# Strategy for accumulation roulette tournament ?

Discussion in 'Roulette Tournament Strategy' started by PlayHunter, Dec 26, 2013.

1. ### PlayHunterActive Member

Those kind of tournaments with re-buys where each player play against the whole field and at the end the top players get 60/80/100 x their starting BR. Particularly the tournament I am talking about was 1000 BR for start with unlimited rebuys. Max allowed bet per number was 25 but for outside bets was 100000. Max number of hands was 50.

There were some big stacks at the end, the winner finished with 96000, second had 80000, and the rest 60-80K. I have tried my luck for 9 re-buys for a total of 10 attempts. I could not score more than 9K (once).

I believe the tournament was -EV, because the prizes were fixed amounts regardless of how many re-buys the players have put in it. - I would just like to know if my strategy (as follow) was correct:

Pick one column and bet the whole 1000, if won pick (the same, or) another column for a 3000 bet. I was hopping to eventually win 4 all-in bets in a row playing columns. - Could I have done something better ?

Last edited: Dec 26, 2013
2. ### gronbogTop Member

The game of roulette has the property that no matter what bets you may scatter around the board, and in what amounts, the house edge for a given spin stays the same for your total bet. Another thing to realize that every bet, inside or out, is equivalent to covering some subset of the numbers of the board with the appropriate fraction of the total bet. For example, betting on black for 1800 is equivalent to covering 18 random numbers on the board with 100 each. Betting a column for 1200 is the same as betting 12 random numbers on the board for 100 each, and so on. You can, however change the variance of the game by spreading the same amount of total bet over more numbers (less variance) or fewer numbers (more variance).

This is where your description of the tournament confuses me. You said it was 25 max on a single number, but said you could bet 100000 on the outside. Betting 100000 on black is equivalent to betting 5555.55 on each of 18 numbers. Betting 1000 on a column is equivalent to betting 83.33 on each of 12 numbers. So it would seem that one can circumvent the inside maximum by betting on the outside. The bet with the highest variance will depend on what you're allowed to bet on two streets, a street, 5 numbers, 4 numbers and two numbers.

In order to optimize your chances of reaching your goal, you need to determine how much you need to bet on a single number in order to reach it. For example, for the goal of 96k, to be within range, you need to be able to bet at least 2667 on at least one number (96k / 36). We already know that you can't do that, so you would need to be able to bet 2667 on each of 2 numbers (so you would need 5334). If either of them hit you would have reach your goal. The optimal bet is to fill as many numbers as possible (or allowed) with bets of 2667. If any one of them hits, you have reached the goal. The more numbers you cover, the greater the chance of a hit.

Having said all of that, until you are within range, you want to increase variance by betting as much as possible on as few numbers as possible. The bet with the highest variance is to bet on a single number, so you would normally want to bet everything on one number. In this case you can only bet 25, so you want to try to bet it all on 2 numbers, 3 numbers, 4 numbers, 5 numbers or 6 numbers -- which ever is the lowest they will allow you to go all in on. If all else, fails, then going all-in on columns would seem to be the next best bet until you are within range.

As it seems that your all-in bets are limited to the outside in this case, you will be making all-in bets on columns until you are within range. You will first be in range when your bankroll is 18 x 2667 = 48006 or greater. Since the progression of your bankroll will be 1k, 3k, 9k, 27k, 81k, this will happen when you have 81k. (Edit: correction, you will first be in range when your bankroll is 12 x 2667 = 32004, but the net effect is the same as this will still happen when your bankroll is 81k). This would normally allow you to cover 30 numbers with 2667 each, giving you a 30/38 chance (or 30/37) chance of reaching your goal from there. Your problem will be that you won't be able to cover exactly 30 numbers with outside bets. You can get close with a combination of outside bets however. If you bet 18 x 2667 on red and 10 x 2667 on even, you can cover 28 numbers. I think that's the closest you will get. Edit: The previous bet is incorrect and will not be sufficient in the case where a black even number is hit. It turns out that you can cover exactly 30 numbers with the proper amounts by betting 18 x 2667 on the first 18 (1-18) and 12 x 2667 on the 3rd 12 (25-36).

You could also go for an intermediate total higher than 32004 and lower than 81k, using the same technique once you've reached 27k, but I have a suspicion that the combination of the final two bets will have the same probability of success regardless of the intermediate total.

Last edited: Dec 30, 2013
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3. ### gronbogTop Member

I made a couple of corrections to my response above. I also calculated the probability of reaching the goal of 96k using different intermediate totals after 27k is reached and, as I suspected, the combined probability of success for the overall sequence is always the same at 0.785% for a double zero wheel and 0.897% for a single zero wheel.

Last edited: Dec 30, 2013
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4. ### S. YamaActive Member

Clarification of betting limits would help.
Usually in roulette tournament when they set max bet limits on a single number then it is twice that on two-numbers bet, triple on a street, etc..

In a game with negative edge for the player, the best strategy is to make high bets to reach the goal as soon as possible, so there is less bets made, because they all carry negative edge.
Imagine a game where casino edge is 10%, chance of winning is 45% and chance of losing is 55%, payoffs are one to one. Your bankroll is \$1 million, the goal is to reach \$2 millions, and the bets are minimum \$1 and no maximum, with unlimited time to play. If you bet \$1, your chances are almost zero, if you bet 1 million on the very first spin/hand your chances of reaching the goal are 45%.
The reason for it is that a lucky person who would reach the goal would have to be so lucky that with his billions of dollars bet he would have to overcome a tremendous cumulative edge each bet carries.
Just the opposite would happen if you would have the same edge, betting \$1 almost guarantees you reaching \$2 millions (however long it would take), while betting \$1 million advances you only 55% of the times.

Let’s take roulette (double zeros) example. Your bankroll is \$1K and the goal is \$36K.
You could bet all-in on a one number and if you win you reached the goal. 1/38th = 2.6316%
You, also, could bet all-in on “an even” chance (red, odd, small, etc.) to turn your brl to \$2k, then \$4K, \$8K, \$16K, then \$32K; now you could cover 32 single numbers with the bet of \$1k and if you hit it you reached the goal. Your chances are 18/38th times 18/38th – do it four times and then times 32/38th = 2.008%
Practically it is relatively a small difference but they are significant proportionally.

PlayHunter, I think you chose a very good strategy. As gronbog wrote, there are differences related to variance (which term may be confusing in this aspect to many people) and slight benefits by trying to get paid on the smallest number of bets.
The objectives are to either reach the goal or bust out, not overshoot the goal, and do it with as little dollar “action” as possible.
Since the smallest group of numbers you can bet with no limits (or 100,000) is 12 (columns or segments) you would have to look for smaller subsets where you would bet them to the limit, keeping in mind that at some point you would have to increase the size of the groups to either reach the goal or bust.
It would be academically interesting to find out if 50 spins is long enough to start betting on one number and then changing to a larger group, keeping the above mentioned priorities.
If you consider \$25 as a unit your starting bankroll consists of 40 units. You could group all your chips on 6 numbers (two neighbor streets) and get all your chips in there. That would be 6 singles, 7 doubles, 2 streets, and 2 squares, and one double street. Two of those numbers bring you to \$6,300 and four to \$5,400. You could continue this or simply go now to the columns/sections till you reach your goal, adjusting the bets to get right where you need to be at the last spin.

Was the tournament -EV?
Using your strategy you reach about 81K 12/38th to the fourth power which is 0.9945% for double zero and 1.1% for single zero. It would be about neutral EV if that score paid you 100 times your starting bankroll; otherwise it was considerably negative with a huge variance.

S. Yama

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5. ### PlayHunterActive Member

Thank you very much Mr. Yama and Gronbog, your clear and detailed examples helps me to much better understand the strategy and the EV of it.

Yes, the tournament was -EV, as top prize paid about 84 times the buy-in, and the game was double zero roulette. Probably I should have passed it.

6. ### gronbogTop Member

One must certainly have luck on their side in most casino tournament formats. However, you can reduce the amount of luck needed by using the appropriate strategic betting strategy. That's where the edge over the other players comes from, and that's what this thread was about.

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