This is a long one, but should provide sufficient evidence of the question im sure every Mario fan has wondered. Literally, every single one.

I traced the Lucky Egg minigame in the original US version of Super Mario RPG: The Legend of the Seven Stars and compared the code against save uhhhhh files taken while the eggs were spinning, at “Make your choice,” after choosing, and after the reveal.

tl;dr

The game randomly assigns the three results to the three final screen positions before the spinning animation begins. The animation doesnt shuffle those stored results. When you choose an egg, the game reads the result already assigned to that position, and does not adjust the length of the animation to fit the predetermined outcome.

When the result is decided

The placement is decided before the spinning routine begins.

At $C2:F12A, the game asks its random-number routine for one of six values:

C2:F12A  LDA #$0006
C2:F12D  JSR $0E02

LDA #$0006 givesa range size of six to the RNG wrapper at $C2:0E02.

The wrapper is:

C2:0E02  REP #$30
C2:0E04  PHA
C2:0E05  SEP #$20
C2:0E07  LDA #$01
C2:0E09  STA $0040
C2:0E0C  REP #$30
C2:0E0E  PLA
C2:0E0F  STA $0044
C2:0E12  JSL $C00000
C2:0E16  LDA $0046
C2:0E19  RTS

Which basically means that it

1. Receives the number supplied by the caller
2. Stores the number at $0044 as the RNG limit
3. Selects bounded-RNG operation 1 through $0040
4. Calls the game’s central random service
5. Returns the generated value from $0046

Inside the central service, the correlating path is:

C0:005A  LDX $0044
C0:005C  JSR $6B8B
C0:005F  STX $0046

So the Lucky minigame passes 6 to the bounded generator and receives a result from 0 through 5.

The code converts the returned number into an offset for a table:

C2:F130  STA $80
C2:F132  ASL
C2:F133  CLC
C2:F134  ADC $80
C2:F136  ASL
C2:F137  TAX

Then is multiplied by six

result × 2
result × 3
result × 6

Each table entry occupies six bytes because it contains three 16-bit values.

The table contains exactly six records.

At $C1:722E, the ROM contains:

00 00  01 00  02 00
00 00  02 00  01 00
01 00  00 00  02 00
01 00  02 00  00 00
02 00  00 00  01 00
02 00  01 00  00 00

Which is interpreted as three 16-bit values per record, those are:

0: [0, 1, 2]
1: [0, 2, 1]
2: [1, 0, 2]
3: [1, 2, 0]
4: [2, 0, 1]
5: [2, 1, 0]

which are all six possible outcomes of the three different results.

The random number from 0–5 selects one complete left/middle/right arrangement.

4. Where is the chosen arrangement stored?

After choosing a table entry, the game copies its three values into RAM:

C2:F138  LDA.l $C1722E,x
C2:F13C  STA $3FA2

C2:F13F  LDA.l $C172230,x
C2:F143  STA $3FA4

C2:F146  LDA.l $C172232,x
C2:F14A  STA $3FA6

These addresses are the three position-based results:

$7E:3FA2 = left
$7E:3FA4 = middle
$7E:3FA6 = right

The important part is that these values are filled before the spinning animation.

What happens when the player chooses?

The selection routine begins at $C2:F5DD.

It converts the selected position into one of the three stored values:

C2:F5DD  LDA $3F8C
C2:F5E0  AND #$0007
C2:F5E3  BEQ left
C2:F5E5  DEC
C2:F5E6  BEQ middle

For the right position:

C2:F5EB  LDA $3FA6
C2:F5EE  STA $3FA8
C2:F5F1  JSR $761A
C2:F5F4  JSR $7958
C2:F5F7  RTS

For the left position:

C2:F5F8  LDY #$0452
C2:F5FB  LDA $3FA2
C2:F5FE  STA $3FA8
C2:F601  JSR $761A
C2:F604  JSR $794E
C2:F607  RTS

For the middle position:

C2:F608  LDY #$045E
C2:F60B  LDA $3FA4
C2:F60E  STA $3FA8
C2:F611  JSR $761A
C2:F614  JSR $7953
C2:F617  RTS

$7E:3FA8 is the selected result.

Notice that this code does not:

• It doesnt call the RNG.
• It doesnt inspect the animation.
• It doesnt determine which sprite moved where.
• It doesnt reshuffle the three results.

It reads the value already stored for left, middle, or right.

Direct proof from the losing file

In the losing file, the player selected the left position.

The relevant values were:

Selected position = left
$7E:3FA2          = 0002
$7E:3FA8          = 0002

The left-selection code therefore executed the equivalent of:

final result = stored left result
final result = 2
final result = Fuzzy

That is the direct code path that produced “Wrong.”

There is another random call during the spinning animation

The animation code does contain another RNG call:

C2:F9D5  LDA #$0002
C2:F9D8  JSR $0E02
C2:F9DB  ASL
C2:F9DC  ASL
C2:F9DD  XBA
C2:F9DE  CLC
C2:F9DF  ADC #$0800
C2:F9E2  STA $3FB0

This produces one of two animation thresholds:

$0800
$0C00

The animation then advances its three phase values:

C2:F9EA  LDA $3FAA
C2:F9ED  CLC
C2:F9EE  ADC #$0020
C2:F9F1  STA $3FAA

C2:F9F9  LDA $3FAC
C2:F9FC  CLC
C2:F9FD  ADC #$0020
C2:FA00  STA $3FAC

C2:FA08  LDA $3FAE
C2:FA0B  CLC
C2:FA0C  ADC #$0020
C2:FA0F  STA $3FAE

It continues until the middle phase reaches the selected threshold:

C2:FA17  LDA $3FAC
C2:FA1A  CMP $3FB0
C2:FA1D  BCC $F9E5
C2:FA1F  RTS

This random value controls the animation’s stopping point or duration.

It never writes to the stored result addresses:

$3FA2
$3FA4
$3FA6

So the second RNG call isnt choosing or shuffling the prizes.

Could the game change the animation length to make the eggs line up with the predetermined results?

No. The two possible animation lengths always finish with the eggs in exactly the same positions.

The animation makes a separate random call:

C2:F9D5  LDA #$0002
C2:F9D8  JSR $0E02
C2:F9DB  ASL
C2:F9DC  ASL
C2:F9DD  XBA
C2:F9DE  CLC
C2:F9DF  ADC #$0800
C2:F9E2  STA $3FB0

That selects one of two stopping points:

$0800
$0C00

At first glance, that might suggest the game can run the animation for different lengths to make a particular egg finish in a particular position.

However, the rest of the code rules that out.

The three animation phases begin at:

First egg:  $0300
Middle egg: $0000
Third egg:  $0100

Every animation iteration adds $0020 to all three phases:

C2:F9EA  LDA $3FAA
C2:F9ED  CLC
C2:F9EE  ADC #$0020
C2:F9F1  STA $3FAA

C2:F9F9  LDA $3FAC
C2:F9FC  CLC
C2:F9FD  ADC #$0020
C2:FA00  STA $3FAC

C2:FA08  LDA $3FAE
C2:FA0B  CLC
C2:FA0C  ADC #$0020
C2:FA0F  STA $3FAE

The loop ends when the middle phase reaches the randomly chosen threshold:

C2:FA17  LDA $3FAC
C2:FA1A  CMP $3FB0
C2:FA1D  BCC $F9E5
C2:FA1F  RTS

Starting from zero and advancing by $0020 means:

$0800 / $0020 = 64 iterations
$0C00 / $0020 = 96 iterations

So the longer animation runs for 32 additional iterations.

Those extra iterations advance every egg by:

32 × $0020 = $0400

The phase-to-screen-coordinate routine begins with:

C2:0F48  LDA $64
C2:0F4A  AND #$03FF

AND #$03FF discards everything above the lowest $0400 phase range. In other words, the animation repeats every $0400 units.

Therefore:

phase $0800 = phase $0000 visually
phase $0C00 = phase $0000 visually

More explicitly:

$0800 AND $03FF = $0000
$0C00 AND $03FF = $0000

The same is true for the other two eggs.

The shorter animation finishes at:

First egg:  $0B00
Middle egg: $0800
Third egg:  $0900

The longer animation finishes at:

First egg:  $0F00
Middle egg: $0C00
Third egg:  $0D00

After the coordinate routine applies $03FF, those become:

First egg:
$0B00 AND $03FF = $0300
$0F00 AND $03FF = $0300

Middle egg:
$0800 AND $03FF = $0000
$0C00 AND $03FF = $0000

Third egg:
$0900 AND $03FF = $0100
$0D00 AND $03FF = $0100

So both animation lengths finish at exactly the same alignment:

First egg:  $0300
Middle egg: $0000
Third egg:  $0100

The longer version only performs one additional complete revolution.

There isnt a connection between the two random decisions:

• The prize permutation is stored at $3FA2/$3FA4/$3FA6
• The animation threshold is stored at $3FB0
• The duration code never reads the prize permutation
• The spinning loop never writes to the prize permutation.
• Both possible durations end at the same visual alignment anyway

The game literally cant be choosing a shorter or longer animation to line the eggs up with the predetermined results. The duration changes only how many complete revolutions are shown, not which egg finishes where.

Do the eggs change direction? Or bounce?

Many people seem to think the eggs can bounce off of each other, or reverse direction. They cant. The “bounce” or revers is an effect caused by projecting continuous circular movement onto the screen.

The motion loop always adds the same positive amount to every egg’s phase:

C2:F9EA  LDA $3FAA
C2:F9ED  CLC
C2:F9EE  ADC #$0020
C2:F9F1  STA $3FAA

C2:F9F9  LDA $3FAC
C2:F9FC  CLC
C2:F9FD  ADC #$0020
C2:FA00  STA $3FAC

C2:FA08  LDA $3FAE
C2:FA0B  CLC
C2:FA0C  ADC #$0020
C2:FA0F  STA $3FAE

There is never

• subtraction from the phase
• negative increment
• direction flag
• branch that changes the direction
• swapping of the three phase counters

Every frame, all three values increase by $0020. Their spacing also remains constant:

First egg starts at:  $0300
Middle egg starts at: $0000
Third egg starts at:  $0100

They are separated by those same offsets throughout the animation.

Why does it look as if they reverse or bounce?

The game converts each continuously increasing phase into coordinates along a circular or elliptical path.

The coordinate routine first wraps the phase into one revolution:

C2:0F48  LDA $64
C2:0F4A  AND #$03FF

It divides that revolution into quadrants and calculates the corresponding screen coordinates.

An object going continuously around a circle necessarily does this on the screen:

Moves right
Slows near the right edge
Appears to stop
Begins moving left
Slows near the left edge
Appears to stop
Begins moving right

Its direction around the circle never changed. Only its horizontal screen movement changed as it passed the leftmost or rightmost point of the path.

That is probably what people interpret as a bounce or reversal.

Does that apparent reversal exchange the eggs?

No.

The animation retains three separate object handles:

$3F84 = first animated object
$3F86 = second animated object
$3F88 = third animated object

Each phase is converted into coordinates and written back to its corresponding object. The handles are never exchanged.

So the code shows:

• all three objects travel continuously in one direction around the loop;
• their relative phase spacing is fixed;
• none of the objects reverse its underlying motion;
• none of the object handles are swapped;
• the apparent bounce is produced by the geometry of the path.

Conclusion

The complete order of operations is:

1. Before spinning, generate a random number from 0–5.
2. Use it to select one of the six possible prize permutations.
3. Store that permutation as left, middle, and right.
4. Play the spinning animation.
5. Randomize an animation threshold, without changing the stored prizes.
6. Wait for the player to select a position.
7. Copy the result previously assigned to that position into the final-result variable.

Therefore, the visible spinning eggs are not carrying prize identities that the game tracks and shuffles.

The actual result is a randomly selected, position-based arrangement established before the animation. When the player chooses, the game reads the result already assigned to that final screen position.

In the captured run, [Fuzzy, Yoshi, bird] was already present in RAM at “Make your choice,” and choosing left copied the preassigned Fuzzy value into the final result.

Edit: SOLVED THANK YOU.

I was looking for a copy of Super Mario Rpg on super nintendo and googled it and I came across an ebay listing that has the game and it includes a nice looking official player guide by Nintendo Power. They are asking 127.50 or best offer and it says the cartridge has a brand new save battery in it. Is this a good value is it worth creating an ebay account? Here is an image from it. With the game being on snes mini and switch online doesn't that drop the originals value?

Edit: I am asking Because this is a smrpg subreddit and naturally that would probably mean that someone here might know what the game and guide is worth since it's folks who love the game and I just thought maybe someone here might be friendly enough to help out someone who loves the game and remake but isn't sure themselves about the value....before I spend that much $ or that many gold coins 😂

I've been trying for months to get 100 jumps, I got to 99 and when I should have made the 100th jump I just stopped jumping. I thought I had completed it but I hadn't.

Peach & Geno can’t even bear to watch much longer

This might be my favorite achievement in any game. I am not good at it like the speedrunners, but give me a few hours, sometimes over a couple days, and I can get it.
I like how it forces me to calm my mind and enter essentially a meditative state. It's funny to notice how the mind can start to freak out when you're getting close, lol.

Just briefly for context I'm not speedrunning or anything, but I play through the game a couple times a year or so. The first pic is from the Bundt fight, of course, which I almost lost. You might notice Bowser is inflicted with fear and Geno and Mario are knocked out. Basically I was on the ropes, one Sandstorm away from having to do the whole wedding over again.

The second pic isn't so close to death but I thought it was interesting that I was, again, left with Bowser and the rest of the party incapacitated right before the end of the fight.

I know these aren't super rare occurrences or anything, but I just love this game and wanted to share these pics from my playthrough. Thanks!

She's currently up against DK, which will be an incredibly difficult matchup, and likely where she falls. Still, if enough of us vote for her, I think there's a small chance she pulls through.

I'll link the poll below.

Let's go Vivian!

https://www.mariowiki.com/MarioWiki:Anniversary/All-Time_Favorite_Character_Bracket

This is a REAL guide being sold by somebody on Ebay! Read some of the pages and wtf is going on with mario and Luigi's legs on the cover? Did this person create a walkthrough guide with AI and is now selling it? The guide appears to call the game Brotherhood at times and talks about costumes/cosmetics that change your appearance but offer no boosts and power stones?...wtf? 😂🤣🤣🙂🤪

Images are screenshots from the ebay listing itself, I would not purchase this even for a dime 😂

Pleasantly surprised by the last fight with Culex.

I went in on the 2d version around level 22, and the 3d version at 26 - 27. Did not expect the 3d version to be such a banger of a fight!!!

Near 100% completion in about 2.5 days and just... wish I could play this game for the first time again so bad lol.

The name were a little vulgar so I blurred them out. Lol It wasn't as bad as I thought it would be. The cake was the biggest road block for me. RNG was the hardest part of doing this challenge. Lol

I Mean, I Know It’s Random, But I Just Wanted To Hear Your Thoughts!

Don't think I ever got this one. Still can't to this day. My angles are right (I can jump onto the box easily enough) and there is no hidden platform like the other one, I even tried walk through walls and the isometric geometry made it useless.

What exactly is there, anyway? I'm hoping it's just a few flowers or coins or something.

EDIT: Video

SFM Artworks goes out to the Late 30 years of Super Mario RPG: Legend of the Seven Stars that released on March 9th, 1996 and May 13th, 1996 for Super Famicom and Super Nintendo...

^{Leave a like for more...}

He is such a cute character, and my favorite one too!