Probability of Drawing A Given Card?
The Blind Eternities forum
Posted on Nov. 9, 2022, 9:32 a.m. by Delphen7
Warning. It's somewhat long.
This is a question that I was confronted with a few days ago, and it's been actually really bothering me.
To begin, let's say you have a 60 card deck filled with unique cards, are drawing 30 of them, and want to know your chances of drawing a given card.
Hypergeometric Calculators (And simple probability) state that since you are drawing 30/60 (or 1/2) of your deck, you have a 50% of drawing your given card.
(I had a friend who's into gambling bring this up, and this is where I got confused) However, say that you draw these cards one at a time. The first time, you have a 1/60 chance of drawing your given card. If that's not it, you now have a 1/59 chance, and so on and so forth until you draw it.
Say that the card is 30th from the top (So the last one you'd draw), but we don't know that. The simple probability tells that we'd have a 50%, but the gambler's chance tells us we'd have a 1/60, then 1/59... then a 1/30.
I asked a statistician friend of mine, and he said that he assumed that somehow these two different methods would have to equal (Because the simple is correct, but the gambler's appears correct too), but didn't actually know how to get them to equal.
Hence the thread. Is it a logical fallacy on the gambler's/my part, or is there some way to get the two probabilities to equal?
Thanks!
sergiodelrio says... #3
The first time, you have a 1/60 chance of drawing your given card. If that's not it, you now have a 1/59 chance, and so on and so forth until you draw it.
This is false. Spoiler: the two methods line up mathematically and give the same result. Here's how:
Let's simplify the problem, cause I'm a lazy person. The problem is now one card out of a deck of four cards. One method is to look at the top 2, but I'm not getting into that since we already agree this method gives the desired outcome.
Now, when you're looking at one card at a time, the chance for the first card are 1/4. Easy. The second card is trickier, since we must account for also the first card in the case that WASN'T our card. If we didn_t do that we'd just pretend to be using a three-card-deck, but we're not. This is a four card deck!
So instead of adding 1/4+1/3 and notice it's not 0.5, we do the real math:
1/4 + (1/3)*(3/4) = 0.5
Since our odds of drawing the card on the first draw are 1/4, the odds of NOT DRAWING IT are 3/4, and we need to adjust our second draw by this factor. If our deck was bigger, we'd have to adjust more draws accordingly.
November 9, 2022 10:40 a.m.
TheOfficialCreator says... #4
Yep. The probability additive property excludes the event P(A and B) for all P(A or B).
November 9, 2022 10:43 a.m.
sergiodelrio That is an excellent explanation, thanks for the response!
TheOfficialCreator That makes sense.
TheOfficialCreator says... #2
It appears that you are both correct.
The above link seems to explain the innate probability fairly well, since using complex probability equations in this case yields the same result as your basic intuition of the situation.
November 9, 2022 9:42 a.m.