How to calculate probability of being dealt certain cards?
July 9, 2008 5:44 PM Subscribe
You're dealt 8 cards face down from a 52 card deck. All values are determined by the number on the card, not the suit. What is the probability of each card being dealt?
The first card you're dealt has a 4/52 chance of being any of the 13 values (2 thru Ace). The 2nd card you're dealt has a 4/51 chance of being any of the remaining 12 values, plus a 3/51 chance of being the same value as the first card.
This is where my retard brain ceases to work. How do I know the odds of the 3rd card, given that the 2nd card may or may not be the same as the 1st card? If the 2nd card has a new value, the third card is 4/50 of the 11 remaining cards, plus 3/50 it's the same value as the 1st card and 3/50 it's the same as the 2nd card. But if the 2nd card is a repeat of the 1st value, there is a 4/50 chance of it being one of the 12 remaining values and a 2/50 chance of it being the same as the 1st and 2nd card.
So you see the dilemma I have calculating the odds for the 3rd card to be dealt until the 8th. I know I'm probably looking at this entirely the wrong way but I've always been a math retard beyond basic arithmetic. Anyone care to explain how this works?
The first card you're dealt has a 4/52 chance of being any of the 13 values (2 thru Ace). The 2nd card you're dealt has a 4/51 chance of being any of the remaining 12 values, plus a 3/51 chance of being the same value as the first card.
This is where my retard brain ceases to work. How do I know the odds of the 3rd card, given that the 2nd card may or may not be the same as the 1st card? If the 2nd card has a new value, the third card is 4/50 of the 11 remaining cards, plus 3/50 it's the same value as the 1st card and 3/50 it's the same as the 2nd card. But if the 2nd card is a repeat of the 1st value, there is a 4/50 chance of it being one of the 12 remaining values and a 2/50 chance of it being the same as the 1st and 2nd card.
So you see the dilemma I have calculating the odds for the 3rd card to be dealt until the 8th. I know I'm probably looking at this entirely the wrong way but I've always been a math retard beyond basic arithmetic. Anyone care to explain how this works?
Are you trying to figure out the total number of eight card sets from a deck of 52 cards, where the suit of the card is disregarded? (in other words, a hand of 2-9 all hearts is treated identically to 2-9 all spades?)
posted by jenkinsEar at 5:54 PM on July 9, 2008
posted by jenkinsEar at 5:54 PM on July 9, 2008
By the way, that's a good general strategy for figuring out probabilities - if you're having trouble envisioning how to calculate the chance of something happening, try to instead envision how to calculate the chance of it not happening. It might be a significantly simpler calculation.
posted by Flunkie at 5:58 PM on July 9, 2008
posted by Flunkie at 5:58 PM on July 9, 2008
What exactly do you want to know the probability of? The chance of a specific number showing up a given number of times? The chance of any given combination of 8 numbers? Does the order of the cards matter?
posted by thrako at 5:58 PM on July 9, 2008
posted by thrako at 5:58 PM on July 9, 2008
What, exactly, do you want to calculate the probability of? Are you trying to determine the probability of each and every possible situation? If so, these are the cases you need to go through:
posted by mr_roboto at 6:00 PM on July 9, 2008
I think that's it.... no guarantees though. You need to calculate the probability for each of these cases separately.
posted by mr_roboto at 6:00 PM on July 9, 2008
Look up something called conditional probability. It is exactly what you use to calculate things like this. Sorry but I don't have time to write it all down here.
posted by number9dream at 6:06 PM on July 9, 2008
posted by number9dream at 6:06 PM on July 9, 2008
What is the probability of each card being dealt?
You seem to mean: before we start dealing cards, how can I calculate the probability that the nth card is of value v. And you could calculate that by weighting and summing the likelihood of each possible series of cards up to card n - 1 and the resulting available cards. But before you start dealing cards, all the possible orders of the cards average out, and the chance that card n is of some selected value v is just 1/13.
The analogy is to Feynman's summing of the possible paths that a particle might take from a lamp to your eye: in reality, it takes them all at once (hence the double-slit experiment) but in effect once you've summed up the paths the result is that it takes the same path that a probabilistic wave function would have selected for it.
posted by nicwolff at 6:28 PM on July 9, 2008
You seem to mean: before we start dealing cards, how can I calculate the probability that the nth card is of value v. And you could calculate that by weighting and summing the likelihood of each possible series of cards up to card n - 1 and the resulting available cards. But before you start dealing cards, all the possible orders of the cards average out, and the chance that card n is of some selected value v is just 1/13.
The analogy is to Feynman's summing of the possible paths that a particle might take from a lamp to your eye: in reality, it takes them all at once (hence the double-slit experiment) but in effect once you've summed up the paths the result is that it takes the same path that a probabilistic wave function would have selected for it.
posted by nicwolff at 6:28 PM on July 9, 2008
What is the probability of each card being dealt?
The OP really needs to clarify but I don't think the question is what the odds are in any position, 3rd, 8th or whatever, but, what are the odds that any particular value will be among the 8 cards dealt. I think that's the question Flunkie answered.
posted by beagle at 6:45 PM on July 9, 2008
The OP really needs to clarify but I don't think the question is what the odds are in any position, 3rd, 8th or whatever, but, what are the odds that any particular value will be among the 8 cards dealt. I think that's the question Flunkie answered.
posted by beagle at 6:45 PM on July 9, 2008
As others have noted, you can't just have a probability, you can only have a probability for a well-defined event (which can itself be a combination of other well-defined events).
To get at a specific sub-part, you write:
This is where my retard brain ceases to work. How do I know the odds of the 3rd card, given that the 2nd card may or may not be the same as the 1st card?
You'd do this with a probability tree. The contingency you notice there, that the odds in the third draw depend on what has gone before, is inescapable.
You'd start with an initial node, out of which come 13 branches, one for each value. The probability of each branch is 1/13.
Then, at the end of the topmost branch, you'd draw another 13 branches representing the second draw. Except this time, the probability of another 1 is 3/51, and the probability of all the others is 4/51.
Then you'd do this again for each of the other twelve first-stage branches.
Then you'd go to the new topmost branch, that corresponds to a 1 and another 1, and draw another 13 branches coming out of *that*. Now, the probability of a third 1 is only 2/50, and the probability of each of the other twelve third-stage branches is 4/50.
What's the probability of three ones in a row like that? 4/52*3/51*2/50=1 in 5525.
Then you'd repeat the exercise for the other 12*13=156 other second-stage branches.
At the end, you'd have a tree with eight layers of branches and 815,730,721 different possible sequences of numbers. For each of these 815 million end nodes, you find the final probability by multiplying through the probabilities at each branch. These final probabilities will vary from end node to end node.
From there, you can define events. Say, "sums to ten." You could do this by looking at each of the 815 million sequences and taking their sum, noting which sum to ten and which don't. Then the probability of having a sequence that sums to ten is just the sum total of all of the end nodes that fit the "sum to ten" event.
posted by ROU_Xenophobe at 8:46 PM on July 9, 2008
To get at a specific sub-part, you write:
This is where my retard brain ceases to work. How do I know the odds of the 3rd card, given that the 2nd card may or may not be the same as the 1st card?
You'd do this with a probability tree. The contingency you notice there, that the odds in the third draw depend on what has gone before, is inescapable.
You'd start with an initial node, out of which come 13 branches, one for each value. The probability of each branch is 1/13.
Then, at the end of the topmost branch, you'd draw another 13 branches representing the second draw. Except this time, the probability of another 1 is 3/51, and the probability of all the others is 4/51.
Then you'd do this again for each of the other twelve first-stage branches.
Then you'd go to the new topmost branch, that corresponds to a 1 and another 1, and draw another 13 branches coming out of *that*. Now, the probability of a third 1 is only 2/50, and the probability of each of the other twelve third-stage branches is 4/50.
What's the probability of three ones in a row like that? 4/52*3/51*2/50=1 in 5525.
Then you'd repeat the exercise for the other 12*13=156 other second-stage branches.
At the end, you'd have a tree with eight layers of branches and 815,730,721 different possible sequences of numbers. For each of these 815 million end nodes, you find the final probability by multiplying through the probabilities at each branch. These final probabilities will vary from end node to end node.
From there, you can define events. Say, "sums to ten." You could do this by looking at each of the 815 million sequences and taking their sum, noting which sum to ten and which don't. Then the probability of having a sequence that sums to ten is just the sum total of all of the end nodes that fit the "sum to ten" event.
posted by ROU_Xenophobe at 8:46 PM on July 9, 2008
The 2nd card you're dealt has a 4/51 chance of being any of the remaining 12 values
What? It has a 4/51 of being each of the remaining values, but 48/51 of being any of them, no?
posted by juv3nal at 9:46 PM on July 9, 2008
What? It has a 4/51 of being each of the remaining values, but 48/51 of being any of them, no?
posted by juv3nal at 9:46 PM on July 9, 2008
Response by poster: I'm looking for the probability that any card value (say a 4), will be repeated in the 8 cards that are drawn. What are the odds of 2 4s occurring in 8 cards? How about 3 4s and 4 4s? I think I have enough advice here to at least get me started.
posted by b_thinky at 11:37 PM on July 9, 2008
posted by b_thinky at 11:37 PM on July 9, 2008
The derivation section of the Wiki page on Poker probability has some useful information that can be easily adapted to this problem. Keep in mind that the formulas they quote exclude higher hands. So, for example, the formula for 3-of-a-kind does not count all the full-houses. This is easy to fix though. First, some notation:
N = 52 is the number of cards in the deck.
n = 8 (in your example) is the number of cards dealt.
k is how many times a rank repeats.
s = 4 is the number of suits.
(a,b) := a choose b = a!/b!(a-b)! is the binomial coefficient of x^b in (1+x)^a.
We must pick k cards from s total of that rank, and then distribute n-k cards from the N-s cards of other ranks. The total number of hands is (N,n). Thus,
P(k) = (s,k)*(N-s,n-k) / (N,n),
or, for your specific example:
(4,k)*(48,8-k) / (52,8).
Note that this counts any and all k-of-a-kind hands, including those which may have multiple cards of another hand.
posted by dsword at 11:36 AM on July 11, 2008
N = 52 is the number of cards in the deck.
n = 8 (in your example) is the number of cards dealt.
k is how many times a rank repeats.
s = 4 is the number of suits.
(a,b) := a choose b = a!/b!(a-b)! is the binomial coefficient of x^b in (1+x)^a.
We must pick k cards from s total of that rank, and then distribute n-k cards from the N-s cards of other ranks. The total number of hands is (N,n). Thus,
P(k) = (s,k)*(N-s,n-k) / (N,n),
or, for your specific example:
(4,k)*(48,8-k) / (52,8).
Note that this counts any and all k-of-a-kind hands, including those which may have multiple cards of another hand.
posted by dsword at 11:36 AM on July 11, 2008
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In general, the chance of something happening is one minus the chance of it not happening.
So what's the chance of none of the eight cards being a three?
It's the chance of drawing a card and it not being a three - i.e. 48 out of 52 - times the chance of drawing a second card (without putting the first back) and it not being a three - i.e. 47 out of 51 - times the chance of drawing a third that's not a three - i.e. 46 out of 50 - times ... times the chance of drawing an eighth and it not being a three - i.e. 41 out of 45.
That is:
(48/52) * (47/51) * (46/50) * (45/49) * (44/48) * (43/47) * (42/46) * (41/45)
Which is about 50.1%.
Which means the chance of at least one card being (for example) a three is one minus that - i.e. about 49.9%.
If this isn't the question you're interested in, please clarify.
posted by Flunkie at 5:54 PM on July 9, 2008