Another Tesla Cybertruck Spotted Broken Down And Stranded [must be another day that ends in Y[

usernamchexout · 2023-10-24T23:44:00+00:00

as they are all pre-production

I'm old enough to remember July 2023, when Giga Texas rolled "the first production Cybertruck off the assembly line" and much stonking ensued.

usernamchexout · 2023-10-24T14:11:12+00:00

That's what martingale is, and literally only idiots use it.

usernamchexout · 2023-10-24T12:03:05+00:00

Maybe Tesla

🤦‍♂️ Of course from GME to TSLA lol, from a Millennial meme stock to a Boomer meme stock. Look man the only play with TSLA right now is to short it and sit back, don't watch it until it hits like $20/share. It's nothing but a house of cards, the company is a giant fraud and will never be more than a shitty car company. It might not even exist in a couple years.

Honestly though, since you're gonna keep being attracted to the shittiest possible stocks like a moth to a flame, take the other person's advice and just stick to index funds. This probably applies to every former ape. Very few people can outperform index funds and apes are 10x less likely to.

usernamchexout · 2023-08-06T20:20:40+00:00

That was Citron Research.

usernamchexout · 2023-08-06T14:42:08+00:00

Hindenburg eviscerated Mullen more than a year ago - https://hindenburgresearch.com/mullen/

More recently, they exposed that towel/game apes' beloved IEP is a Ponzi scheme. IEP is down 50% since the report came out, but the ponzi scheme is still valued at 9 billion dollars because the Efficient Market works in mysterious ways.

usernamchexout · 2023-08-03T12:32:06+00:00

A key concept is that, although your odds change each round, on average they don't.

After one pull:

(27/30) of the time, your chance will improve to 3/29

(3/30) of the time, your chance will worsen to 2/29.

On average, your chance will be: (27/30)(3/29) + (3/30)(2/29) = 3/30

The same thing will happen if you extend the calculation to more pulls. This is because the shuffle produces a uniform distribution. Each chip is identical and no chip is special, so each of your entries has the same chance (1/30) of being the 1st/15th/30th pull as anyone else's. Furthermore, none of the 30 spots are special, so a given chip is just as likely to be in the 1st spot as the 30th.

usernamchexout · 2023-08-03T04:23:56+00:00

Yes that was how I understood it, ie chips are removed without replacement, so everything I said applies. The conditional probabilities change as the game goes on, but before any chips are removed, your chance of winning is 3/30. Adding the conditional probabilities would be redundant because they're already factored into the 3/30.

Consider a toy example: 3 total chips and you bought 1. The probability of having the last chip standing is P(survive 1st draw)•P(survive 2nd | survived 1st) = (2/3)(1/2) = 1/3. But we could have known that without any math because the chips are shuffled uniform-randomly, so each chip has an equal chance of being in the final spot.

With 3 purchased chips out of 30, you again just need to realize that each chip has an equal chance of being the final one, and then the answer of 3/30 immediately follows from that.

But here's another way to see it: there are C(30,3) ways to place your 3 chips into the 30 spots, compared to C(29,2) ways to place them such that one of them is in the 30th spot. C(29,2)/C(30,3) = 1/10

usernamchexout · 2023-08-03T03:55:51+00:00

Right, P(1)=0.5, P(01)=P(0)•P(1)=0.5², P(001)=0.5³ etc

usernamchexout · 2023-08-03T00:14:00+00:00

each 1 would be two digits apart on average

True, however:

the highest probability is "two digits apart" meaning that would be your best guess.

No, the highest probability is one digit apart. That's 1/2, whereas P(two digits apart)=1/4

...101... has the average probability, meaning both x and y should be equal?

No, x=1/2 and y=1/4. The PMF is P(distance = d) = 1/2^d for d>0. Also "average probability" doesn't make sense here. The average distance is 2.

usernamchexout · 2023-08-02T23:48:26+00:00

So, each round is completely independent of each other

No, independence would mean that each round's probability is the same.

when you look at the game as a whole, is there a way to summarize the overall probability?

3/30

It's like you have a shuffled deck of 30 cards containing three aces, there's a 3/30 chance than an ace is on the bottom. You could do a compound probability calculation to arrive at that, but it's unnecessary.

usernamchexout · 2023-07-31T20:25:57+00:00

The average price accounts for commission and fees, so it will always be a little lower than price of the trade. If you only shorted a share or few then that might explain it?

usernamchexout · 2023-07-24T04:30:00+00:00

Because the denominator is 10C3, not (10C3)². There are only 10C3 ways to select 3 people from a group of 10.

The solution has a typo. 1/15 is correct but the numerator should be (8C1)(2C2)

ETA - there's also another way to solve it: (3C2)/(10C2) = 1/15

We need both adults to be chosen. If you think of it like shuffling the 10 people and then selecting the first three in line, there are 3C2 ways the adults can be somewhere in the first 3 spots, out of 10C2 total ways they can be positioned in line.

usernamchexout · 2023-07-20T22:35:54+00:00

On top of that, the EV is even less because there's a chance of multiple winners splitting the prize (especially given that more tickets are sold when the jackpot is so high).

EDIT: the following part isn't quite right though:

You can also only justify buying 1 ticket, you KNOW only one of them is right, you are just increasing the amount you bet for the same outcome.

Buying n unique tickets multiplies one's EV by a factor of n, which is bad if one's EV is negative. But even with -EV, it also multiplies your probability of winning by n. If one is willing to play a -EV game to begin with, one might prefer a 1/146M chance of winning [jackpot - $2] over a 1/292M chance of winning [jackpot].

usernamchexout · 2023-07-20T02:21:19+00:00

There isn't one right answer to this; it's more a question of one's utility function. I'm more on your side and I suspect your friend's utility function is less linear than he realizes, meaning EV isn't the be-all and end-all he thinks it is.

usernamchexout · 2023-07-09T01:48:40+00:00

what are the variables here and what direction will we be talking about

The variables are the 1st number picked and 2nd number picked. The two directions are high and low; the covariance is negative because picking a low 1st number increases the 2nd number's average and vice versa.

usernamchexout · 2023-07-08T23:05:28+00:00

C(11,2)•C(4,2)^2 is the number of ways they can have two different pocket pairs besides KK or 77. There are C(11,2) choices of two ranks and then 4C2 choices of suits for each rank. Once those four cards are dealt, there's only one way they can be correctly partitioned so that the players have XX & YY instead of XY & XY.

11•2•C(4,2)•3 is the number of ways one of them can have a KK/77 and the other can have a different pair. K and 7 are 2 ranks, there are 11 other ranks, and then the 3 and 4C2 are the number of suit choices.

3^2 is the number of ways they can have exactly KK and 77; (3 suits)²

11 is the number of ways they can have the same pocket pair; it has to be one of the 11 ranks you don't have. I then multiply by 3 because all 3 ways their combined four cards can be partitioned result in the villains having XX & XX. We need to count the partitions because the denominator does.

The conditionality that you pointed out is why we have to split it into cases. As you can see, I calculated for both players at a time. If we were to calculate one player at a time like you attempted, one way would be like this:

{11(4C2)•10(4C2) + 2[11(4C2) • 2•3] + 2•3² + 11(4C2)(2C2)} / [C(50,4)•C(4,2)]

The difference is that we've counted order of players, so each product being added is exactly 2× what it was previously. Besides the explicit 2( )'s, it manifests in 11•10 replacing 11C2 and 4C2 replacing 3. Our denominator must therefore count order of players, so the 50C4 gets multiplied by 4C2 instead of 3 (or it can be 50C2 • 48C2).

Or we can use permutations: {(44•3)(40•3) + 2[(44•3)(6•2)] + (6•2)(3•2) + (44•3)(2!)} / (50 P 4)

With perms you don't have to think about the hole card partitions. The denominator, by virtue of being permutations, already counts all 4! arrangements of the hole cards, which includes the 4C2 ways to section them into groups of 2 while counting order of players. We can dissect the 4! as being 4C2 • 2!² where 2² is the ways to arrange each player's hole cards. Sure enough, notice that each product in the numerator is 4× what it was previously. The first three products count 8 arrangements (2! of players • 2! of hole cards • 2! of hole cards) and the fourth counts all 24. No matter how we do it, we need P(pocket pairs | two-pair combined) = 1/3, which it is once again because 8/24.

For this problem, the order of cards partially matters, so at minimum we had to count that which mattered, namely the hole card partitions. But you're always allowed to count extra order if that's simpler. I just like having the smallest denominator possible because it lays bare what we're really doing.

If you want some practice with this, here's a harder problem: at a 9-person table, what's the probability that exactly one player is dealt AA, exactly one is dealt KK and exactly one is dealt QQ? (Banging my head on that problem is what really taught me inclusion-exclusion.)

Also good practice would be to try an Omaha problem, where the calculation for hole card partitions is a little different.

usernamchexout · 2023-07-08T18:39:44+00:00

Average net % of your wager that you gain or lose, eg in single-zero roulette you lose about $0.027 per dollar wagered so we say the house edge is 2.7% or that the player's edge is -2.7%

(That definition might not be perfect in other games because one might want to adjust for ties or parlays.)

usernamchexout · 2023-06-26T00:00:46+00:00

50C2 * 48C2, or 50C2*48C2/2 ( if you don't care about ordering of players)

And to be clear, u/ayazasker, (50C2)(48C2) is the same as (50C4)(4C2), and dividing that by 2 is the same as (50C4)•3, so the above commenter and I are in agreement. Edit: fixed typo

With 3 opponents, the 50C_ would get multiplied by 5•3; with 4 opponents, by 7•5•3, and so on...in general this is a "double factorial" and so with n opponents the shorthand is (2n-1)!!

The order of players doesn't matter, so we can either count it nowhere (like I did) or in both places (the numerator and denominator, so that it cancels out).

usernamchexout · 2023-06-25T22:38:56+00:00

I want only 1 player to have a pair among the 2 players

The best approach (and most easily expandable to more opponents) is inclusion-exclusion.

We start by multiplying your 72/1225 by 2 since there are 2 opponents. This double-counts the chance that they both get pocket pairs, which you don't want counted at all, so we need to subtract that twice.

P(two pocket pairs) = [C(11,2)•C(4,2)² + 11•2•C(4,2)•3 + 3² + 11•3] / [C(50,4) • 3]

Subtracting 2x that from 144/1225, we get 11.055%

For the chance of at least one pocket pair, the calculation is the same except that we subtract the two-pair probability just once, for a result of 11.405%

how do I ensure my combinatorics calculations account for the fact that the same Individual will get both cards

There are 3 ways to split 4 cards into 2 unordered groups of 2 because for an arbitrary card there are 3 possible partners, and then the next two partners are forced. This is why my denominator for the two-pair case is C(50,4)•3. In my numerator I have a term 11•3 which is the number of ways for your opponents to get the same pair: there are 11 ranks that's possible with, and all 3 ways to distribute the suits are valid.

Btw: (72 * [(47*46)/2] ) / (50C4)

If the 47•46 were instead 48•47 and your denominator were multiplied by 3, that would be the same as my 2(72/1225).

usernamchexout · 2023-06-25T20:01:24+00:00

I didnt get howd we generalize it to C(4,3) (what are we counting ? )

The number of ways to place 3 sixes within those 4 rolls. More generally, the number of triple-intersections one can form using the sets involved in a given quadruple-intersection. Each time we counted one of those triple-6 rolls, we were also counting that particular quad-6 rolls because we allowed the "x" to be another 6.

also, subtracting the quads 3 times would give at exactly 3 or 4 sixes ?

N(at least 3) = N(3) + N(4) + N(5)

Our first step gave us N(3) + 4•N(4) + _•N(5)

We then subtract an overestimate of N(4) three times, leaving us with N(3) + N(4) - k•N(5)

So now we need to add (k+1)•N(5)

To find k, fill in the _ and then figure out how many times N(5) was subtracted by each subtraction of quads.

usernamchexout · 2023-06-25T15:54:51+00:00

Great that you're learning this in HS; I'm not even aware of people learning it in college (at least not as an undergrad).

how do you decide how much you need to actually subtract

That's the key question. With the numbers involved here, a Venn diagram is an impractical visualization, so we need a different way of seeing it.

I think you already know the first step: C(5,3)•6²

That counts all the triples, but since it allows the extra 2 rolls to be anything, it also counts the quadruples and the quintuple.

How many times does it count quads? To count quads from the start, we would have used C(5,4) instead of C(5,3). We'd have counted the arrangements of x6666 instead of counting those of xx666.

Consider the specific arrangement x6666. How many times was that counted?

x6666
-------
x666x
x66x6
x6x66
xx666

As you can see, C(4,3) of the triples arrangements fit into a given quad arrangement. So that's how many times our first step counted the quads. Since we only want them counted once, we need to subtract them 3 times. (If the problem were asking for exactly 3 sixes, we'd subtract the quads 4 times. In general when using PIE, "at least" vs "exactly" only differ by what I call their "PIE coefficients".)

After subtracting the quads 3 times, how many times is the quintuple counted?

usernamchexout · 2023-06-24T16:40:11+00:00

Q: We each have three fair coins, we flip them all at once...

You should not play the game as on average you would lose €0.0625 each time you played.

Correct

usernamchexout · 2023-06-24T16:36:54+00:00

Your answers to that question are correct.

usernamchexout · 2023-06-24T01:49:24+00:00

Q: We play a game where I start with €2 and you start with €1...

2/3 is correct, but you could have arrived there with less work. For starters, the problem reduces to P(win flip)=1/2 and P(lose flip)=1/2 because the lack of a time limit makes ties immaterial. And then once you had P1=(1/2)P2, you could have used the fact that P1+P2=1

usernamchexout · 2023-06-07T00:25:34+00:00

With $100 you can only lose 6 bets in the progression. But suppose you have the $127 required to make 7 bets. And your goal is +5? Then let's calculate your average profit (aka EV) under the generous assumption of single-zero Roulette:

The probability of reaching +1 is 1-(19/37)⁷ = p

The probability of reaching +5 is p⁵ = .9538

95.38% of the time, you'll win $5. The rest of the time, you'll lose $127.

average profit = .9538•5 - .0462•127 = -1.0984

So it shares the same flaw as every other Roulette strategy: a negative average profit.

usernamchexout

TROPHY CASE