A variant coin toss game with weighted value

Question

This is a coin toss game with weighted values for the outcome. If you get Head, you get 100 dollars; if you get Tail, you get 80 dollars. You have 50/50 chance getting H/T.

• The player needs an entrance fee of 10 Gems to play the coin toss game.
• Each coin costs 50 Gems.
• In each round, the player can purchase 1 coin. Coin(s) that are not purchased remains on the table and can be purchased in the next round.
• In rounds 1, 2 and 3, the player gets to toss 2 coins.
• Round 4 and round 5 are similar but the player can only toss 1 coin.
• If the player refuses to take any coin in a round, the game ends, and the player gets whatever they purchased previously. (This means in each game, the player can toss the coin up to 8 times)
• After going through all 5 rounds, the player can decide to purchase whatever remains on the table then leave the game.

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Here is an example for one gameplay with 500 Gems:
Enter game:
Gems remain: $500-10=490$

Round 1:
Toss 2 coins -> HT.
Purchased coin(s): H
Coin(s) left on the table: T
Gems remain: $490-50=440$

Round 2:
Toss 2 coins -> TT
Purchased coin(s): HT
Coin(s) left on the table: TT
Gems remain: $440-50=390$

Round 3:
Toss 2 coins -> HH
Purchased coin(s): HTH
Coin(s) left on the table: TTH
Gems remain: $390-50=340$

Round 4:
Toss 1 coins -> T
Decided not to purchase any coin.
Leave game.
Purchased coin(s): HTH
Coin(s) left on the table: TTHT
Final earning: $2*100+1*80=\$280$
Gems remain: $340$

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For a limited amount of Gems，what is the best strategy for maximizing the earning? (eg: How many rounds should the player play in each game? And in what condition should the player end/stay in the game?)

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The tricky part of the question is due to the entrance fee. Otherwise, it would be best to keep restarting the game and purchase only for the coins with Head.
My initial thought would be to calculate the expected value of different total toss in one game.
For example:
If the player toss only once in the game, then
$EV = (0.5 * 100 + 0.5 * 80)/(50+10) $
If the player toss twice in the game, then
$EV = [0.25 * (80 + 80) + 0.5 * (100 + 80) + 0.25 * (100 + 100)] / (50 * 2 + 10)$
Then we choose to play N rounds in one game such that the expected value is highest.
But I am not sure if this is the right approach.
As for the final strategy, I am thinking that we should play all 5 rounds regardless of the outcome of the toss, then purchase all the coins with a Head on the table. However, a friend of mine says it would be better to play only 4 rounds in each game. Why?

Answer 1

You seem to be making it more complicated than it is. You only want to buy heads. As the entrance fee is half than the difference in value between heads and tails and (flipping two coins) you have $\frac 34$ chance of at least one head you should leave and restart any time you can't buy heads. You should only buy one head each round until the last. That will let you get through a round where you flip tails. At the end, buy all the heads you can afford.