# Roulette Tournament: Any thoughts?

Discussion in 'Roulette Tournament Strategy' started by toonces, Mar 10, 2005.

1. ### S. YamaActive Member

Process

I just realized how awkward and difficult it is for me to describe this methodology. It is much easier to think over the process than to explicate it using words. But I promised to do it. So, here we go.

Roulette carries high casino edge of 5.26% (2/38) on all but one bet (group bet for 0, 00, 1,2, and 3).
For good tournament players that’s nothing to worry about, just the opposite. Let weaker players bet, play and pay the high vigorish. Effect of casino edge is proportional to all money bet and is a fixed part of the game that we can not do much about.
The most important aspect of the tournament is to employ as much money from the bankroll as needed without overbeting.
In a simple situation where leader bets on only one group of numbers to catch him we have a two step procedure and two main models of play.
1- Bet on maximum groups/numbers that are not part of your opponent bet, for the amount that winning it alone is sufficient to overcome the opponent if he loses his bet. And the rest of your bankroll should be used to make bets on smaller groups or individual numbers from the group the opponent has bet, for the amount that winning it covers the competitor winning his bet.
2- Bet the same way, a minimum more than the opponent has bet, plus the difference you were trailing, plus the amount you intend to bet on other groups/numbers that are not part of his bet. And as many groups/numbers that are not part of his bet, for the amount that winning it alone is sufficient to overcome the opponent if he loses his bet.

This would work great if we could make bets in infinitesimal increments. With bets limited to rounded numbers we need to make a second step – comparing which method overbets our goal amount the least and if holding some unbet money makes the necessary gains closer to payouts for bets in rounded amounts.

[In the teaser we had two goal numbers to reach for. More than \$3,660 if the opponent loses, and more than \$3,840 if the opponent wins.
Here are the inclusive returns for the following bets on individual numbers: \$100-\$3,600 \$105-\$3,775 \$110-\$3,960.
\$105 bet wins \$115 more than we need for the low number and \$110 bet wins \$120 more than we need for the high number. They would be overbets.
We should try to bet first possible (in the teaser it had to be dividable by \$5) lower amount and keep enough money to make up for smaller wins. Winning \$3,600 (bet of \$100) and keeping \$65 gets us over the low goal number, and winning \$3,775 and keeping \$65 gets us over the high number. ]

Keeping unbet money may but doesn’t have to increase our chances.
The amount of unbet money would always have to be smaller than the amount needed to cover yet another single number.

S. Yama

2. ### S. YamaActive Member

another one

Okay, March Madness is over, time to get back to a roulette wheel.

Your bankroll is \$1,000,000; you have to bet first. Your only opponent has at least \$200 less than you do but you don’t know exactly how much less. The minimum bet required is \$100, no maximum restrictions. \$5 increments on all bets.

For those who don’t play roulette every day here is payoff table and bets min/max for the teaser:
1 number ---------- 35 TO 1 …... Min \$5, max 100
2 numbers --------- 17 TO 1 …... Min \$5, max 200
3 numbers --------- 11 TO 1 …... Min \$5, max 300
4 numbers ---------- 8 TO 1 …... Min \$5, max 400
5 numbers ---------- 6 TO 1 …... Min \$10, max 600 …(this is 0, 00, 1, 2, 3 group bet)
6 numbers ---------- 5 TO 1 …... Min \$10, max 700
12 numbers --------- 2 TO 1 ..…. Min \$25, max 1,500
18 numbers --------- 1 TO 1 …... Min \$25, max 3,000
24 numbers --------- 1/2 TO 1 ... Min \$25, max 5,000

PS
To truly understand “mechanics” of tournament bets assume that your opponent have no restriction on his bet, which means he could bet a fraction of a dollar.

S. Yama

Last edited: Apr 2, 2005
3. ### instagatorNew Member

This seems too simplistic. Bet \$100 on each number. I'll get back \$3600 and lose \$200 forcing him(her)(it) to win a bet at a minimum disadvantage of 5.6%.
IG

4. ### S. YamaActive Member

Good try

Betting \$100 on each number? Good try.
It does seem like a very good play but it is not as golden as it glitters.

For convenience let’s call this \$1,000,000 bankroll –X. If you bet \$100 on all individual numbers, no matter where the ball lands, you end up losing \$200. Seems like a good deal as your lead is \$200 and your opponent needs to make a winning bet. Your bankroll will be X-200. Your opponent could bet 37 individual number for \$Y and keep at least one cent unbet to beat you in 37 out of 38 spins.
We remember that payoff for a single number is 35 to 1 and player gets back his original wager. Your opponent has to bet 1/36 of your end bankroll on individual number to tie you. \$Y is (X-200)/36.
To achieve this her, his, or its bankroll needs to be 37*(X-200)/36+1cent.

Consider bet of \$50 on every individual number. No matter what the spin you end up losing \$100, so your bankroll would be X-100. It is easier for an opponent to beat bankroll of X-200 than X-100.

Of course betting \$50 on all individual numbers is not the best play, but what is?

S. Yama

5. ### instagatorNew Member

If X-100 is better than X-200, then X-10 would be better than X-100. ???
IG

6. ### LimoExecNew Member

x-10 would be better than x-20

7. ### S. YamaActive Member

getting closer

Yes, x-10 is even better.
The best move is to bet minimum required. In our case \$100.

Now, how would you bet that C-note?
There are two possible answers, one theoretical, and the other with specific bets.
Come on, give it a try.
It is part of conceptual tournament strategy thinking.

S. Yama

8. ### ProspectMember

Theoretically, we should bet \$100 / 38 = \$2.63 on each number. No matter which number shows, we would lose \$5.26. Notice how it coincides with the house edge.

In reality, we need to bet in increments of \$5. So, I am not sure how to explain this part. But, we need to find some kind of common multiplier of 5. For example, 5 can be 2 * 2.5; 3 * 1.66; 4 * 1.25; etc.

So, that means it is possible to bet \$2.50 per number by making split bets across the table. This will take \$95 to cover the table. So, no matter what number shows, we would lose \$5. Then, bet the remaining \$5 anywhere to meet the \$100 minimum. The worst case scenario is that we would lose that too for a total lost of \$10.

9. ### S. YamaActive Member

On the right track.

Prospect, that was exactly the idea behind the teaser.

One must bet minimum and spread it across all number as evenly as possible .

In a real casino, your bet of \$5 on all split numbers and the remaining \$5 anywhere would be great one.
But who will come with the perfect bet? How to make this remaining \$5 “working” on the broadest selection of numbers?

***
If anyone is interested why we should spread it evenly here is the explanation using some examples.

If you leave out two numbers and bet the minimum \$100 by placing \$50 on complementing groups of 18 numbers (red, even, first 18, etc.) and only leaving out two zeros, you would get very close to a perfect bet.
Your opponent would need to have (x-100)/36 (because you lose whole 100 if any zero comes) times two plus 1 cent to overcome you if any of zeros come. Then he would need to have enough bankroll for thirty five bets of (x-50+50)/36 (you win one and lose one bet if it is not zero).

Let’s make another bet, like Prospect suggested, and bet all number \$2.50 (\$5 on split numbers) and the remaining \$5 bet on a group of four (this is the biggest group allowing single \$5 bet – check betting requirements and limits a few messages above). Note that if a number from the group of four is called you win \$35, and if any of the other thirty four numbers come you lose \$10.
Now, the opponent will skip one number from the group of four and bet the other 37. He needs 34 bets of (x-10)/36 plus 1 cent, and three times (x+35)/36.

If you do all the calculation the first bet is still slightly better than the second one.

What’s the bottom line?
If your opponent skips one of the numbers that you have bet then it makes the amount you’ve bet on this number go to waste.
In the first case the opponent could skip one of the numbers where you have bet \$50 on 18 numbers and annihilate (50/18=2.778) and in the second example it was \$5/2 plus \$5/4 for a total waste of 3.75.

There is still a better bet that would require your opponent to have whooping 70 cents more in his bankroll to catch your \$1,000,000.

S. Yama