Quote:
Originally Posted by fergus lee
.I think we are owed an apology.
|
Sorry. I only look in this forum once every few months because it is usually full of idiotic babble. Good to see some numbers finally being introduced to back up somebody's claims here.
However, I don't see anything like a conclusive argument in your favour.
If Andy and Bob were playing independently of each other than they would each have a 20/37 chance of winning each time. But if they're playing together then the probability of both of them winning definitely isn't 400/1369, because there are only a few numbers that can come up for them both to hit. I'm using this diagram to get my info. You might be able to memorise all this stuff but I've never even played roulette so I don't. I think this is the same as the English version except we have no 00, which I will also ignore.
Here are their strategies. Andy is always betting the following numbers:
Andy- "Blackevens and redodds":
1,2,3,4,5,6,7,8,9,10,12,20,21,22,23,24,25,26,27,28
Bob is betting on the column and dozen of the last winning number. As each number is random he has a 1/9 chance of playing each of the following strategies each turn (although what does he do if the last number was 0? I'm assuming he bets the same dozen and column as the last non-0 winner then).
1,2,3,4,5,6,7,8,9,10,11,12,13,16,19,22,25,28,31,34----14
1,2,3,4,5,6,7,8,9,10,11,12,14,17,20,23,26,29,32,35----14
1,2,3,4,5,6,7,8,9,10,11,12,15,18,21,24,27,30,33,36----14
13,14,15,16,17,18,19,20,21,22,23,24,1,4,7,10,25,28 ,31,34----11
13,14,15,16,17,18,19,20,21,22,23,24,2,5,8,11,26,29 ,32,35----9
13,14,15,16,17,18,19,20,21,22,23,24,3,6,9,12,27,30 ,33,36----10
25,26,27,28,29,30,31,32,33,34,35,36,1,4,7,10,13,16 ,19,22----9
25,26,27,28,29,30,31,32,33,34,35,36,2,5,8,11,14,17 ,20,23----9
25,26,27,28,29,30,31,32,33,34,35,36,3,6,9,12,15,18 ,21,24----10
The numbers on the left are the ones he bets on, each 1/9 of the time, and the ones on the right after the dashes are the total numbers shared by both his and Andy's choices. I'm not entirely sure I have counted them all correctly, but even one error won't make much difference in the end run I don't think. It does look a bit weird with there being 2 x 10 and 1 x 11. It would be more mathematically harmonious with 3 x 10, but then roulette is a strange thing indeed, as if designed by God to appear beatable but not really be beatable.
So we can work out the probability of them both winning as follows:
P(both win) = (3/9 x 14/37) + (3/9 x 9/37) + (2/9 x 10/37) + (1/9 x 11/37)
= 42/333 + 27/333 + 20/333 + 11/333
=100/333.
OK now I've done that it appears that that is actually a slightly higher probability than you thought it was, which goes in your favour. But that seems to have been just a waste of my time because my main issue was with your inference from this probability to there being an edge.
Ok so I propose to do the following: work out the probabilities that just one of them wins and the probabilities that neither of them win, and look at the actual expected value of this strategy (which I propose is exactly the same as
any roulette system, i.e. negative.
So here's my method. Take the first strategy that Bob will be using (the first one in my list)
1,2,3,4,5,6,7,8,9,10,11,12,13,16,19,22,25,28,31,34----14
They have 14 numbers matching. Therefore there are 6 numbers they each have uniquely, and therefore 12 more numbers where only one of them wins. So now like I did before I'll calculate how likely it is only one of them wins:
P(1 wins) = (3/9 x 12/37) + (3/9 x 22/37) + (2/9 x 20/37) + (1/9 x 18/37)
= 36/333 + 66/333 + 40/333 + 18/333
= 160/333
This means that they should both lose 73 times out of 333.
So, now let's consider it in terms of expected value. They're both making 20 bets a turn, so let's say they each bet £1 on each number. If any of their numbers win they receive £36, and they're betting £20, between them £40 a turn, so we need the strategy to be worth more on average than £40 for it to be profitable.
Expected Value = (100/333 x 72) + (160/333 x 36).....+[73/333 x 0]
I put the last bit in brackets because it doesn't add anything to their expected value, it's just there for completeness.
Expected Value = 7200/333 + 5760/333
= 12960/333
= £38.92
So if my calculations are correct it is a losing strategy. I'm now going to masturbate my own ego and show that it's exactly as profitable (i.e. not profitable) as any roulette strategy.
Suppose we have one person betting the whole £40 on any random number (or column, dozen etc, always getting the same odds), they have a 1/37 chance of winning £1440.
Expected Value = 1/37 x 1440
= £38.92
It's actually exactly the same number as 37 x 9 = 333 and 1440 x 9 = 12960.
Therefore your strategy does not work. It may
appear to work because over 3/4 of the time Andy and Bob will be raking some prize and only losing outright 1/4 of the time. However nearly half the time (and this represents 2/3 of the time they are "winning") they rake in £36 having bet £40, so they lose money even when one of them wins, even if it seems like they're doing well because they're being paid. The person who bets £40 on one number is going to appear to lose more because they very rarely win anything. If Andy and Bob sat at a roulette table with £4000, after 100 turns we would expect them to have £3892 left, the same as we would expect the £40-bettor. The difference is that the £40-bettor has very high variance. In the 100 turns he'll probably cash in 0,1,2 or 3 times, and it isn't that unlikely for him to lose the whole £4000 (6.5% of the time), whereas it'd be practically impossible for Andy and Bob to lose the whole lot (0.00000000000000000000000000000000000000000000000 000000000000000001224% of the time).
The thing with roulette is, regular, bet-on-one-number-strategy roulette has a high variance and that makes it better for the player if anything. The guy who bets £40 on one number a turn for 100 goes, I suspect (total estimate here), will win some money about 30% of the time I imagine, which is when he gets lucky. But with your system I reckon Andy and Bob will win less than that because they're basically giving away a tiny amount every turn. You would have to be insanely lucky to win with this system over about 10,000 goes, whereas the £40-1-bettor wouldn't have to be as lucky I don't think (still very lucky). I'm not sure about anything I said in this last paragraph though. But the rest of the post is basically a big QED against your system.
Therefore in conclusion:
