MidasIn short: There is no single true number, only conditional arithmetic: if each next level is reached with chance p, all N levels fill with chance p^(N−1), while exposure grows as 2^N − 1.
This applies to martingale grids and DCA-style bots with growing safety orders on exchanges and platforms such as Binance, Bybit, OKX, 3Commas, Veles and Pionex, and to martingale grid expert advisors on MT4 and MT5. The values of p below are inputs picked to show a shape. They are not measurements of gold, bitcoin or any other instrument.
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What is the probability that a martingale grid blows up?
There is no single true number, only conditional arithmetic: if each next level is reached with chance p, all N levels fill with chance p^(N−1). Here p is the chance that, once level k has filled, price goes on to level k + 1 before the basket closes in profit. A basket is one deal: the first order plus any added orders, closed together.
The ladder has a size rule as well. With a multiplier of 2 the orders are 1, 2, 4 … 2^(N−1) units and the total is 2^N − 1. In general it is (m^N − 1) / (m − 1), so ten orders at m = 2 are 1,023 units and at m = 1.5 about 113 units.
How the numbers move with p and N
Full-ladder chance = p^(N−1), because the first order opens the basket and N − 1 further levels must be reached. A blow-up is read here as "all N levels fill and the last one is the most the account can carry".
| Orders N | Total exposure (units) | p = 0.5 | p = 0.7 | p = 0.9 |
|---|---|---|---|---|
| 6 | 63 | 3.1% (1 in 32) | 16.8% (1 in 6) | 59.0% (1 in 1.7) |
| 8 | 255 | 0.78% (1 in 128) | 8.2% (1 in 12) | 47.8% (1 in 2.1) |
| 10 | 1,023 | 0.20% (1 in 512) | 4.0% (1 in 25) | 38.7% (1 in 2.6) |
| 12 | 4,095 | 0.05% (1 in 2,048) | 2.0% (1 in 51) | 31.4% (1 in 3.2) |
Example: p = 0.7, N = 10 gives 0.7^9 = 0.0404, about 4.0%, one basket in 25. Change p to 0.5 and it becomes 0.5^9 = 1/512. One input moves the result about twentyfold (4.04% against 0.195%), and nobody knows p in advance.
Why do rare events dominate?
Because size grows as fast as chance falls: at p = 0.5 every level adds the same expected exposure, so the deep levels weigh as much as the first. Level k has chance 0.5^(k−1) and size 2^(k−1); their product is 1 unit at every level.
Take the fair coin: stake 1, double after each loss, at most N rounds. You win +1 with chance 1 − 2^−N and lose 2^N − 1 with chance 2^−N. Expected value = (1 − 2^−N) − 2^−N × (2^N − 1) = 0. For N = 10 that is +1 with chance 1,023/1,024 and −1,023 with chance 1/1,024. Many small wins, one huge loss, nothing gained on average.
For p above 0.5 the deepest level weighs most. The expected units at level k are (2p)^(k−1); at p = 0.7 that is 1, 1.4, 1.96 … 20.7 at level 10. Only for p below 1/m do the deep levels weigh less than the first.
The second reason is payoff. Say a basket wins g on average and a blow-up costs the deposit D. The break-even blow-up chance is g / (g + D). With g = $1 and D = $1,000 that is 1/1,001, about 0.1%. At p = 0.5 and N = 10 the chance is 1/512, about 0.2%, so the average basket is (1 − 1/512) × 1 − (1/512) × 1,000 = −$0.96. The inputs are invented; the shape is not: a small win against a large loss needs a very small blow-up chance. Adding orders to get there does not come free, since a fixed deposit spread over more doubling orders shrinks every earlier order and the win with it.
How does the risk add up over many baskets?
The chance of at least one blow-up in n baskets is 1 − (1 − q)^n, where q is the chance per basket. Small per basket does not mean small over time.
| Per-basket chance q | n = 10 | n = 100 | n = 365 | n = 1,000 |
|---|---|---|---|---|
| 0.195% (p = 0.5, N = 10) | 1.9% | 17.8% | 51.0% | 85.8% |
| 4.0% (p = 0.7, N = 10) | 33.8% | 98.4% | about 100% | about 100% |
This assumes the same ladder at the same size every time, with no stop and no change in the account.
What does this arithmetic leave out?
- p is not constant and levels are not independent. Price moves cluster; after a deep fall the chance of the next level can be higher or lower than at the start.
- "Fill all levels" is not "lose the account". With a bigger deposit the account survives past the last level; with leverage it can end before it.
- Stops, take-profit rules, fees and funding change the loss and the win and are not in the formula.
- No estimate of p is given here. Any estimate from history carries sampling error, and the future may differ.
This is arithmetic of assumed inputs, not a forecast. Trading with leverage carries risk: if the deposit is too small, a stop-out is possible. Size your deposit with a safety margin. Nothing here is individual investment advice.
Check your own numbers
- Grid bot risk calculator for exchange and platform grid bots.
- DCA bot calculator (safety orders) for base and safety orders.
- Gold grid calculator for MT5 gold grid EAs.
- Midas Grid emulator replays a gold grid EA's settings on real XAUUSD ticks since January 2025: a history replay, not a probability. It does not test crypto bots.
Related: martingale grid on gold: the stop-out arithmetic, how much deposit a grid bot needs, is grid trading safe.
Related: Martingale 2.0: why a doubling grid exits after a pullback of about $2 shows how the lot rule of a grid EA shortens the way back to break-even.
FAQ
What is the probability that a martingale loses?
It depends on p and N. If each next level is reached with chance p, the full ladder of N orders fills with chance p^(N−1): about 4.0% for p = 0.7 and N = 10, 0.20% for p = 0.5. These are examples of the formula, not measurements.
Does martingale always lose in the end?
Not by arithmetic alone. A finite ladder with a fixed blow-up chance q per basket has a chance 1 − (1 − q)^n of at least one blow-up in n baskets, which grows toward 100% as n grows. In a fair game the expected value stays 0; the strategy only reshapes it into many small wins and one large loss.
How many orders can a martingale grid have?
As many as the deposit can fund: the total is the first order × (m^N − 1) / (m − 1), so with m = 2 every extra order doubles it. Ten orders are 1,023 units, eleven are 2,047.
Is a multiplier of 1.5 safer than 2?
The chance of reaching the last level does not change, since it depends on p and N. The size at risk does: ten orders at ×1.5 total about 113 units against 1,023 at ×2. With the same first order, a basket that closes on its first order wins the same; baskets that add orders win less, the deeper the more, because less position sits in the deep levels.
Can I find the real probability for gold or bitcoin?
Not as a single number. It depends on the instrument, the step, the period and the volatility, and any estimate from history is uncertain. The Midas Grid emulator replays gold grid EA settings on real ticks to show what would have happened, which is a history, not a probability.
Is there a safe martingale?
No setting makes it safe. A smaller multiplier, more orders or a wider step change the numbers in the formulas, but a basket can still run through every level.