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M2theK's Achievements


Lepton (1/13)



  1. Thanks @mathematic and no worries. I also figured it out a little bit different, but both ways are redundant, just the notation is a bit different. My lack of knowing binomial coefficients was the problem here, since I didn't understand your approach first, but no its clear. 48 - 3 = 45 # Number of non-aces in the deck ((n),(k)) binomial coefficient (n over k) ((45),(3)) # Number of possibilities to draw no ace (ignoring order!) divided by the total number of ways to draw 3 cards ((48),(3)) is (((45),(3))) / (((48),(3))) # Probability of drawing no ace P = 1 - (((45),(3))) / (((48),(3))) = 17.96% # probability of success The universal equation I was looking for is therefore: P = 1 - (((D-T),(H))) / (((D),(H)))
  2. I am not sure why I am asking the impossible when D=48, T=3, H=3? There must an equation to correctly calculate the probability that can be universalized to whatever value I choose for T and H as long as T > 0; H > 0; D ≥ T, D ≥ H. If I violate these assumptions, no equation can, of course, calculate the impossible. But it's like saying x/y is not legit, because in case of y=0 there is no result. I just would like to know a way to quickly calculate the odds in other legit cases, for example, D=48, T=12, H=5. But I never said there is one ace left in the deck: D=48, T=3, H=3 with the aces was just an example. I am more interested in more complicated cases such as D=75, T=14, H=3. Hence may assumptions that there must be a universal way to calculate these probabilities.
  3. When the deck has 48 cards and 3 aces are left, the chance of randomly drawing an ace with one card draw is 3/48 = 1/16 = 6.25%. Why then is the result in both cases (H=1 or 3) identical? I don't understand! There should be a difference in the probability...
  4. 1/16 = 6.25% is the chance when drawing one card, true! But H (number of cards drawn next turn) = 3. Hence, the probability must be larger (~ 17%). According to the HTML source code that uses the equation posted above [1], the probability is 17.95%. Btw: In your example, it also correctly calculates the chance of 6.25% (D=48, T=3, H=1). The problem is that, as @Bignose correctly mentioned, the equation should not work, because the term ((D-T-H)!) gets negative in case of D=30, T=30, H=3. But still, the source code correctly calculates the chance = 100% using the mentioned formula! Hence, I would like to know (1) why this is possible (see source code), and (2) what would be the correct universal equation, independent of the number input for D, T, and H, as long as the pre-conditions as described above are considered (T > 0; H > 0; D ≥ T, D ≥ H)? In other words, I am looking for an equation, where I can put any kind of numbers in, as long as I adhere to the pre-conditions.
  5. I am aware that the equation is incorrect, hence my question: what is the correct equation to calculate the probabilities?
  6. I want to convert a stochastic formula into a proper equation describing the probability of a specific card draw in a card game. Example: A and B are people playing cards. They each have a 52 card deck and four cards in hand. A has an ace, a ten, and 2 fours. If A draws 3 more cards from his deck, and he wants to draw one more ace, what is the probability? That is, there are: 48 cards in the deck (52 card deck, minus the 4 cards in A's hand); 3 target cards (4 aces in the deck, minus the one in A's hand); and 3 cards to be drawn. [1] The original formula: probability=1−((D−T)!−(D−T−H)!)(D!−(D−H)!−1))=17.96%, where: D is the current deck size, T is the number of target cards in the deck, and H is the number of cards to be drawn in the next turn. An attempt to notate a proper equation, see attachment. However, the original equation seems to be notationally incorrect, since the term ((D-T-H)!) can become negative in case of: D = 30, T = 30, H > 0, but factorials are only defined for non-negative integer numbers. Further conditionals are: T > 0; H > 0; D ≥ T, D ≥ H. Question 1: What is the correctly denoted equation to prevent the term ((D-T-H)!) from becoming negative? Any solutions to this problem? Thanks in advance! Source: The equation was derived from a HTML source code provided by Scott Gray (see the full HTML source code at (http://www.unseelie.org/srccgi/ScottsGamingCgi.pdf, p. 2-3). Note: The equation in HTML source code actually works in the above-described case (see also http://www.unseelie.org/cgi-bin/cardco.cgi?deck=30&target=30&hand=3), but is obviously not functional as denoted in [1].
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