American roulette and European roulette accept the same Fibonacci progression, but the cost of a long series differs between them. The bet follows the row 1, 1, 2, 3, 5, 8. A loss shifts one step forward. A win shifts two steps back, not to the start. Growth is slower than doubling, so the table ceiling arrives later, but one win also does not return the whole ladder.
The row is built so that each number equals the sum of the two before it. The payout of an even chance equals the bet. So a win on the current number closes the two previous steps and leaves the earlier losses in the minus. This is the main difference from martingale, where one successful spin zeros the cycle to plus one chip. The cashier can accept a coin so you can play with crypto, but the rate does not enter the numbers of the row.
European roulette removes one losing pocket: 19 instead of 20. The row therefore reaches a bet of 89 less often, and the numbers 1, 1, 2, 3, 5 themselves do not depend on the makeup of the wheel.
The ladder and the sum of a loss in American roulette
Below is the row for a start of 1 unit. “Sum” is how much has already been staked if every step up to and including this row has lost. The twelfth step is 144. The thirteenth is 233. At a teaching maximum of 500 the row still fits, whereas doubling hits the ban on a bet of 512 already on the tenth step.
| Step | Bet | Sum of all losses | What a win of this step closes |
|---|---|---|---|
| 1 | 1 | 1 | the step itself, the cycle at zero or in the plus by 1 |
| 2 | 1 | 2 | both of the first steps |
| 3 | 2 | 4 | steps 2 and 3, the first step stays in the minus |
| 4 | 3 | 7 | steps 3 and 4 |
| 6 | 8 | 20 | steps 5 and 6, that is 5 and 3 |
| 8 | 21 | 54 | steps 7 and 8, that is 13 and 8 |
| 10 | 55 | 143 | steps 9 and 10, that is 34 and 21 |
| 12 | 144 | 376 | steps 11 and 12, that is 89 and 55 |
An example. The series reached step 6, a bet of 8, and 12 has already been lost before it (1+1+2+3+5). A win of 8 takes the last 5 and 3 and leaves the first 4 units in the minus. The pointer returns to step 4, and the bet is 3 again. The cycle is not finished. The next loss moves it forward again.
How much the same depth eats in American roulette
On the eighth step of Fibonacci the bet equals 21, and the sum of the losses is 54. Doubling at the same step number stakes 128 and already has 255 on the table, if one counts the previous seven lost bets plus the current readiness. From this comes the practical sense of the row: the bankroll for an average series is smaller, but climbing out of the hole requires several wins, not one.
| Depth | Fibonacci bet | Doubling bet | Lost by Fibonacci up to this step | Lost by doubling before this step |
|---|---|---|---|---|
| 6 | 8 | 32 | 20 | 31 |
| 8 | 21 | 128 | 54 | 127 |
| 10 | 55 | 512 | 143 | 511 |
The tenth step of doubling already cannot be staked on many tables. The tenth step of this row is 55, and it is usually inside the limit. Length is what one pays for that: to erase twelve losses, six wins are needed in the right places, because each one removes only two steps. Six wins in a row from step 12 take the pointer to the start. Six wins diluted by losses can leave the series alive for tens of spins.
The probability of reaching a large step
A loss of the color on a wheel of 38 pockets has probability 20/38. Ten losses in a row, after which the bet becomes 89, are (20/38)¹⁰, about 0.16%, roughly 1 case out of 613. On a wheel of 37 pockets there are 19 losing pockets, and the same depth is about 0.13%, roughly 1 out of 785. French roulette with half the stake returned on zero does not use these shares: the rule acts on the size of the loss, not on whether the spin was counted as a step forward. If zero took only half, the progression can no longer be applied by a whole step without a separate rule, which the RoulCoin table does not have.
| Losses in a row | Next bet of the row | 38 pockets | 37 pockets |
|---|---|---|---|
| 8 | 34 | 0.59%, about 1 out of 170 | 0.48%, about 1 out of 207 |
| 10 | 89 | 0.16%, about 1 out of 613 | 0.13%, about 1 out of 785 |
| 12 | 233 | 0.045%, about 1 out of 2 200 | 0.033%, about 1 out of 3 000 |
The figures of the last row are rounded to a denominator that is easy to read. The exact American estimate for twelve losses is (20/38)¹², that is about 1 out of 2 216. For a decision about the bankroll the order of magnitude is enough: a series goes deeper than the tenth step rarely, but the sum of 143 by the tenth step is already not theoretical.
One win here erases two steps, not the whole minus. While the pointer has not returned to one unit, the session is not in the plus by the size of the row.
How to keep the pointer
A player in England sees the same 38 pockets on the RoulCoin table as a player anywhere else: 18 cells of the chosen color against 20 unfavorable ones. The pointer is convenient to keep as the number of the step, not as memory. A loss is plus one to the number. A win is minus two, but not below step 1. The bet is taken from the row, not “about that much.”
First the row is walked on teaching chips: the guest balance for this is virtual and does not go to withdrawal. Playing online without signup does not replace writing ten steps on paper. Playing online for money is worth doing after checking step 10 against the maximum in the account.
Three rules, without which the row comes apart:
- After a win, step back exactly two steps, even if one wants to “lock in” a large bet.
- Do not raise the unit of the row in the middle of a series. Changing 1 to 5 multiplies the whole later minus by five.
- Stop the pointer on a step chosen in advance, for example on the 10th, even if the limit still allows 89.
Where the row lies over a short distance
Five cycles, each of which closed in two or three steps, look like a steady plus of several units. The sixth, which reached a bet of 55, takes 143 if it is broken on the loss of this step. A short session almost does not show these 143. So a trial run makes sense not up to the first plus, but up to a depth appointed in advance.
After a win, can one step back only one step?
One can, but that is already a different method. One step back does not close the two previous numbers of the row, because the current bet equals their sum, not one of them. The minus starts to pile up even on the alternation “loss, win.”
Does the row suit a bet on a Dozen?
The payout of a Dozen is 2 to 1, and there are 12 winning pockets out of 38. The row is calculated for a payout of 1 to 1. A win on a Dozen is larger than the bet, so it closes not two steps but a different sum. These rules cannot be mixed in one series: the pointer stops matching the money.
Where does the pointer stand after two wins from step 8?
Step 8 is a bet of 21. The first win returns to step 6, the second to step 4. Four numbers of the row are taken off, and the earlier minus remains. These two spins do not bring the series to one unit.
Can 20 be staked instead of 21?
No. 21 equals the sum of 13 and 8, that is the two previous steps. A win of 20 does not close both of them, and stepping back two steps stops agreeing with the money on the table.
On which step does the row hit a maximum of 500?
The fourteenth step is 377, and it is still less than 500. The fifteenth is 610, and the teaching table will no longer accept it. The break here arrives on a bet of 610, not on a “round” depth of ten steps.
The Fibonacci row is useful to keep next to other schemes: the shared playing strategies cancel neither the 20 losing pockets out of 38 nor the bet ceiling. Before live units it is logical to play online for free and take the pointer to the tenth step at least once on teaching chips.