# Hypergeometric distribution for different probabilities for each draw

The object of my exercise is to calculate the probabilities of different draws from a set, without replacement, but each draw does not have the same probability as the others.

If we use Powerball for example, there are $59$ balls and we pick $5$ of them.

Using the Hypergeometric distribution, we can calculate the probability of matching $3$ of the $5$ balls picked.

This is done under the assumption that all balls have equal probability of appearance.

What if the probability of the balls were not equal but were different for each ball? This cannot be described by the Hypergeometric distribution.

How can I calculate the probability of success for a draw, for balls that I chose, given the frequency distribution or probability for the specific balls I choose?

The draws are done without replacement, as in the original hypergeometric distribution

I believe that Fisher adjusted the weights on the singular events to ensure that the individual probabilities remained equal or 'as if they were independent' so that the 'without replacement' effect did not negate the difference in these singular probabilities.

By saying that it negates it, I mean in the sense, that inevitably, if one samples the urn a great deal of times, and there is only one ball physical ball identical to itself, then the the frequency may eventually just mirror the combinatorial proportions of the urn. With each ball having relative frequency =$\frac{1}{n}$ where $|n|$ is the cardinality of the urn. So that on first appearances, it would appear that the singular probability differences make no real difference (in the long run)

I just want to make sure that I understand the question.

I presume you wanting to calculate the probabilities wherein the single case probabilities of each 'individual ball', or 'definite description probabilities' are not equal, where one samples, say from an urn, without replacement.

This being in contrast to the case, where to say an urn, which contains 50 balls, where one ball, $B_50$, for example, is marked out as the 'success ball', whilst the other $49$ balls are marked out as 'failure' balls, where Ball $50$ serves as rigid desig-nator which uniquely denotes that specific entity. Where 'failure' in (former) below, denotes the disjunctive reference class, consisting of the dis junction of every failure ball being drawn (balls $1-49$). These (former) probabilities may always be different, but I presume you dont just want that. .

**(former)**So whilst the 'probability of success' may be less then the probability of drawing a 'failure' ball, $$Pr(\text{failure})=PR(\neg\text{success})=\frac{49}{50}>Pr(\text{success ball})=\frac{1}{50}$$

Thus the 'probability of non-success (in one trial) is different from the probability 'drawing a 'success, (ball 50)' may be distinct from 'probability of success'.

(latter) The probability ( drawing any given specific ball may (failure or not) may be equal, or rendered equal, to the probability of drawing any other specific ball (failure or not)).

For example if one samples the entire urn for example.Where any given specific ball is going to be drawn, once and only once at some point in time, and if one just uses frequency data, the specific events would appear, on first sight to be equally probabable

I presume you want that, or the possibility that, the (latter) singular specific ball drawing probabilities to be different from one another?. That is, that $(A)$ may not be the case, and not merely that $(B)$ holds below as if often does.

Where by 'specific' or unique failure ball, I mean the ball described under its maximally specific reference class description, that unique denotes that physical ball, such as (ball $49$).

So that if a failure ball is drawn and it not THAT failure ball (say $48$ instead of $49$), it does not count as getting THAT failure ball $(49)$.

I presume you wish to consider the case where these singular 'de re' probabilities referrent statements such as in $(B)$ are not equal; and that perhaps all of the individual physical balls $(1-50)$ have distinct probabilities.

$(B)$ PR( A Specific 'failure ball'= 'the probability of drawing the specific success ball').

This is arguably going to hold, in some standard hyper-geometric cases; when one samples the entire urn at at least.

Where this to be distinguished from $(A)$ the probability of getting

$(A)$ $[[\text{b_1 is drawn} \vee \text{b_2 is drawn} \vee.....\vee \text{b _49 drawn}] \land [\text{b_50 is not drawn}]]\leftrightarrow_p [\text{b_1 is drawn} \vee \text{b_2 is drawn} \vee........\vee \text{b_49 drawn}]\leftrightarrow_p [\text{b_50 is not drawn} ]\leftrightarrow_p [\neg \text{b_50 is drawn} \, /\,\neg \text{success} \, / \, \text{failure} ]$

I distinguish the equivalence relation with a p subscript as matter of the fact that the scenario, necessitates the equivalence, and it not a logical equivalence as such, except in the context of the sample space (except perhaps the last implication). But enough of this philosophical pedantry.

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As a hyper-geometric distribution (may, but not may not always) determine that $(B)$ is false, particularly if you draw all $50$ balls! The frequency distribution for any given specific ball will be $\frac{1}{50}$.

Even if the single case probabilities are different. This is as the non-independence, induced by non-replacement urn models tends to negates probabilistic advantage of higher probability events.

As they are more likely to be drawn first; so, when one samples a great deal of urn (the cardinality of drawings, start begin to are significant fraction of the cardinality of balls in the urn). As the conditional probability of the lower probability events are going to increase (over their prior probability) by a higher degree.

This is in fact an issue for infinite frequen-tism and in some single case approaches (even though its a bit off topic ) to the Monty hall problem.

see (for the former, more relevant issue)

If those are outcomes kinds when rendered maximally specific are not possible on that single case; consider the distinction between this trial produces its 'heads outcome' and this trial produces 'its' 'tail outcome', wherethe $\text{relative frequency of heads} > \text{tails}$.
After all, these 'other heads' outcomes, are not possible on your trial, $t1$, and the only way you that 'a heads outcome' can occur on $t1$, is if this maximally specified outcome '$t1$ heads' occurs, then the fact that the relative frequency of heads is higher may not be relevant.
Particularly if you know that your trial is '$t1$'. It depends on the situation of course