Random Thoughts on Roulette

[Dice Man]

It depends what kind of definition of probability we are using as to whether past events have a probability. Whatever definition we use it won't make any difference to this discussion because we are concerned with the logic of subjective probabilities here, that is the relationship between theory (e.g. "Bob won") and evidence (e.g. "0 did not come up" or "I just saw Bob win")

By saying that the past is certain and the future uncertain you're making a lot of metaphysical speculations. We have just as much reason to think that the future is as certain as the past, it's just that we don't have knowledge of the future as we have of the past.

"Dependent" means that they depend on each others bets before at leat one of them can win.Thus you are arguing that the one cannot win without the help of the other.

No that is not what "dependent" means in the context of probability. Here are two examples you'd get in GCSE maths and earlier about the difference between dependent and independent events. Suppose we have a bag with 10 balls in it, 4 red and 6 blue. If we want to know the probability of taking a red one out followed by a blue one where we replace the ball after each pick then we calculate it as 4/10 x 6/10, 24% likely. We can do this simple calculation because they are independent events. Us taking a red or blue ball on the first pick has no effect on the probability of taking a red or blue ball on the second pick because the first ball is replaced. This is the sort of calculation you are doing with your 20/37 x 20/37 theory of the probability that both Andy and Bob win.

If we do not replace the first ball then we work it out as 4/10 x 6/9 and the probability of getting the second ball is dependent upon the probability of getting the first ball. They are not entirely dependent in the sense you mean, whereby it's impossible to get a blue ball out second unless we get a red ball out first or anything like that. Probabilistic dependence isn't a matter of all-or-nothing.

Imagine the roulette example is like the balls-example. There is a bag with 37 balls in it. Andy has written his name on 20 of them and so has Bob, and they will have both written their names on some of the balls. This time we are only taking one ball out of the bag, but suppose that we only get information about it a piece at a time. So, the probability that the ball has got Adam's name on it is 20/37. We now want to know whether it also has Bob's name on it. We know that Adam and Bob randomly share 20 out of 36 numbers, not 37, because it is explicitly stated that neither of them bets on 0. For any one number that Adam has his name on, Bob has a 20/36 chance of also having his name on it. So the probability that they both have their name on it is 20/37 x 20/36 which is 400/1332, which is 100/333. How convenient that number is? It's the same figure that took me about half an hour to get when I first replied to you, not realising at the time that I could get it with just 4 numbers and 3 operations.

You might think that 20/37 x 20/36 is a bit strange here, because we are only taking one ball out of the bag, not two. But effectively we are taking two balls out of the bag. 20/37 is the probability of taking one of Andy's balls out of the bag. What we could then do to work out the conditional probability that Bob's name is also on it is to replace that ball (stay with me here...), but we go into the bag and remove the 0 ball (the only ball with nobody's name on it). We could then take a new ball out of the bag to determine the probability of Bob's name being on the original ball. Hence the 20/36 figure.

[fergus lee]

Your reasoning is flawed Diceman because in no case have Andy and Bob 20 numbers in common.The most they have in common is 14.So your 20 numbers scenario is- how shall I put it ? Delusional.

Your previous "analytical" speculation is answered, not by Maths but by simple arithmetic."If Andy wins then it is likely that Bob also wins".The zero is irrelevant because ,since neither Andy or Bob bet it they both lose when it comes up -and in all cases.They thus have only 36 numbers where they can win.If Andy wins and "it is likely that Bob also wins"then the winning number must be one of only 14.So how is BOb "likely to win also" when there are 22 numbers suggesting that it is UNLIKELY that he also wins ?

Your reluctance to admit that their bets are independent is holding up proceedings so why do youdo it ?Is it because you really do know Probability Theory and know where I am heading and are reluctant to admit that you might posibly be wrong?

Whatever your answer I would like SteveA or Danny C to come in here and tell me to carry on without Diceman or whether Diceman has exposed me as a delusionist and should stop here.

Otherwise I am not prepared to carry on with this nonsense

[Dice Man]

Your reasoning is flawed Diceman because in no case have Andy and Bob 20 numbers in common.The most they have in common is 14.So your 20 numbers scenario is- how shall I put it ? Delusional.

I have never said that they do. What they do have is 20 numbers shared out of a set of 36. So they have between 4 and 20 numbers shared as you say. Because of Bob's randomish strategy we can never know how many they do share but our best estimate is that they share 20x20/36 balls, just over 11.

At the moment I think we both agree that the probability of just Andy winning is 20/37 (or just Bob winning: it doesn't matter which order you do them in, so I'll just do them alphabetically). Then you want to say that Bob is also 20/37 to win even given that Andy has already won, and I want to say it is 20/36 given that Bob has won, so that's where we are. How am I saying that they share 20 balls here? We're replacing the ball that Andy just won on (because they can both win on the same ball), and we're taking out of the bag the one ball that Bob definitely cannot win on. He has an equal probability of winning on the 36 remaining balls, and that equal probability is 20/36, the ratio of the balls he has bet on divided over the total balls he could have bet on.

Your previous "analytical" speculation is answered, not by Maths but by simple arithmetic."If Andy wins then it is likely that Bob also wins".The zero is irrelevant because ,since neither Andy or Bob bet it they both lose when it comes up -and in all cases.They thus have only 36 numbers where they can win.If Andy wins and "it is likely that Bob also wins"then the winning number must be one of only 14.So how is BOb "likely to win also" when there are 22 numbers suggesting that it is UNLIKELY that he also wins ?

Yes. Suppose that Andy has already won. Then if Bob wins as well the winning number must have been 1 out of X, where X could be between 4 and 14 inclusive. There are many correct ways of doing this calculation, all of which get my answer. We could do it this way instead. The probability that Andy wins is agreed upon to be 20/37. So then suppose that Andy wins. So let's open up the bag and remove all the balls that aren't Andy's winners. So now there are 20 balls left in the bag all with Andy's name on, because now we want to work out the probability that one of Andy's winners is also Bob's winner. So we know that they share between 4 and 14 balls so there are definitely between 4 and 14 Bob-balls in the bag now. But as mentioned, the average number of balls they will share is given by 20x20/36 which is 100/9. So now to work out the probability that Bob wins given that Andy has won we answer it with 100/9 /20, i.e. "about 11" balls over 20, which surprise surprise, is 20/36.

Also my premises were of the form "If Andy wins then it is more likely that Bob wins" not just "If Andy wins then it is likely that Bob wins" as you presented it (which is also true but trivial). By "If Andy wins then it is more likely that Bob wins" I mean the following:

The prior probability of Andy winning, P(A) = 20/37.
The prior probability of Bob winning, P(B) = 20/37 as well.
The conditional probability of Bob winning, given that Andy won, P(B¦A) = 20/36.
That's why I'm saying, if Andy won, it's more likely that Bob won, because the probability has increased from 20/37 to 20/36.

Your reluctance to admit that their bets are independent is holding up proceedings so why do youdo it ?Is it because you really do know Probability Theory and know where I am heading and are reluctant to admit that you might posibly be wrong?

To be perfectly honest, I'm not sure how your theory will get an edge, even if their bets were independent. But I'm only saying that they are dependent because they are. It means nothing to me to admit I am wrong. I have already done it a few times in this thread. A few posts ago you asked me only if their probability of both winning was 20/37 x 20/37 and I agreed, only to rescind it the following post as I realised my original calculation was correct precisely because the bets are dependent. Since then I've just tried to explain how they're dependent in the simplest way I can.

So far I've given 3 methods at arriving at the same answer, the one I used at the very start which took ages and accounted for every ball that Bob might bet on and how many he shares with Andy, which worked out as P(both win) = (3/9 x 14/37) + (3/9 x 9/37) + (2/9 x 10/37) + (1/9 x 11/37). The second is the much simpler P(both win) = 20/37 x 20/36, and the third is the nearly-as-simple P(both win) = 20/37 x (20x20/36) / 20, and it's fairly easy to see how the last two get the same answers, but they all do.

You can't just start at the assumption that the bets are independent. I know that they look independent on first appearance but they really aren't. My 4-premise argument shows why they are not. Have a look at it again now that I have clarified what I meant by "If Andy wins then it is more likely that Bob wins".

Whatever your answer I would like SteveA or Danny C to come in here and tell me to carry on without Diceman or whether Diceman has exposed me as a delusionist and should stop here.

I'm hoping you'll just see the truth and won't need to leave the debate labelled as a delusionist.

[maxormark]

Seriously though, it's all just a matter of pure chance> Play long enough and all outcomes are equally likely say the mathematicians.

Of course, along the way, any particular sequence of numbers is equally possible and certainly any sequence of colours. Maths won't win you anything, but luck might; and we all have experience of luck, good and bad.

Real good luck comes in choosing to bet at the exact time when that series comes up and, in my own experience, some people just have it. Skill, if it exists, comes in knowing when the luck is about to run out.

[sappaper]

Seriously though, it's all just a matter of pure chance> Play long enough and all outcomes are equally likely say the mathematicians.

Of course, along the way, any particular sequence of numbers is equally possible and certainly any sequence of colours. Maths won't win you anything, but luck might; and we all have experience of luck, good and bad.

Real good luck comes in choosing to bet at the exact time when that series comes up and, in my own experience, some people just have it. Skill, if it exists, comes in knowing when the luck is about to run out.

I think maths are IMPORTANT to play roulette.
It's a matter of chance, that's correct but i believe there's more because almost everything in this world is based on maths

[RubyRoulette]

Wow this is a complicated and interesting thread - it is all down to chance and luck

[fergus lee]

I am inclined to agree with sapaper maxomark.People who rely entirely on luck are the ones who are most likely to wind up as problem gamblers because they tend to believe that their bad luck must change and go overboard with their bets.They also are the ones who inclined to believe that the croupier is so expert that he deliberately doesn't bring up their numbers.

We who " fiddle withn the numbers " know and accept that we may lose.What really matters is that your winnings exceed your losses.I take some comfort from the fact that LUCK bettors also reject the mathematicians argument that you MUST lose.I find that I am "luckier " since using the numbers .Obviously you would expect me to say something like that judging by my previous posts !

I seem to have made a mistake in thinking that Andy and Bob's bets were entirely independent.Not because of Diceman's ridiculous answer but because they do indeed share some numbers .However,I am not so sure that they NEED to be independent to justify my argument.
In my next post I shall restate my justificatioin for betting the suggested numbers and number the points so that someone can point out exactly where I am wrong, if I am wrong.

What do you mean - you are bored to tears ?

[Dice Man]

I seem to have made a mistake in thinking that Andy and Bob's bets were entirely independent.Not because of Diceman's ridiculous answer but because they do indeed share some numbers.

That's not the reason at all. If Bob chose 20 random numbers to bet on equally distributed between 0 and 36 inclusive then Bob winning and Andy winning would be independent. They're currently dependent because they both share a number that they never bet on. Take it to the extreme and it should be more obvious. Suppose that Andy always bets on 1-20, and Bob bets on 20 random numbers between 1 and 21. It would be very unlikely for just one of them to win (2/37), and if you knew that one had won it would be highly probable the other had won too. They would certainly be dependent. Andy and Bob's actual strategy is the same, just only very slightly dependent.

[fergus lee]

Two fictitious mates Andy and Bob, decide to spend an evening playing roulette in a casino.

They have different " systems ".Andy always bets all of te BLACk / EVEN and RED/ ODD numbers while Bob bets all of the numbers in the same column and dozen as the last winning number.Thus, they each bet twenty numbers on each spin

( 1 ) Noting that they sometimes both win on the same spin they stop betting and work out the probability that it might be worthwhile betting on only those numbers that they have in common on each spin.

( 2 ) They do this by multiplying their individual probabilities which are 20/37

(3 )Multiplying their two 20/37 chances results in 400/1369 so they are likely to win only 400 of their 1369 bets

( 4 ) To find out whether or not this is likely to be a paying proposition they first multiply their expected 400 wins by 36 since this is the return on each win ( 35+1 )This results in the figure of 14400.
From these winnings, however, they must deduct the number of chips they have bet on their 1369 bets.THey divide the 14400 by 1369 .This shows that they can only make a bet if the numbers in common are ten or less.

( 5 )There are tweny numbers that fit this scenario and sixteen which do not.(The zero is a loser in all cases )Wishing to bet every spin they search for a way to bet on the sixteen numbers.

(6 ) They settle for still choosing Andy's BLACk/EVEN and RED / ODD numbers but assuming that the same column and column will NOT win.

( 7 ) The calculation is now 20/37 multplied by 16/37 which results in 320 / 1369 .Thus ,they find that bettng where there are 8 or less numbers in common is likely to be a winning proposition.

(8 ) THe mathematicians' reasoning that you cannot win is by calculating that making 37 bets of one chip on only one number is 1/37 x 36/1 so is a losing proposition .

(9 ) Their reasoning is the same as mine but looks at the problem from a different angle / perspecive.My calculations counter their calculations so it cannot be conclusively proved hat you cannot win at roulette . ( In the LONG RUN , of course ! )

[fergus lee]

Line ( 6 ) should, of course read "column and DOZEN "