How Do You Calculate Average Cost After Multiple ETH Buys?

Short answer: Average cost = total invested ÷ coins held, where coins held = Σ(amount ÷ buy price). It is amount-weighted, not a simple average of the buy prices.

Steps

Break each buy into an amount and a price: coins from that buy = amount ÷ price.

Add all the coin amounts to get coins held, then divide total invested by coins held to get the average cost.

The same method applies to buys of different sizes: the larger buy carries more weight in the average.

Formulas

Coins from buy iamount i ÷ price i
Coins heldsum of coins across all buys
Average costtotal invested ÷ coins held
Unrealised P&Lcoins held × current price − total invested

Worked example (recompute it yourself)

Three buys: 1,200 USDT at 2,600, 800 USDT at 2,400, 1,500 USDT at 2,200:

Coins per buy = 0.46153846 / 0.33333333 / 0.68181818, so coins held = 1.47668998 ETH and total invested = 3,500 USDT.

Average cost = 3,500 ÷ 1.47668998 ≈ 2,370.17 USDT, below the simple average of the three buy prices, 2,400 USDT.

At a current price of 2,471 USDT, position value = 1.47668998 × 2,471 ≈ 3,648.90 USDT, an unrealised gain of about +148.90 USDT (+4.25%).

Notes

The calculation above excludes fees; including buy fees raises the effective average cost slightly.

Average cost only says where your cost sits. It is not advice to keep buying: averaging down lowers the cost but increases position size.

FAQ

Why is it below the simple average of the three prices?

Because the amounts differ: the 2,200 buy is the largest and acquired the most coins, pulling the weighted cost down. The weighted average here is 2,370.17 versus a simple average of 2,400 USDT.

Is the method the same for equal amounts?

Yes. Both use "amount ÷ price" per buy, summed into coins held, then divided into total invested. The only difference is that uneven amounts show a bigger weighting effect.