t500 bonus spin strategy

[The_Mormegil]

I think I have gathered enough evidence (read: I farmed enough Gold) to say that r = 4.

4? That would be a really weird choice. I trust your claim that r>3 but r=4 would require at least a month of farming by the community to confirm r<5.

That r >3 is proven. That r = 4 is not proven, but I think my data is sufficient: every spin actually gives you about 12 spins to check it out and in a full day of farming (about 15 spins, more or less) I never got a fifth rare. However, I encourage the community to farm more to confirm this.

[Rubik87]

I was doing the calculations from below (instead of from above) assuming there are exactly 3 different rares in the spins, i am discouraged to continue now that i read there are 4 (maybe 1 is often repeated).

Anyway, doing all the cases up to 4 spins left I found the first interesting thing: in the case
ABC
112
it is better to spin A or B (29/81 chance of winning something) than C (28/81 chance).

[The_Mormegil]

I was doing the calculations from below (instead of from above) assuming there are exactly 3 different rares in the spins, i am discouraged to continue now that i read there are 4 (maybe 1 is often repeated).

Anyway, doing all the cases up to 4 spins left I found the first interesting thing: in the case
ABC
112
it is better to spin A or B (29/81 chance of winning something) than C (28/81 chance).

Hmmm... intuitively true: you are going to spin A or B anyway, and by spinning C beforehand you are eliminating the chance of it being the right one to start with.

[OldTrees]

I was doing the calculations from below (instead of from above) assuming there are exactly 3 different rares in the spins, i am discouraged to continue now that i read there are 4 (maybe 1 is often repeated).

Anyway, doing all the cases up to 4 spins left I found the first interesting thing: in the case
ABC
112
it is better to spin A or B (29/81 chance of winning something) than C (28/81 chance).

When R=3, I get 29/81 for either choice. This held true for R>3.
Please double check you didn't miss any options.

[Jenkar]

Just a question. Translation in english, from maths?

[The_Mormegil]

@Jenkar: still working.

I was doing the calculations from below (instead of from above) assuming there are exactly 3 different rares in the spins, i am discouraged to continue now that i read there are 4 (maybe 1 is often repeated).

Anyway, doing all the cases up to 4 spins left I found the first interesting thing: in the case
ABC
112
it is better to spin A or B (29/81 chance of winning something) than C (28/81 chance).

When R=3, I get 29/81 for either choice. This held true for R>3.
Please double check you didn't miss any options.

Spin C:
1) A1B1C1 = r-2 / r
--- this leads to a win on 1/r^2 cases.
2) A1A1B1 = 2 / r
--- this leads to a win on 1/r cases.

Total: r-2 / r^3 + 2 / r^2 = 3r - 2 / r^3
Total with r = 3: 7/27
Total with r = 4: 10/27

Spin A/B:
1) A0A2B1 = 1/r
--- this leads to a win on 1/r cases.
2) A0A1B2 = 1/r
--- this leads to a win on 1/r + r-1 / r^2 cases.
3) A0B1C2 = r-2 / r
--- this leads to a win on 1/r * (1/r + r-1 / r^2) cases.
Total: 1/r^2 + 1/r^2 + r-1 / r^3 + r-2 / r^3 + (r-2)(r-1) / r^4 = r^2 + r^2 + r^2 - r + r^2 - 2r + r^2 - 2r - r + 2 / r^4 = 5r^2 - 6r +2 / r^4
Total with r = 3: 29 / 81
Total with r = 4: 58 / 256

They are not identical. However, with r = 3 spinning A/B is better, while with r = 4 it's definitely worse.
Conclusion: since r = 3 has been proven false, spinning C is better.

[Chromatophore]

Spin C:
1) A1B1C1 = r-2 / r
--- this leads to a win on 1/r^2 cases.
2) A1A1B1 = 2 / r
--- this leads to a win on 1/r cases.

Total: r-2 / r^3 + 2 / r^2 = 3r - 2 / r^3
Total with r = 3: 7/27
Total with r = 4: 10/27

There are some minor errors.
As stated we arrive in case 1 with probability (r-2)/r and case 2 with probability 2/r

1) A1B1C1.

a) Chance is 2/r that we end up  at
                  A1A0B1     (we can match either remaining reel)
Now we need another match.  Chance is 1/r.

b) Chance is (r-2)/r that we end up at
                  A1B1C0
Now we need to match the first two reels to the final reel on consecutive spins.  Chance is (1/r)^2 = 1/r^2

2) A1A1B1.  We spin the last reel.  Chance of an immediate win is 1/r.  Otherwise --with probability 1-1/r = (r-1)/r-- we land on
                 A1A1B0
Chance of a win from here is (we need to spin each of the first two reels and match the final reel each time) (1/r)^2= 1/r^2

Chaining all the conditional probabilities together, our total chance of success starting from A1B1C2 is

[(r-2)/r]*[(2/r)*(1/r)+{(r-2)/r}*(1/r^2)]+(2/r)*{(1/r)+[(r-1)/r]*(1/r^2)}

Factoring out 1/r^4 from the above expression and expanding gives

1/r^4*[2*(r-2)*r+(r-2)^2+2r^2+2*(r-1)]
=(2r^2-4r+r^2-4r+4+2r^2+2r-2)/r^4

Collecting like terms yields

(5r^2-6r+2)/r^4,

Spin A/B:
Total: 1/r^2 + 1/r^2 + r-1 / r^3 + r-2 / r^3 + (r-2)(r-1) / r^4 = r^2 + r^2 + r^2 - r + r^2 - 2r + r^2 - 2r - r + 2 / r^4 = 5r^2 - 6r +2 / r^4

which agrees with this result.

I await a proof or counterexample to the following conjecture:

Conjecture: Given any number of rares and any number of reels, if the reels are all different it makes no difference to our chance of success which reel we spin first.

[The_Mormegil]

All true. About the conjecture... It may be true. There are numerous such cases in probability, after all, so... But I would like to have proof of that.

[Rubik87]

Please, Zanz, tell us how the special spin works (r=4? r=5? can cards be repeated?) so we can try to proove what the best strategy is, for fun.

[OldTrees]

@Jenkar: still working.

I was doing the calculations from below (instead of from above) assuming there are exactly 3 different rares in the spins, i am discouraged to continue now that i read there are 4 (maybe 1 is often repeated).

Anyway, doing all the cases up to 4 spins left I found the first interesting thing: in the case
ABC
112
it is better to spin A or B (29/81 chance of winning something) than C (28/81 chance).

When R=3, I get 29/81 for either choice. This held true for R>3.
Please double check you didn't miss any options.

Spin C:
1) A1B1C1 = r-2 / r
--- this leads to a win on 1/r^2 cases.
2) A1A1B1 = 2 / r
--- this leads to a win on 1/r cases.
A1A1B0 = r-1 / r can still win on 1/r^2
Total: r-2 / r^3 + 2 / r^2 = 3r - 2 / r^3
Total with r = 3: 7/27
Total with r = 4: 10/27

-snip-
29/81
58/256

[The_Mormegil]

@Jenkar: still working.

I was doing the calculations from below (instead of from above) assuming there are exactly 3 different rares in the spins, i am discouraged to continue now that i read there are 4 (maybe 1 is often repeated).

Anyway, doing all the cases up to 4 spins left I found the first interesting thing: in the case
ABC
112
it is better to spin A or B (29/81 chance of winning something) than C (28/81 chance).

When R=3, I get 29/81 for either choice. This held true for R>3.
Please double check you didn't miss any options.

Spin C:
1) A1B1C1 = r-2 / r
--- this leads to a win on 1/r^2 cases.
2) A1A1B1 = 2 / r
--- this leads to a win on 1/r cases.
A1A1B0 = r-1 / r can still win on 1/r^2
Total: r-2 / r^3 + 2 / r^2 = 3r - 2 / r^3
Total with r = 3: 7/27
Total with r = 4: 10/27

-snip-
29/81
58/256

Yes, Chrome already got this mistake above.

[OldTrees]

Update:

The correct strategy

For no pairs spin the slot with the most spins remainingFor 1 pair spin the unpaired slotWhen a slot is locked spin the other slots to match it

This strategy has been tested for any number of rares from 3 to 40

If anyone can provide means of converting a table of X and Y values into a polynomial that would help.

Currently I have only seen evidence of 4 rares in the spins so the probability
Starting with
AAA: 1
AAB: 0.719127655
ABC: 0.617195129

Since the chance of starting with
AAA is (1/r²) or 1/16 if r=4
AAB is (3)(r-1)(1/r²) or 9/16 if r=4
ABC is (r-1)(r-2)(1/r²) or 6/16 if r=4

For a total chance (assuming R is 4) of
0.698457479
(if R is dependant on the number of rares in the game and not a constant then it will get rare fast.)
Total chance if R is 20
0.072648547