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# Cards, Suits, and Basic Probability

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FastLearner
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Joined: 12/31/1969

My math skills are only so-so, and for some reason I can't seem to wrap my head around this fairly simple-seeming math problem.

I have a deck of cards in 5 suits. For each suit there is one of each number card. For example if there were 9 cards in each suit then there'd be a 1 of suit A, a 2 of suit A, etc. through a 9 of suit A. Then there'd be a 1 of suit B, etc. Like normal cards, basically, but with five suits and no face cards.

The question I have is this: How many numbered cards need there be of each suit so that in any given hand of 7 random cards from the deck you'd have an equal chance of being dealt a 3-card set* or a 3-card run** (or for that matter a 4-card set or a 4-card run)? I want the odds of a set and run in a random set of 7 cards to be identical. How many of each suit need there be? Is it possible for them to be equal? How the heck would I figure it out?

Thanks in advance, math gods/goddesses.

(* A set is cards in different suits with the same number, like three of a kind in poker.)
(** A run in cards is cards in a series within the same suit, like a small straight flush in poker.)

phpbbadmin
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Joined: 04/23/2013
Let me take a whack at it.

Well. On second hand, Let me not. I know how to calculate the odds of drawing them based upon just three cards, but not 3 out of 7. Instead visit this page, it will probably help:

http://www.mathwizz.com/statistics/help/help3.htm

Good luck!
-Darke

FastLearner
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Joined: 12/31/1969
Cards, Suits, and Basic Probability

Awesome, thanks, I'll take a look now.

I realized after sleeping that there may be no way for them to come out even, but I still need to figure out the odds of a set or a run for a given hand size and given number of cards and suits. At this point my ideal deck is 1 through 9 of five suits... how can I figure the odds? How about if there were, for example, two each of 1 through 5 in five suits?

My brain can't seem to figure out how to break the problem down without simulation (which I could do but seems like cheating).

Oracle
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Joined: 06/22/2010
Re: Cards, Suits, and Basic Probability

My degree is in math. I can almost feel the cobwebs lifting, and I get a chance to talk math-speak again :).

The probability of getting getting a hand is the number of ways to get that hand divided by the total number of possible hands.

Since the total number of hands is constant for a fixed deck, we can ignore the probability of getting a certain hand and just deal with the number of ways to get that hand.

Let the total number of ranks be x.

To get a 3 of a kind or better (I'm assuming a 4 of a kind counts as a 3 of a kind), we have xC1 choices for the rank, 5C3 choices for the specific suits in that rank, and (5x-3)C4 for the other 4 cards. This is:

xC1 * 5C3 * (5x-3)C4 (1)

To ge a run of 3 or better, we have 5C1 choices for the suit, (x-2) choices for the low card in the run, (5x-3)C4 choices for the other 4 cards. This is:

5C1 * (x-2) * (5x-3)C4 (2)

Let (1) and (2) be equal so there will be the same number of ways to get each hand.

xC1 * 5C3 * (5x-3)C4 = 5C1 * (x-2) * (5x-3)C4

Solve for x.

xC1 * 5C3 = 5C1 * (x-2)
x * 10 = 5 * (x-2)
10x = 5x - 10
5x = -10
x = -2

So unless I made an arithmetic error, there is no positive number of ranks that will let the probabilities be the same.

Jason

Time to get my d6 :)

FastLearner
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Joined: 12/31/1969
Cards, Suits, and Basic Probability

Excellent, thanks! I'm working on a little script to do the calculations for various entries and this will help a lot.

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