Determining individual probability from given group probability

Gabrosin

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I'd like to solve for the individual probability of an event given a group probability, and my math skills are a little rusty. Is it possible to easily solve this problem, and if so, how would I go about it?

I'm going to draw five cards from a non-standard deck of cards. I'm told that my probability of drawing at least one heart among those five cards is 95%; drawing at least one diamond, 50%; drawing at least one club, 10%; drawing at least one spade, 1%. How can I determine the probability of drawing a heart/diamond/spade/club when drawing just one card?

Thanks in advance for any assistance!
 
Hi Gabrosin,

Take hearts as a case study. "At least one heart" means one or more. So the probability of drawing at least one heart in 5 draws is (where the symbol P(A) means "probability of event A occurring"):

P(1 heart OR 2 hearts OR 3 hearts OR 4 hearts OR 5 hearts) = P(1 heart) + P(2 hearts) + P(3 hearts) + P(4 hearts) + P(5 hearts)

The righthand side comes because these are mutually-exclusive events (if you drew exactly one heart, it can't be true that you also drew three hearts). Probabilities of mutually-exclusive events add together.

But this requires computing 5 separate probabilties, and it's a pain. What's easier is to recognize that drawing at least one heart is the exact opposite outcome of drawing NO hearts. These two events are mutually-exclusive, and they are the only two possible outcomes. Either you drew some hearts (one or more) or you drew no hearts (zero). So these two outcomes have to add up to a 100% chance of occurring:

P(at least one heart) + P(no hearts) = 1

or P(no hearts) = 1 - P(at least one heart)

P(no hearts) on a given draw is much easier to compute.

Assume you have n cards, and h of them are hearts. Then the probability of no heart on a single draw is (1 - h/n). What is the probability of no hearts in 5 independent draws? Hint: probabilities of independent events multiply together.
 
What does / mean?

I interpreted '/' as a sort of exclusive OR in this context. Meaning that, I assumed the OP was asking, what's the probability of drawing a card of a particular suit, for each of the suits? I don't mean that the events are XOR'd and we want to compute that probability. I mean that the phrases in the question are being asked as "one or the other, but not both." So the slash allows the OP to several questions compactly: what's the probability of drawing a heart? What's the probability of drawing a spade? Et cetera. Hope that is clear.
 
Assume you have n cards, and h of them are hearts. Then the probability of no heart on a single draw is (1 - h/n). What is the probability of no hearts in 5 independent draws? Hint: probabilities of independent events multiply together.
I am a little confused by the term "independent draws." If we draw one card, note its suit, replace it, and repeat that process four more times, we would clearly have independent draws. But that is not an equivalent to drawing five cards without replacement.

I have not worked this out, but we have four unknowns (the number of clubs, diamonds, hearts, and spades in the non-standard deck). In other words, we do not know what h and n are so I cannot see (maybe I have not had enough coffee) how we just multiply some numbers together. I shall work on this because it does not seem like a class exercise. It probably will turn into a set of four equations in four unknowns, but they may not be linear, the arithmetic may turn ugly, or both.

Of course, I may have missed something obvious in what j-astron was saying. Getting coffee and then going to work on this problem.
 
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Gabrosin.
1)There cannot be an exact answer to your problem. To solve the equations set up we will need to find fifth roots giving irrational answers. We therefore cannot find integers for the number of cards of each suit.
2) Because you have set the probability of each suit independently our deck will contain cards not included in the four suits.
3) Because the portion of spades is so low the deck will contain thousands of cards.

If you accept an approximate answer you can "short cut" the calculation. I have a solution with probabilities very close to those you specify. Meanwhile follow the advice given by others but look for the short cut to avoid complicated expressions requiring advanced techniques or estimating to evaluate.
 
Gabrosin.
1)There cannot be an exact answer to your problem. To solve the equations set up we will need to find fifth roots giving irrational answers. We therefore cannot find integers for the number of cards of each suit.
2) Because you have set the probability of each suit independently our deck will contain cards not included in the four suits.
3) Because the portion of spades is so low the deck will contain thousands of cards.

If you accept an approximate answer you can "short cut" the calculation. I have a solution with probabilities very close to those you specify. Meanwhile follow the advice given by others but look for the short cut to avoid complicated expressions requiring advanced techniques or estimating to evaluate.
Jonathan

You seem have got a lot further than I have, at least so far.

With respect to your first point, a quintic can have integer roots so such a problem may be solvable in integers. Whether this one is, I don't know.

With respect to your second point, is it necessarily true that the probabilities are mutually inconsistent in the absence of a fifth suit? If they are consistent and there is no fifth suit, I suspect we may be in the situation of having four unknowns and only three independent equations, which will entail an infinite number of solutions..
 
"a quintic can have integer roots". Maybe But I'll be surprised if anyone finds four (or five) integer solutions for this problem.
"is it
necessarily true that the probabilities are mutually inconsistent". The whole point of this is to find the proportion of each suit in the deck. If these add to less than unity there must be unsuited cards to make up the difference.

"I suspect we may be in the situation of having four unknowns and only three independent equations" Not with the way I tackled it.
Should I post my numeric solution (with no working)?
 
we will need to find fifth roots giving irrational answers. We therefore cannot find integers f

This is simply wrong: \(\displaystyle x^5 = 32\) has an integer solution.

Maybe But I'll be surprised if anyone finds four (or five) integer solutions for this problem.


A form of proof previously unknown to me. Because you have not given any solution to the original problem, you cannot know whether there is an integer solution to it. The fact that you have found an integer solution to a different but similar problem suggests that some problems of this type do have integer solutions. What criterion distinguishes those that do and those that do not.


is it necessarily true that the probabilities are mutually inconsistent". The whole point of this is to find the proportion of each suit in the deck. If these add to less than unity there must be unsuited cards to make up the difference.

And the proof that the suit proportions in the original problem necessarily add to less than unity is? This would require finding all answers to the original problem. If it is a quintic, there must be at one real answer but could be five. If we find one, we can find the other four.

"I suspect we may be in the situation of having four unknowns and only three independent equations" Not with the way I tackled it.

If the proportions necessarily add to unity and there are only four suits, then there cannot be four independent equations.​
 
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JeffM You're going round in circles. I'll leave this thread for you to provide an answer.
 
What does / mean?

Sorry, I should have phrased it as "the probability of drawing a spade". I want to calculate each of the four probabilities, not some division-operation involving them.


j-astron said:
What's easier is to recognize that drawing at least one heart is the exact opposite outcome of drawing NO hearts. These two events are mutually-exclusive, and they are the only two possible outcomes. Either you drew some hearts (one or more) or you drew no hearts (zero).

Thank you. I've used the "look at the inverse probability" before and I don't know why it didn't occur to me to do it here. So I can look at the probability of not drawing a spade, then take the fifth root of that to determine the probability of not drawing a spade in a single draw, then flip that again to get my answer. Not as painful as I thought it might be!

Jonathan said:
"a quintic can have integer roots". Maybe But I'll be surprised if anyone finds four (or five) integer solutions for this problem.

The genesis of the problem is reverse-engineering the probability of a digital event and the actual percentages are non-integer, so having a non-integer solution for this is fine by me.

JeffM said:
I am a little confused by the term "independent draws." If we draw one card, note its suit, replace it, and repeat that process four more times, we would clearly have independent draws. But that is not an equivalent to drawing five cards without replacement.


I believe these are independent events, which would be equivalent to drawing the card and then replacing it in the deck before drawing again, so that's how I aimed to phrase it.

Thank you all for the discussion!

 
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