3
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If we play the game with 7, 5, 3 and 2:

727272...

If we play the game with 2 and 9:

292929...

If we play the game with 9, 3 and 5:

953953953...

If we play the game with 3, 9 and 5:

359359359...

If we play the game with 5 and 9:

595959...

If we play the game with 7, 5, 2 and 7:

77272727...

NOTE: (sorry for adding this so late) what matters is how you write digits in handwriting, and this needs to be included in your answer.

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  • $\begingroup$ May you please add more examples? $\endgroup$
    – PotatoLatte
    Jun 10 '18 at 13:47
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The rule is:

Start with the first (left-most) number. Then: if the last number that you were on was odd, move left once (wrapping around to the right-most if you were on the left-most number), and if the last number that you were on was even, move right once (wrapping around to the left-most digit if you were on the right-most number).
Note: see the next spoiler for a handwriting justification.

Accounting for the added note:

Based on OP's comment/hint on this post: enter image description here
When written out, 2, 3, 5, 7, and 9 end on a "free end". The number 2 ends with the pencil stroke going right, and left-leaning for 3, 5, 7, and 9. The reason why the other digits don't get represented are because the stroke ends for 1 and 4 by going down (without going left or right), and the stroke ends inside the digit for 6, 8, and 0 (which could make it a little ambiguous).

Interestingly:

There are only a few "effective" ways that this game can end. If all digits are even, you will continuously loop rightward. If all digits are odd, you will continuously loop leftward. Otherwise, you'll eventually reach a 2-cycle alternating between an odd and even number.

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  • $\begingroup$ Good. To understand what I wrote about the ways of writing number, you might want to think back about the instructions of proper ways of writing during elementary school. But, it is true that the rule I was thinking of can coincide with odd and even, because other digits are unable to be used. $\endgroup$
    – O0123
    Jun 11 '18 at 7:24
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If all of the numbers are odd, you alternate either from smallest to largest or from largest to smallest, based on the first number. If at least one of the numbers are even, you alternate the biggest and smallest number based on which one comes first.

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Wrong

If you are given an even number of numbers, you alternate between the maximum and minimum numbers, based on which one comes first in the list of numbers. If you are given an odd number of numbers, you repeat the numbers from smallest to biggest based on the maximum and minimum numbers and where they come in the list.

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  • $\begingroup$ Whether the number of numbers are even or not doesn’t matter. $\endgroup$
    – O0123
    Jun 10 '18 at 13:41
  • 2
    $\begingroup$ @VincentMiaEdieVerheyen: This is a perfectly valid answer: it's the puzzle setter's responsibility to ensure that there are no alternate solutions. $\endgroup$
    – Deusovi ♦
    Jun 10 '18 at 13:44

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