Yes, you should not play a game with negative expected return.
The option to stop playing at any point in the interim is a real option, but it is not the baseline case here.
If you have to play 1000 rounds, and your expected return is 0, then assuming no time value for money, and no discount for risk aversion, you should be willing to pay $0 to play this game, since your expected winnings are 0.
If there is risk aversion, then someone will need to pay you to play this game, or - if flipping equal heads or tails is defined to win $1, then you’d be willing to pay $0.90 or something to play this game (where the discount reflects risk aversion). If 1000 rounds of the game takes a material length of time, you’re likely to discount further to reflect time value of the money spent.
But let’s go back to the game where you are not risk averse and you play 1000 rounds and have the chance to stop at any point in the interim. Now in this game, the end value at 1000 rounds is still zero, but there is also a decent chance that you will have a positive balance in the interim, which you can lock in by stopping, and which you can try to neutralize by continuing.
What this means is that someone who forces you to play 1000 rounds would have to give you better odds than someone who doesn’t, which is perhaps why you see this as the baseline case.
I guess the way to value this option would be to run a binary tree, which allows you to exercise if the current value is worth more than the risk-neutral probabilities of flipping the coin again. If the risk-neutral probabilities equal the ordinary probabilities (which they might, if the RFR=0), then it’s very possible that they’d sum out to the same. I’d bet a tree that only has 10 flips or so would be able to do it.
I am not 100% sure, but it seems logical that a real option like that is worth paying for.