hi folks,

let's say someone plays a session of roulette in which there is a strong tendency for blacks. instead of betting on blacks he bets on the 1st and the 2nd col. he assumes that if there is trend for black there should also be a trend for these 2 col.

is this assumption based on the believe that conditional probability sort of favors his bet selection?

oh, I know that this has been tried before and doesn't work  . I'm just trying to understand the idea and theory behind it.

cheers

hans

Quote from: Marshall Bing Bell on April 14, 2013, 09:54:17 PM
If you knew with about 75% certainty that black will show next spin then yes you would have an advantage with the 2 column bet. IMO.

What is knowing with 75% certainty?

Conditional probability just means that you're restricting the possible "sample space" from which you select outcomes. In a sense all probability is conditional, but often the condition is assumed and not explicitly stated. So for example, when you say that the probability of red hitting is 18/37, what you're really saying is that the probability is 18/37 GIVEN THAT the wheel is unbiased, spins are independent, and any other factors which might have a bearing on the probability of red hitting.

In mathematical notation, "GIVEN THAT" is denoted by a vertical line "|". So the probability of red coming up can be expressed as P(R|X), where X is "the wheel is unbiased" etc.

Another way of thinking about what conditional probability is is to ignore the outcomes which you're not interested in. So if you wanted to know the probability that an even number will hit, given that the 2nd dozen hits - P(E|2nd Dozen) - then you can ignore all outcomes which are NOT in the 2nd dozen, and it's easy to work out what the probability is: there are 6 even numbers in the 2nd dozen so the probability is just 6/12, or 1/2.  A conditional probability is just the probability of an event when the event is contained in a subset of the universal set (the set which contains all possible outcomes).

Now suppose you want to know what the probability is of red hitting, given that the last 10 spins were black. This is a conditional probability because of the phrase "given that", so you could write it like this:

P(R|BBBBBBBBBB) = ?

Now this seems to be quite a different kind of problem from the previous one, where the concern was with members of certain class (the even numbers) being restricted to another class (the numbers in the 2nd dozen), and yet it is still a "conditional" probability because we're asking how the event to the right of the "|" symbol limits or affects the event to the left of the "|". In this case, because we know that spins are independent, that is, the event of BBBBBBBBBB does not affect the event of R, because on each spin there are still 18 blacks that could possibly hit (unlike in Blackjack, for example, where cards are REMOVED from the deck) we can see that

P(R|BBBBBBBBBB) = P(R)

This is just an expression of the fact that the wheel has no memory.

In general, if any events X and Y are independent, then P(X|Y), (the probability of X given that Y has occurred) is just equal to the probability of the event X. Y is irrelevant to the probability of X.

Quote from: HansHuckebein on April 14, 2013, 12:56:59 PM
is this assumption based on the believe that conditional probability sort of favors his bet selection?

I suppose you could put it like that, but the conditional probability doesn't favour the trend of black continuing, which is the most important thing.

thank you bayes for your detailed explanaition.