Probability of subevents

Question

How to compute the probability of sub-events given that we only have the events probabilities?

An example to clarify, drawing cards: We consider two cards at a time, assume we are given the following: Probability that we have two hearts is $p_{hh},$ one heart and one spades is $p_{hs},$ heart and clubs $p_{hc}$ and heart and diamonds $p_{hd}.$ And similarly for all other pairs we can consider. But the question is: how from these probabilities (describing events of pairs of cards) can we compute the probability of having a card of hearts? (i.e. now we're looking into single card events, I know the answer in this example is trivial because we just have 13/52, but I'm trying to learn how it can be computed as a subevent given that we know the probability of a larger event. I imagine a simple union of the four events will not give the correct answer, or does it? I mean $p_h = p_{hh}+p_{hs}+p_{hc}+p_{hd}$ is very likely wrong.

Answer 1

An example to clarify, drawing cards: We consider two cards at a time, assume we are given the following: Probability that we have two hearts is $p_{hh},$ one heart and one spades is $p_{hs},$ heart and clubs $p_{hc}$ and heart and diamonds $p_{hd}.$ And similarly for all other pairs we can consider. But the question is: how from these probabilities (describing events of pairs of cards) can we compute the probability of having a card of hearts? (i.e. now we're looking into single card events, I know the answer in this example is trivial because we just have 13/52, but I'm trying to learn how it can be computed as a subevent given that we know the probability of a larger event. I imagine a simple union of the four events will not give the correct answer, or does it? I mean $p_h = p_{hh}+p_{hs}+p_{hc}+p_{hd}$ is very likely wrong.

Nope, it is correct.   However, $13/52$ is wrong.

If $p_{hd}$ is the probability of selecting one heart and one diamond in either order when selecting two cards from a standard deck, and so on, then $p_{hh}+p_{hs}+p_{hc}+p_{hd}$ is the probability of selecting at least one heart when selecting two cards.

Where as $p_{hs}+p_{hc}+p_{hd}$ is the probability of selecting exactly one heart.

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$$p_{hh} = 1/17, p_{hd}=p_{hs}=p_{hc}=13/102\\~\\ p_{hh}+p_{hs}+p_{hc}+p_{hd} = 15/34\\~\\ p_{hs}+p_{hc}+p_{hd} = 13/34$$

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The probability of not selecting any hearts is ${19}/{34}$

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$$p_{hh}=\dfrac{\binom{13}{2}}{\binom{52}{2}} = \frac 1{17}\\p_{hs}=\dfrac{\binom{13}{1}\binom{13}{1}}{\binom{52}{2}} = \frac{13}{102}\\p_h = \dfrac{\binom{52}{2}-\binom{39}{2}}{\binom{52}{2}} = 1-\frac{19}{34} = \frac{15}{34}$$

Answer 2

First of all, I think it really depends on the nature of your experiment. For example, drawing two cards are not independent, while for some other experiments we may have independent experiments. But here is an approach. You can have the following relations.

$P(hh)=P(h_1)\times P(h_2\mid h_1)$

Where $P(h_2\mid h_1)$ is the probability of getting a heart in the second draw, given we have drawn a heart in the first draw.

Likewise, you can have

$P(hs)=P(h_1)P(s_2|h_1)+P(s_1)P(h_2|s_1)$

You have $10$ such equations, if you do not have any orders in double draws. Then, you can obviously see that you need conditional probabilities like $P(h_2|s_1)$ and there are $4^2=16$ possible conditional probabilities. adding the single probabilities, you have a total of $20$ unknowns. From an information theoretic point of view, you do not have enough information to find single draw probabilities. However, having certain conditions, like independency of drawings, you might be able to get rid of some unknowns and get what you want.