t500 bonus spin strategy

[OldTrees]

The Elegant complete solution is long and in progress.

[Hodari]

At least we're lucky that Schroedinger's Cat isn't rare.  Otherwise we'd have to consider the chance that it simultaneously IS and ISN'T in the box after each spin which would further complicate things

[OldTrees]

Nvm. It quickly became brute force.

Here is my work so far if someone else wants to pick it up.

White = 0 spins left
Red = 1 spin left
Yellow = 2 spins left
Green = 3 spins left
r = number of rares

Rules:
When a slot becomes locked (white), convert all other slots to match that slot if possible.
When a pair exists ABA spin to match the pair because the worst case scenario ABA is still able to try matching the paired slots to the single. This is not the case in the reverse ABC which also has to reserve 1 spin to be able to try to match the locked slot.

Probabilities:
AAB
(1/r) chance of rare
AAB
(1/r) + (r-1)(1/r)(1/r)
=(1/r²)(r+r-1)=(1/r²)(2r-1)
AAB
(1/r) + (r-1)(1/r)(1/r) + (r-1)(1/r)(r-1)(1/r)(1/r)
=(1/r³)(r²+r²-r+r²-2r+1)=(1/r³)(3r²-3r+1)

ABC
(1/r)(1/r)
=(1/r²)
ABC
(1/r²)(2r-1)(1/r)
=(1/r³)(2r-1)
ABC
(1/r³)(3r²-3r+1)(1/r)
=(1/r⁴)(3r²-3r+1)
ABC
(1/r²)(2r-1)(1/r²)(2r-1)
=(1/r⁴)(4r²-4r+1)
ABC
(1/r³)(3r²-3r+1)(1/r²)(2r-1)
=(1/r⁵)(6r³-9r²+5r-1)
ABC
(1/r³)(3r²-3r+1)(1/r³)(3r²-3r+1)
=(1/r⁶)(9r⁴-18r³+18r²+6r+1)

ABC
(2/r)(1/r) + (r-2)(1/r)(1/r²)
= (1/r³)(4r-2)
ABC
Either 1st r=3
2(1/r) AAB (1/r) + (r-1)(1/r)(1/r²)
(r-2)(1/r) ABC (1/r³)(4r-2)
=(1/r⁴)(2r²+2r-2) + (1/r⁴)(4r²-10r+4) =(1/r⁴)(6r²-8r+2)
OR 2nd r>4
(1/r) ABA (1/r)
(1/r) ABB (1/r) + (r-1)(1/r)(1/r) = (1/r)(2r-1)
(r-2)(1/r) ABC (1/r³)(2r-1)
=(1/r⁴)(r²) + (1/r⁴)(2r³-r²) + (1/r⁴)(r²-5r+2) =(1/r⁴)(2r³+r²-5r+2)
The chance of a rare from a AAB is the chance of AAB + (1-AAB) (AAB)

Also the chance of starting with
AAA is (1/r²)
AAB is (3)(r-1)(1/r²)
ABC is (r-1)(r-2)(1/r²)

[Hodari]

The chance of a rare from a AAB is the chance of AAB + (1-AAB) (AAB)

It's more complicated than that actually.  From an initial start of AAB and following the most obvious strategy for example, you can win by spinning the third reel and having it match.  Or you can win if it doesn't match the first time but does on your second or third spin with that reel.  After that, Reel 3 is locked so if you still haven't matched, then you still have a chance to win if you can get the other 2 reels to match whatever it ended up on(and I actually did win once this way already) on any of the remaining spins with those reels(each possible combination of number of spins taken must be accounted for here as well).  So yeah if you want to fully analyze it, you have to look at each spin, figure out all remaining choices and follow each branch there until 1 of 2 conditions is met:
1. You win
2. 2 reels are locked with no more remaining spins and these 2 do not match

[OldTrees]

The chance of a rare from a AAB is the chance of AAB + (1-AAB) (AAB)

It's more complicated than that actually.  From an initial start of AAB and following the most obvious strategy for example, you can win by spinning the third reel and having it match.  Or you can win if it doesn't match the first time but does on your second or third spin with that reel.  After that, Reel 3 is locked so if you still haven't matched, then you still have a chance to win if you can get the other 2 reels to match whatever it ended up on(and I actually did win once this way already) on any of the remaining spins with those reels(each possible combination of number of spins taken must be accounted for here as well).  So yeah if you want to fully analyze it, you have to look at each spin, figure out all remaining choices and follow each branch there until 1 of 2 conditions is met:
1. You win
2. 2 reels are locked with no more remaining spins and these 2 do not match

Open the spoiler. You repeated some of what I said.
Ex:
The chance of getting a rare from AAB (1 pair, 1 spin left on 1st and 2nd slots, 2 spins left on 3rd slot)
Is the chance of getting a rare from AAB (1 pair, 3rd slot has 2 spins left) [=(1/r) + (1-1/r)(1/r)=(2r-1)/r²]
Plus (1 - the chance of getting a rare from AAB) [=1-(2r-1)/r²=[(r²-2r+1)/r²]
Times the chance of getting a rare from AAB (1 pair, 1st and 2nd slots have 1 spin left each) [=1/r²]
For a total of (2r-1)/r² + (r²-2r+1)/r² x (1/r²) = (2r³-2r+1)/r⁴

Edit: A more complete reference table
AxAyBz means 1 pair (AAB) with x,y,z spins left
A2A1B1 means 1 pair (AAB) with 2,1,1 spins left

r still means the number of rares in each slot

AAB
xyz  (chance of winning a rare starting at xyz)
001 (1/r)
110 (1/r²)(1)
002 (1/r²)(2r-1)
210 (1/r³)(2r-1)
111 (1/r³)(r²+r-1)
003 (1/r³)(3r²-3r+1)
310 (1/r⁴)(3r²-3r+1)
220 (1/r⁴)(4r²-4r+1)
211 (1/r⁴)(r³+2r²-3r+1)
112 (1/r⁴)(2r³-2r+1)
320 (1/r⁵)(6r³-9r²+5r-1)
311 (1/r⁵)(r⁴+3r³-6r²+4r-1)
221 (1/r⁵)(r⁴+4r³-8r²+5r-1)
212 (1/r⁵)(2r⁴+r³-5r²+4r-1)
113 (1/r⁵)(3r⁴-2r³-2r²+3r-1)
330 (1/r⁶)(9r⁴-18r³+18r²+6r+1)
321 (1/r⁶)(r⁵+6r⁴-15r³+14r²-6r+1)
312 (1/r⁶)(2r⁵+2r⁴-9r³+10r²-5r+1)
222 (1/r⁶)(2r⁵+3r⁴-12r³-3r²-6r+1)
213 (1/r⁶)(3r⁵-6r⁴+10r³-10r²+5r+1)
331 (1/r⁷)(r⁶+9r⁵-27r⁴+36r³-12r²-5r-1)
322 (1/r⁷)(2r⁶+5r⁵-21r⁴+29r³-20r²+7r-1)
313 (1/r⁷)(3r⁶-11r⁴+19r³-18r²+6r-1)
223 (1/r⁷)(3r⁶+r⁵-15r⁴+23r³-19r²+7r-1)
332 (1/r⁸)(2r⁷+8r⁶+63r⁴-48r³+7r²+4r+1)
323 (1/r⁸)(3r⁷+3r⁶-26r⁵+50r⁴-49r³+27r²-8r+1)
333 (1/r⁹)(3r⁸+6r⁷-44r⁶+99r⁵-111r⁴+55r³-3r²-3r-1)
       
ABC
111
(1-2/r)(1/r²)+(2/r)(1/r)
=(1/r³)(3r-2)
211  (please double check this math. This is the third time I did this and it keeps alternating being equal and not)
1st Slot: (1-2/r)(1/r³)(3r-2)+(2/r)(1/r³)(r²+r-1)
=(1/r⁴)(5r²-6r+2)
3rd Slot: (1/r)(1/r)+(1/r)(1/r²)(2r-1)+(1-2/r)(1/r³)(2r-1)
=(1/r⁴)(5r²-6r+2)
221 (weird. Both paths are the same?)
1st Slot: (1-2/r)(1/r⁴)(5r²-6r+2)+(1/r)(1/r⁴)(r³+2r²-3r+1)+(1/r)(1/r⁴)(2r³-2r+1)
=(1/r⁵)(8r³-14r²+9r-2)
3rd Slot: (2/r)(1/r²)(2r-1)+(1-2/r)(1/r⁴)(4r²-4r+1)
=(1/r⁵)(8r³-14r²+9r-2)
311 (again both paths are the same. Perhaps this is caused by no pairs?)
1st Slot: (1-2/r)(1/r⁴)(5r²-6r+2)+(2/r)(1/r⁴)(r³+2r²-3r+1)
=(1/r⁵)(7r³-12r²+8r-2)
3rd Slot: (1/r)(1/r)+(1/r)(1/r³)(3r²-3r+1)+(1-2/r)(1/r⁴)(3r²-3r+1)
=(1/r⁵)(7r³-12r²+8r-2)
222
(1-2/r)(1/r⁵)(8r³-14r²+9r-2)+(2/r)(1/r⁵)(2r⁴+r³-5r²+4r-1)
=(1/r⁶)(12r⁴-28r³+27r²-12r+2)
321 (all equal again. I checked with excel so I will only simplify 1)
1st Slot: (1-2/r)(1/r⁵)(8r³-14r²+9r-2)+(1/r)(1/r⁵)(r⁴+4r³-8r²+5r-1)+(1/r)(1/r⁵)(2r⁴+r³-5r²+4r-1)
2nd Slot: (1/r)(1/r⁵)(r⁴+3r³-6r²+4r-1)+(1-2/r)(1/r⁵)(7r³-12r²+8r-2)+(1/r)(1/r⁵)(3r⁴-2r³-2r²+3r-1)
3rd Slot: (1/r)(1/r²)(2r-1)+(1/r)(1/r³)(3r²-3r+1)+(1-2/r)(1/r⁵)(6r³-9r²+5r-1)
=(1/r⁶)(11r⁴-25r³+24r²-11r+2)
322
1st Slot: (1-2/r)(1/r⁶)(12r⁴-28r³+27r²-12r+2)+(2/r)(1/r⁶)(2r⁵+3r⁴-12r³-3r²-6r+1)
3rd Slot: (1/r)(1/r⁶)(2r⁵+2r⁴-9r³+10r²-5r+1)+(1-2/r)(1/r⁶)(11r⁴-25r³+24r²-11r+2)+(1/r)(1/r⁶)(3r⁵-6r⁴+10r³-10r²+5r+1)
=(1/r⁷)(16r⁵-46r⁴+59r³-72r²+14r-2)

[Xenocidius]

At least we're lucky that Schroedinger's Cat isn't rare.  Otherwise we'd have to consider the chance that it simultaneously IS and ISN'T in the box after each spin which would further complicate things

I loled.

On-topic: So this is all assuming that there are only 3 rares chosen for spins. Is this true?

[Chromatophore]

On-topic: So this is all assuming that there are only 3 rares chosen for spins. Is this true?

For my part, yes.  OldTrees was working with r rares.

I got to the rare spin maybe 3-5 times and i didn't see more than three different cards.

[Captain Scibra]

Eh, I actually had the inside being the odd one the 1st two times, and spinning the outside got me the rare the second time, would have the first if I didn't do the obvious thing.  This may have just been coincidence though.

[Hodari]

On-topic: So this is all assuming that there are only 3 rares chosen for spins. Is this true?

I had been assuming that ANY rare could be chosen on any spin.  If it is indeed only 3 then that would significantly change things and probably would make it much closer in terms of which strategy would be better.  I would still guess that the obvious strategy would be the best bet in that case but it would be far less clear than for n being around 17(12 weapons, 3 types of shard,  pharaoh, squid) like I had assumed

[The_Mormegil]

Uhm...
Best strategy for AAB: spin the B first, then the rest.
Best strategy for ABC: spin the following
1 ) A
2 ) A'AB -> B; A'BC -> B
3 ) A'AB' -> B'; A'A'B -> B; A'B'C -> C
4 ) A'AB'' -> B''; A'A'B' -> B'; A'B'C' -> A'
5 ) A'AB''' -> A; A'A'B'' -> B''; A''B'C' -> B'; A''A'B' -> B'
6 ) A'A'B''' -> A'; A'B'''C' -> A'; A''B''C' -> C'; A''A''B' -> B'; A''A'B'' -> B''
7 ) A'A''B''' -> A''; A'B'''C'' -> A'; A''B'''C' -> A''; A''A''B'' -> B''; A''A'B''' -> A''
8 ) A'A'''B''' -> drop; A'B'''C''' -> drop; A'B'''B''' -> A'; A''A''B''' -> A''
9 ) A'B'''B''' -> A'; A''A'''B''' -> drop; A''B'''B''' -> A''

Using the above, and p=1/r, q=(r-1)/r, s=(r-2)/r:
AAA: p^2 * 1
AAB: 3pq * (p + pq + p q^2 + p^2 q^3 + 2 p^2 q^4 + 3 p^2 q^5 + 2 p^2 q^6 + p^2 q^7)
ABC: 6qs * (2p(p + q(p + q(p + q(p^2(1 + 2 q + 2 q^2 + q^3))))) + s(p(p + q(p + q(p + q(p^2(1 + 2 q + q^2))))) + p(p + q(p + q(p^2(1 + 2 q + 2 q^2 + q^3)))) + s(2p(p + q(p + q(p^2(1 + 2 q + q^2)))) + s(2p(p + q(p + q(p^2(1 + q)))) + s(p(p + q(p + q(p^2))) + p(p + q(p^2(1+q))) + s(2p(p + q(p^2)) + s(2p(p) + s(p^2))))))))

And that's pretty ugly but correct. I hope. I'd put it in Maxima or something to see what the hell comes out of it... but I don't have Maxima anymore.

EDIT: also, r is 18 right now: 12 weapons, 3 shards, miracle squid and pharaoh.

[OldTrees]

@The Mormegil
"A'BC -> B"
Evidence? As seen with my calculations on A'B''C'' the correct choice is dependent on the number of rares in the slots per special spin.

PS:has anyone seen more than 3 rares in a special spin?

[The_Mormegil]

@The Mormegil
"A'BC -> B"
Evidence? As seen with my calculations on A'B''C'' the correct choice is dependent on the number of rares in the slots per special spin.

PS:has anyone seen more than 3 rares in a special spin?

Not really evidence, as I specified, the strategy is not proven. Anyway, I chose to split spins as much as possible when you have 3 different rares, because if you don't you'll end up having less chances. Basically, I think it can be proven than if you end up with ANY couple you SHOULD spin the third one as much as possible. If you admit this, splitting spins is the best strategy as it maximizes your chances of having a couple (spinning only one of those will end up allowing only for ONE couple to be formed instead of two) and it maximizes the possible spins left when you actually obtain a couple. Anyway, I think it's easier to formulate two possible strategies and verify what happens rather than trying to prove it analytically.