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Keefer Roulette System
The Keefer Roulette System is a betting system promoted in a book by the same name. It basically a modified nine-step Martingale. Much like the Martingale, you can expect it to usually win small, but lose it all back and more when you reach the maximum number of losses. Like all betting systems, not only doesn't it overcome the house edge, it doesn't even dent it.
How it Works
This system involves betting opposites. For this reason, it is meant to be played with a two-person team. However, one person could do bet both sides, although he would look like a fool betting opposites.
- The system is meant to be played on even money bets where opposite betting is allowed. Roulette, with opposite bets like red and black, is how I shall explain and analyze it.
- The player should decide what his unit size, or minimum bet, is. The player should have a bankroll equal or greater to 384 units. This bankroll will be shared between what I will call player 1 and player 2.
- To start, player 1 bets one unit on any even money bet (red, black, odd, even, 1 to 18, 19 to 36).
- If the bet in step 3 wins, repeat it until it loses. If it loses, then player 1 will bet two units on the same thing and player 2 will bet one unit on the opposite.
- Going further, with every spin, if a bet loses, then that player who made the bet should increase it by one step in the following progression: 1, 2, 3, 6, 12, 24, 48, 96, 192. If a bet wins, then the player who made it should drop that bet to one unit.
- If a 192-unit bet loses, then the session is over.
The hope behind this system is that the player will win enough between 192-unit bet losses to more than cover them. When the player loses nine hands in a row, counting the previous consecutive losses, that losing player will have lost 1+2+3+6+12+24+48+96+192 = 384 units. The book incorrectly argues that between 384 unit losses the player will win on average 1,733.4 units. I will explain later the flaw in the author's math.
Note that the two players will never bet the same amount at the same time.
It is assumed the two players share a common bankroll and may give each other money if one can't cover a bet.
As I will show later, it is ridiculous for the two players to make opposite bets. One can accomplish the same thing at less losses or more wins by just deducting the smaller bet from the larger one. This will result losing less, specifically double the smaller bet, every time the outcome of the bet is a zero.
Following are my results of a random simulation of over 39.7 billion sessions played. A session was defined as playing until nine consecutive losses were attained for the player on either side. The simulation was performed in Visual C++ with a Mersenne Twister random number generator.
|Field||Single Zero||Double Zero|
|Win from streak||-384.00||-384.00|
|Win outside streak||296.55||233.29|
|Expected win per session||-87.45||-150.71|
|Expected loss per session without offsetting bets||-64.53||-113.02|
|Probability net profit||27.14%||18.87%|
|Probability net push||0.09%||0.08%|
|Probability net loss||72.77%||81.05%|
|Average count of bets per session||413.48||340.19|
|Units bet per session||3236.11||2863.66|
|Average units bet per spin||7.83||8.42|
Following is an elaboration on each field.
- Win from streak — This is how much the player can expect to win from each streak of nine losing bets. It is the sum of how much the player wins (a negative number for a loss) at each step in the nine-step progression. -(1+2+3+6+12+24+48+96+192) = -384. This does not count the nine unit win betting the opposite way.
- Win outside streak — This is how much the player can win per session not counting the 384 loss from the nine consecutive losses. For example, in the single zero game, the player can expect to win 296.55 units outside of the 384 unit loss from losing nine consecutive times.
- Expected win per session — This the sum of the two above statistics, representing the expected win (negative for a loss) per session. In the case of single-zero roulette, for example, the player can expect to lose 87.45 units per session.
- Offsetting loss — This is how much the player can expect to lose due to offsetting bets when the ball lands in zero. For example, if the player bet 1 unit on red and 6 on black, and the ball landed in zero, the player would lose 7 units. However, the player could cut those loses by eliminating one units from each side, to five units. In the case of single-zero roulette, the player will lose 22.91 units per session due to such offsetting bets when the ball lands in a zero.
- Expected loss per session without offsetting bets — This is how much the player could expect to win per session (negative for a loss) if he didn't make offsetting bets.
- Probability net profit — Probability a session shows a profit. In the case of single-zero roulette, this is 27.14%.
- Probability push — Probability a session push. In the case of single-zero roulette, this is 0.09%.
- Probability loss — Probability a session shows a loss. In the case of single-zero roulette, this is 72.77%.
- Median win — This means that at least half of sessions will finish equal or less, as well half equal or more, than this figure. In the case of single-zero roulette, this figure is a loss of 176 units. This should not be confused with the mean/expected/average win per session above.
- Average count of bets per session — Average number of bets made per session. In the case of single-zero roulette this is 413.48.
- Units bet per session — Average total amount bet per session. In the case of single-zero roulette this is 3236.11 units.
- Average units bet per spin — Self explanatory. For example, in single-zero roulette, the player will average betting 7.83 units per spin.
- Overall return — Ratio of total amount won (negative for a loss) to the total amount bet. This, not surprisingly, equals the theoretical house edge for the game. In the case of single-zero roulette, this is 1/37 = 2.70%.
Why Keefer's Math is Wrong
On page 11 of Keefer's book begins the section "The Mathematics Behind the System." Here the author tries to explain why for every time the $5-unit player loses $1,875 dollars due to a string of nine consecutive losses, the player will otherwise win $8,667, for a net profit of $6,792.
He begins by correctly saying the probability of nine consecutive losses in double-zero roulette on an even-money bet is (20/38)9 = 0.3099%, or 1 in 322.69. Note that this is for betting one side only, as opposed to both sides in the case of the Keefer system, but let's ignore that and assume we are betting just one side.
The next statement is "This means that out of 322 nine-spin series, we should lose nine consecutive spins on a random wheel 1 time. 322 × 9 = 2,898 spins."
This is not true. It seems to imply that if a particular bet does not set off a series of nine consecutive losses then the player gets eight more safe spins. He doesn't. The expected number of losing bets until a winning one is an infinite series of (20/38)0 + (20/38)1 + (20/38)2 + (20/38)3 + (20/38)4 + (20/38)5 + ... = 1/(1-(20/38)) = 38/18 = 2.111.
The average number of spins per session can be expressed as 1/(20/38)9 × (38/18) = 322.69 × (38/18) = 679.12. This can also be verified, as I have done, by a Markov Chain or a simulation.
With that error of 2,898 spins per session in the one-sided situation, the rest of his math falls apart.
Let's consider a simplified version of this system that bets on one side only, for example black. After every winning bet, the player will reset with a one unit bet next time. There are nine possible outcomes for the next sequence of bets until the next reset, whether it be because of a win or nine consecutive losses. The probability of losing exactly x times before a win, in double-zero roulette, is (20/38)x × (18/38). The probability of nine consecutive losses is (20/38)9.
Next, the player will net a one-unit win if the outcome after a reset is a win on the first or second spin. If the player goes 3 to 8 spins before a win, then the net outcome will be a push. The outcome for losing nine consecutive spins is a loss of 384 units.
Knowing the probability of every possible outcome and how much it wins, we can put together the following return table.
Return Table — Double-Zero Roulette
|Losses||Probability||Step||Total Bet||Win||Expected Bet||Expected Win|
|9 or more||0.003099||192||384||-384||1.190005||-1.190005|
The right two cells in the bottom row show that for every time the player starts over with a bet of one, he can expect to bet 8.873254 units and win -0.467013 units. Dividing the expected win to the expected amount bet is -0.467013/8.873254 = -5.26%. Not surprisingly, this equals the theoretical win on every bet of 2/38 = -5.26%.
That the player does this twice at the same time means only that he will lose on average twice as much money. The odds on each side do not change.
I hope I have shown that not only doesn't the Keefer Roulette System have a player advantage, the disadvantage is exactly equal to the house edge in roulette. This is true of EVERY betting system. Thus, if you must use one, please use one of the free ones I list under Internal Links below.
Have you been burned by another betting system you paid for? Did it add insult to injury when the seller didn't refund your money? How about warning others through me to not buy it. I prefer it be able to be played on roulette, baccarat, or craps. It must also be clear how to use it. If it meets those requirements, I'll be happy to put it to a fair analysis and publish my results.
- The truth about betting systems
- Oscar's Grind betting system.
- Labouchere betting system.
- Fibonacci betting system.
- Martingale betting system.
- Anti-Martingale betting system.
- D'Alembert betting system.
Discussion about the Keefer Roulette System in my forum at Wizard of Vegas.
Written by: Michael Shackleford