0.8125/1.625 = 0.40625 = 0.203125 = 0.1015625= 0.05078125 0.025390625.....compute...compute...compute....

Either it happens or it doesn't. 50/50

>>16920587
It's 1/3. It's pretty easy to visualize it with say ten red and ten black cards. It would be really difficult to pick all 10 reds out from that scenario but it would be lot easier to get 2 of the same color if you just picked 2.

>you should be able to solve this
>homework thread

>>16920609
>Total quantity of cards = 52
It's explicitly 4

This is fucked up.

This one panel of Yu-Gi-Oh is way smarter and more insightful than the entirety of Hunter x Hunter. I have no idea what the fuck people see in the latter series compared to this.

>>16920730
It intentionally misdirects you in panel three so the reveal gets to be more clever

>>16920737
I wouldn't call it misdirection. It's the naive assumption that you may have. It intuitively makes sense that if you pick 2 out of 4 that the chance is 50/50 and the 4 combinations supports that idea. It doesn't even make the obvious mistake of there being 3 combinations which I would say would be misdirection.
It's the sort of set up you would use when teaching kids really.

>>16920737
But even that is legitimately clever. It manages to pull off something genuinely tricky, yet actually insightful about probability and combinatorics that the reader is left free to reason about, on a single page. HxH conversely takes entire chapters to either fluff on about absolutely basic shit, or fake out the reader with bullshit that's deliberately written to not be solvable until you get some 'reveal' that shows the characters solving a problem in a copout way that isn't actually reasonable given the information they actually had.

This at least actually teaches you how to think about shit.

>>16920750
>the 4 combinations supports that idea
But there's six combinations of four unique cards. That's the misdirection.

File: four aces enumerated.jpg (97.3 KB)

97.3 KB JPG

Okay, I figured out the flaw in Jounouchi (Joey)'s reasoning. The way the game is phrased tricks him into reducing the problem to blacks and reds, which leads him to incorrectly assume the chances of BB, RB, BR, and RR are all the same. In reality there are only two ways to get each of the winning combinations (BB, RR) while there are four ways to get each of the losing combinations (RB, BR). 4 ways to win vs. 8 ways to lose = 4/12 chance to win = 1/3. He enumerated the possibilities incorrectly, whereas Yugi's argument skipped enumeration altogether.

>>16920757
There are 4 combinations, 2 combinations are just twice as likely.

>>16920775
You have spades, hearts, diamonds, clubs
If order doesn't matter we have the following unique combinations
SH
SD
SC
HD
HC
DC

SC and HD win, the rest don't

If you're just going to count colours then there are three combinations because red-black = black-red
If each card is counted as unique then there's six
But counting four combinations is mixing the two

>>16920587
Let [math]D_1[/math] denote the first draw and [math]D_2[/math] denote the second draw.

Since you only draw one card each draw, the intersection [math]P(R\cap B)=0[/math].

For a total of 4 cards, we have 2 red and 2 black.

First draw:
[math]P(D_1) = P(R \cup B)=P(R)+P(B)=\frac{2}{4}+\frac{2}{4}=1[/math].

Second draw:
If we draw black on the first draw, then there is 1 black card of 3 total cards, so the probability of drawing a black card is as follows
[math]P(D_2=B)=P(D_2=B|D_1=B)P(D_1=B)=\frac{1}{3}\times 1=\frac{1}{3}[/math]

Similarly for red, we have
[math]P(D_2=R)=P(D_2=R|D_1=R)P(D_1=R)=\frac{1}{3}\times 1=\frac{1}{3}[/math]

So on your first draw you have a 100% chance to draw either color, making the probability of winning the game [math]\frac{1}{3}[/math] regardless of your first draw.

>>16921131
looks solid

>>16921131
I simply don't look at the first card. Thanks for that extra 16.66% chance of winning.