(Taken from a mobile game)

You have a wheel of fortune with a total of 30 fields. You start with the following fields (name, size, minimal size, effect):

  • "Diamond Field" (D), size 2, min. size 1, get 10 diamonds
  • "Coin Field" (C), size 2, min. size 1, get 1.000 coins
  • "Unlucky Field" (U), size 26, min. size 5, get nothing

After every turn, you can select a field which will then increase its size by 1 and decrease another field by 1. If you select a diamond or coin field, the unlucky field will decrease.

If one of the fields reached its minimal size, the 3rd one will decrease. If both others are at their minimum, nothing happens. Example:

State  D    C   U
(n)   2D, 23C, 5U  -> Select C -> D will decrease, because U is already at the minimum
(n+1) 1D, 24C, 5U  -> Select C -> D and U are minimal -> nothing happens
(n+2) 1D, 24C, 5U

The costs for turning the wheel are the following:

Turn      1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 ...
Cost [D]  0  1  1  1  2  2  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 10 10 ...

So the turns 21-n will all cost 10D.

You get 10D per day to spend and want to get as many coins as possible. The diamonds that you don't spend can be kept for future days. At the end of a day, the wheel and the cost will reset.

What is the best tactic to convert your diamonds into coins in the long run?

  • $\begingroup$ I assume there's no limit on turns per day? And probability of wheel giving an item is given by "size"/30? $\endgroup$ Apr 29, 2020 at 14:59
  • $\begingroup$ Also you start with 0C 0D ? $\endgroup$ Apr 29, 2020 at 15:00
  • $\begingroup$ Also can you clarify exactly what and how many choices are that for changing field size. The way I see I see it you have 6 options, 3 choices for which field to increase and for each of these, 2 choices for which to decrease. Which of these 6 choices are legal? $\endgroup$ Apr 29, 2020 at 15:06
  • $\begingroup$ @ghosts_in_the_code The limit are the diamonds that you have available, so 10D per day. You always start with 2D, 2C and 26U. If you increase D or C, U will decrease. Increasing U is pretty much useless. $\endgroup$
    – izlin
    Apr 30, 2020 at 6:55
  • $\begingroup$ Increasing U isn't necessarily useless because it's the only way to transfer area from C to D or vice versa. So is it allowed? Also you start with zero coins and zero diamonds on day 1 right? (Apart from the 10 diamonds you get on day 1) $\endgroup$ Apr 30, 2020 at 8:33

1 Answer 1


You play each day 10 turns.
You pay 21 dia total, which is 2.1 pay per turn.

On wheel you can win 2+2 + 9 changes for diamond/coins. So total 4+5+6+…+13 for dia+coins. That is 85 fields for you out of 300 total. 2.833 average win, which needs 2.1 to be dia wins. Rest is coins - or 733 coins per day. You basicly almost always choose dia fields.
(You can achieve that for playing for example with 1000 diamonds on start and when you have 1000+ diamonds, you choose coin on wheel, and when you have below 1000 diamonds, you choose dia on wheel, until you have over 1000.)

  • $\begingroup$ "At the end of a day, the wheel and the cost will reset." After each day the size of the fields will reset, so you will always start with 2D, 2C and 26U. So the prize you would have to pay each day to get to 12D and 14C would be 111D and this is much higher than the daily 10D. $\endgroup$
    – izlin
    Apr 29, 2020 at 12:19
  • $\begingroup$ @izlin missed "After every turn, you can select a field which will then increase its size by 1" $\endgroup$
    – Jan Ivan
    Apr 29, 2020 at 13:05

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