If we approximate the problem by assuming each card has 1/13 chance for each value 1 to 13, we can compute the expected return based on the number of cards remaining, like in the dice problem towr mentioned.
With the last card, the expected return is (1+13)/2 = 7.
With 2 cards remaining, we keep the card if the value is higher than that. The return is (7*7 + 6*(7+13)/2)/13 = 8.6154. (7 chances to continue and get 7, 6 chances to stop and earn (7+13)/2 on average).
With 3 cards remaining, you keep anything above 8.6. The return is (8*8.6154 + 5*(9+13)/2)/13 = 9.5325.
With 4 cards remaining, you keep 10 or more, the return is (9*9.5325 + 4*(10+13)/2)/13 = 10.13791534. That is what you can earn on average.
If you take into account that the cards don't repeat it changes these values. But I don't think it changes them to the point where the strategy should be changed. The average return should be slightly higher due to the fact that when you discard cards, these are likely to be low values, so the return after discarding a card is often higher than computed.
Just for the last card, if you are holding a 7, you can check whether the first 2 cards average to more than 7. If yes, you should better keep your seven. If not, you should draw the last card.
But If I do the same calculations with the rule that you always keep a value 10 or above, I find a result of 10.0068. I think the real value is slightly higher.
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