t500 bonus spin strategy

[Chromatophore]

If i have two of the same should i spin the other one three times and hope it matches up?  Or is there some better strategy?

[SteppingStone81]

If i have two of the same should i spin the other one three times and hope it matches up?  Or is there some better strategy?

definitely best strat

[YoungSot]

If i have two of the same should i spin the other one three times and hope it matches up?  Or is there some better strategy?

If you already have all rares, then yes. If not, then spin for one you don't have.

If you spin the third one 3 times and still don't get what you want, then you can try to spin the first two to match the result of the third one.

[Chromatophore]

I'm going to assume any rare is as good as any other for the purposes of this.

I'll use the following format for the current state of the reels:

A B C
3 3 3

would mean 3 different cards, 3 spins on each reel.

A A B
1 2 2

would mean two matching cards, 1 and 2 spins left on each of those, with one non matching 2 left on that one.

I still have a bit of doubt that the most intuitive strategy is the best.  Say the first two reels match and you spin the last one twice.  So we have

A A B      (i)
3 3 1

Now we can spin the last one again.  That gives us 1/3 chance of a rare, 2/3 chance of being in a

A A B      (ii)
3 3 0

From state (ii) our only chance is to spin the first two reels and hope for a match.  The probability of failing on either reel is (2/3)^3, so to succeed is [1-(2/3)^3]^2 which comes out to just under a half.

This strategy gives us a chance of .663 or just under 2/3.
Our other option (from state i) we could spin one of the matching reels.  This would give a 1/3 chance of

B A B     (iii)
2 3 1

1/3 chance of

C A B     (iv)
2 3 1

1/3 chance of

A A B    (v)
2 3 1

State (iii) has a high probability of a rare if we spin the second reel 3 times (chance of no rare is (2/3)^ 3, so chance of rare is 19/27 = .703).
In state (v) we can re-adopt our original strategy (with one less spin on the first reel).  This gives us a chance of 1/3 + 2/3 * (1- (2/3)^2)*(1-2/3)^3) = .594
Unfortunately state (iv) looks worse than state (v) and if this pans out, the best we could be looking at is (.703 + .594*2)/3 = .630 which isn't as good as our first strategy.

So the intuitive strategy looks best, but i still have a bit of doubt because there could be an improvement to the strategy at (iii) or (v) or maybe (iv) could work out somehow to be better than (v).

[Hodari]

Look at it this way:  Each reel is spun independently of the others.  In order to get all three reels to match, first you have to somehow get 2 of them to match and then spin the third one to match them.  It should be obvious that if you already have 2 matches, you're better off keeping it that way. 
If you have 2 that match and spin one of those 2, there are 3 possibilities:
1.  You get the same card again in which case you're no better off than before and now have one less spin to work with
2.  You get a card which matches the third(originally unmatched) card from your original spin in which case you now have 2 matched cards and one unmatched, just a different card than before..again, you're no better off and now have one less spin to work with.
3. You get a card which doesn't match either of the other 2 in which case you're even worse off than before..and again have one less spin to work with
In other words, you end up worse off no matter what you do.  The ONLY time you should do this is if you're trying for a specific rare(and probably even the one you want was originally the third unmatched card)

If all 3 are different cards initially, then it doesn't matter which one you spin, so if one of them is a rare you want, go for that, otherwise, just pick one at random or go by how much they sell for or whatever other method you want
And of course, if all 3 are the same initially...you win

[Chromatophore]

2.  You get a card which matches the third(originally unmatched) card from your original spin in which case you now have 2 matched cards and one unmatched, just a different card than before..again, you're no better off and now have one less spin to work with.

This is where you are wrong and the reason that the proper strategy is not completely clear.  In this situation you are better off than before because you have two more spins to work with, not one less.  You are now in

B A B
2 3 1

instead of

A A B
3 3 1

with 3 spins on the unmatched reel instead of just one.

Also

3. You get a card which doesn't match either of the other 2 in which case you're even worse off than before..and again have one less spin to work with

Its not clear that this is worse off than before.  It seems worse but only calculation could verify.

[Hodari]

2.  You get a card which matches the third(originally unmatched) card from your original spin in which case you now have 2 matched cards and one unmatched, just a different card than before..again, you're no better off and now have one less spin to work with.

This is where you are wrong and the reason that the proper strategy is not completely clear.  In this situation you are better off than before because you have two more spins to work with, not one less.  You are now in

B A B
2 3 1

instead of

A A B
3 3 1

with 3 spins on the unmatched reel instead of just one.

You still have less TOTAL spins in that case, though yes you do have more on the unmatched reel.  You're still much better off just spinning the unmatched one initially though rather than  spinning one of the others to try and reach that position.


Assume there are N total rares and chance of spinning them is all equal.
Spinning the unmatched reel gives you a 1/N chance of matching on that spin and winning right away.  You still have 2 more chances to spin again and even if those don't work, you have 3 spins each on the other 2.
Spinning one of the matched reels gives you a 1/N chance of it matching the other card which would probably be the only scenario in which you could possibly be better off. 

The odds of either of these 2 scenarios are the same..in the first you already won, in the second you merely achieved what is possibly a slightly better position than you would have had if you simply spun the unmatched reel and lost on that spin.  In ANY other scenario on the second spin you're worse off.  Should be obvious without needing to fully calculate every possible outcome(or even having to have analyzed it this far) that you're best bet is to just spin the unmatched one

[Chromatophore]

Just because something is intuitive does not make it obvious or correct.

[Sevs]

Wrong section there is now an "arena" section

[Chromatophore]

What i don't get is if you have 3 different reels, one has 1 spin left, one has 2 spins, one has 3.

It seems to me that it would be better to spin the one with the most spins left, so that you gain more information about which of the other reels you have to spin, but i guess it doesn't matter.

[tyranim]

what i do:
in the situation that all are different, or the ABC situation, i would just keep the one that i want and spin the others until i get what i want unless i fail hard.
in the situation that all but 1 are the same, the AAB situation, i would spin B until it becomes A.
BUT in the AAB situation, if i have 6 copies of A already, i will spin the first A until its something that i dont already have a bunch of... or until its B. then match it with the other 2

[Chromatophore]

I finished the calculation for

A A B
3 3 1

Spinning the one reel yields a rare 1451/2187 with best strategy

Spinning one of the others gives a rare 1387/2187 so about 3% worse.

This is under the assumption that there are 3 rares on each reel and each has an equal chance of occuring on any spin.  (Quite possibly not correct).

My method was pretty brute force.  If anyone has an elegant solution i'd be interested.