hey, i do not know this math...

[quesytion]

hey, i do not know this math... so probabilities may actually be the wrong section, but it seems to me the best place for me to put the problem to be read.

simple back-ground, a library of cards, that contains 60 cards, some cards are identical, and for example come in sets of 4

how do you calculate odds.. for cards... like.. if you have gone through 10 of 60 cards...
what are the odds, that one card is in those 10... (for a card that occurs four times in sixty cards).




if the question doesnt make sense, throw me some questions...

 

[Steven G]

If you choose 10 cards out of 60 and you want to know the chance that one card is in the 10, then it is certainly true. In fact, there are 10 cards in the set of the 10 cards.

 

[Dr.Peterson]

hey, i do not know this math... so probabilities may actually be the wrong section, but it seems to me the best place for me to put the problem to be read.

simple back-ground, a library of cards, that contains 60 cards, some cards are identical, and for example come in sets of 4

how do you calculate odds.. for cards... like.. if you have gone through 10 of 60 cards...
what are the odds, that one card is in those 10... (for a card that occurs four times in sixty cards).

if the question doesnt make sense, throw me some questions...

Here is my interpretation of what you probably mean:

You have 60 cards. Of those, 4 are the particular type you are interested in (like looking for one of the four aces in a standard deck). You want the probability that at least one of the 10 you pick is that type.

We can ignore which cards are identical, and imagine you write a number from 1 to 60 on each card, and the 4 special ones are 57, 58, 59 , and 60.

There are C(60,10) ways to choose 10 cards. There are C(56,10) ways to choose 10 cards that don't include any special cards. So the probability that you pick none of the specials is C(56,10)/C(60,10) = 0.4723; and the probability that you pick at least one is therefore 1 - 0.4723 = 0.5277

 

[quesytion]

[QUOTE="You have 60 cards. Of those, 4 are the particular type you are interested in (like looking for one of the four aces in a standard deck). You want the probability that at least one of the 10 you pick is that type."[/QUOTE]

yes, the way the numbers are related to each other is described as my situation

 

[quesytion]

so a 53% chance... half of the games i play, i will draw one of four cards, in my first 10 cards of play. (opening hand 7 cards, first 3 turns 3 cards)

i wonder how to convert those brackets C(52,10)/C(60,10) - Google Search to a percentage... this is the case, where i have one of eight cards, in my first 10 cards of play....

 

[Dr.Peterson]

so a 53% chance... half of the games i play, i will draw one of four cards, in my first 10 cards of play. (opening hand 7 cards, first 3 turns 3 cards)

i wonder how to convert those brackets C(52,10)/C(60,10) - Google Search to a percentage... this is the case, where i have one of eight cards, in my first 10 cards of play....

I think you're saying, not unexpectedly, that you are unfamiliar with the notation I used, C(52,10); it's also written as \(_{52}C_{10}\) or \({52}\choose{10}\), and is called "combinations". You can read about it here: https://www.mathsisfun.com/combinatorics/combinations-permutations.html