A semi popular system to beat the roulette table is called the cancellation system.
probability/cancellation New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
A semi popular system to beat the roulette table is called the cancellation system. Basically, to play this system, you write the numbers 1 through 10 on a piece of paper. Make an even money (18/38 expectation of success) bet in an amount equal to the first and last number on your list. Your first bet is 10 + 1, or 11 units. Your unit is $1 and the house limit (highest bet) is $500. If successful (win), cross out the first and last number and bet again using your new first and last number as your bet size. If failure (loss), append the amount lost as your new last (highest) number. Continue playing until you have crossed out all the numbers and won $55. The theory is that since you cross out two numbers for a win and append only one number for a loss, you will eventually cross out all the numbers and win $55. The theory says that this happens even though you expect to win only 18 times out of every 38. 1) I know that roulette is a negative expectation game and that I can't combine negative expectation bets to make a positive expectation. 2) My friend says he uses this system to win $110 every day (he lives in Vegas). 3) I have been absolutely stumped by the math - I "know" that there will eventually be a big loss (caused by being unable to make the next bet because of the house limit) that wipes out all the wins and more but I can't seem to figure out how to calculate it. The puzzle I present is as follows: A) How many times do you succeed, on average, for each failure? B) How much will the average failure cost? C) Bonus: do the same calculations for a European Roulette wheel where your expectation of a win is 18 times in 37 trials. [ Editor: posted by Tom Turcich. ]
Show solution
I ran a simulation of 100000 samples. I stopped each sample when the next bet would exceed $500, or when the player had won $55. I found that the player won 93147 times, and lost 6853 times. You will win about 13.6 times for every time you lose. After 100000 trials, the player was $6617944 in the hole, for a session expectation of $-6.62 [ Editor: $6617944 / 100000 sessions = $66.18 per session, not $6.62. The other figures agree with the larger value: 93147 wins x $55 = $5123085, and 6853 losses x $1713.27 = $11741000, a net loss of about $6618000. ] The average of only the losing sessions was $-1713.27 The worst losing session was $-6236.00 For a single-0 roulette wheel (ignoring en-prison rules which would help even further), the house edge on the game is 1/2 of the double-0 wheels, and winning sessions using this system tend to be a little shorter so the total action played by the player is smaller, which also eats into the house edge. On a 100000 trial run, I got 95140 wins, and 4860 losses. The player will win approximately 19.6 times for each loss. Total outcome for the 100000 trial run was $-2743167, for a session expectation of $-2.74 [ Editor: likewise $2743167 / 100000 = $27.43 per session; check: 95140 x $55 - 4860 x $1641.12 = $5232700 - $7975843 = -$2743143. ] The average losing session was $-1641.12 The worst losing session was $-5835.00 As another point of interest, I also tried increasing the maximum allowed bet to $1000. In this case, the player won more frequently, but the losing sessions were even larger, and the net result was an even worse total outcome, which is to be expected since the house makes its edge on the action, and the total action is increased. If your friend is still using this system, his losses will soon catch up to him. He obviously has not been using this system for long. -- Stephen H. Landrum <slandrum@3do.com>