 My Math Forum Correctly denote card draw probability equation
 User Name Remember Me? Password

 Probability and Statistics Basic Probability and Statistics Math Forum

October 24th, 2018, 04:07 AM   #1
Newbie

Joined: Oct 2018
From: Germany

Posts: 4
Thanks: 0

Correctly denote card draw probability equation

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.

 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?

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 Perl Card Calculator Page!), but is obviously not functional as denoted in .

Any solutions to this problem? Thanks in advance!
Attached Images Snap_2018.10.19_17h32m38s_001.jpg (14.2 KB, 2 views) October 24th, 2018, 10:03 AM   #2
Senior Member

Joined: May 2016
From: USA

Posts: 1,310
Thanks: 551

Quote:
 Originally Posted by M2theK 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.  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? 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 Perl Card Calculator Page!), but is obviously not functional as denoted in . Any solutions to this problem? Thanks in advance!
$P(1\ ace) = \dfrac{\dbinom{45}{2} * \dbinom{3}{1}}{\dbinom{48}{3}} = \dfrac{45!}{2! * 43!} * \dfrac{3!}{1! * 2!} * \dfrac{3! * 45!}{48!} = \\ \dfrac{45 * 44}{2} * \dfrac{3}{1} * \dfrac{3 * 2}{48 * 47 * 46} = \dfrac{45 * 44 * 3 * 3 * 2}{2 * 48 * 47 * 46} \approx 17.17\%.$

Does that answer make sense? Well, we could compute it a different way.

$P(1\ ace) = \\ \left( \dfrac{3}{48} * \dfrac{45}{47} * \dfrac{44}{46} \right ) + \left( \dfrac{45}{48} * \dfrac{3}{47} * \dfrac{44}{46} \right ) + \left( \dfrac{45}{48} * \dfrac{44}{47} * \dfrac{3}{46} \right ) =\\ \dfrac{3 * 45 * 44 + 45 * 3 * 44 + 45 * 44 * 3}{48 * 47 * 46} =\\ \dfrac{3(45 * 44 * 3)}{48 * 47 * 46} \approx 17.17\%.$

Here is another way that we could check. Using the first method we get:

$P(no\ aces) = 82.04\% \\ P(1\ ace) = 17.17\% \\ P(2\ aces) = 0.78\% \\ P(3\ aces) = 0.01\%$

$82.04\% + 17.17\% + 0.78\% + 0.01\% = 100.00\%.$

And we know that sum must be right. October 27th, 2018, 01:45 AM #3 Newbie   Joined: Oct 2018 From: Germany Posts: 4 Thanks: 0 That looks great, thx a lot! Especially the control is very convincing. But honestly, I am still a bit confused. 1. The discrepancy between the 17.17% (your calculation) and 17.96% (script) is not huge, but still... I don't understand why the script is giving the wrong result, even though it is always accurate when I put numbers in that can be easily checked with mental arithmetic (e.g., D=48, T=3, H=1 -> 3/48 -> 1/16 -> 6.25%) 2. My school math is a bit rusty, to say the least, and I can't entirely follow your equations. Would you mind putting the equation(s) in a form that is denoted with D, T, and H instead of the number example? I am not sure which 3 is used (T or H) in the equation. Thanks again! October 27th, 2018, 01:53 AM   #4
Newbie

Joined: Oct 2018
From: Germany

Posts: 4
Thanks: 0

Quote:
 Originally Posted by JeffM1 $P(1\ ace) = \dfrac{\dbinom{45}{2} * \dbinom{3}{1}}{\dbinom{48}{3}} = \dfrac{45!}{2! * 43!} * \dfrac{3!}{1! * 2!} * \dfrac{3! * 45!}{48!} = \\ \dfrac{45 * 44}{2} * \dfrac{3}{1} * \dfrac{3 * 2}{48 * 47 * 46} = \dfrac{45 * 44 * 3 * 3 * 2}{2 * 48 * 47 * 46} \approx 17.17\%.$ Does that answer make sense? Well, we could compute it a different way. $P(1\ ace) = \\ \left( \dfrac{3}{48} * \dfrac{45}{47} * \dfrac{44}{46} \right ) + \left( \dfrac{45}{48} * \dfrac{3}{47} * \dfrac{44}{46} \right ) + \left( \dfrac{45}{48} * \dfrac{44}{47} * \dfrac{3}{46} \right ) =\\ \dfrac{3 * 45 * 44 + 45 * 3 * 44 + 45 * 44 * 3}{48 * 47 * 46} =\\ \dfrac{3(45 * 44 * 3)}{48 * 47 * 46} \approx 17.17\%.$ Here is another way that we could check. Using the first method we get: $P(no\ aces) = 82.04\% \\ P(1\ ace) = 17.17\% \\ P(2\ aces) = 0.78\% \\ P(3\ aces) = 0.01\%$ $82.04\% + 17.17\% + 0.78\% + 0.01\% = 100.00\%.$ And we know that sum must be right.

That looks great, thx a lot! Especially the control is very convincing.

But honestly, I am still a bit confused.

1. The discrepancy between the 17.17% (your calculation) and 17.96% (script) is not huge, but still... I don't understand why the script is giving the wrong result, even though it is always accurate when I put numbers in that can be easily checked with mental arithmetic (e.g., D=48, T=3, H=1 -> 3/48 -> 1/16 -> 6.25%)

2. My school math is a bit rusty, to say the least, and I can't entirely follow your equations. Would you mind putting the equation(s) in a form that is denoted with D, T, and H instead of the number example? I am not sure which 3 is used (T or H) in the equation.

Thanks again! October 31st, 2018, 03:27 AM #5 Newbie   Joined: Oct 2018 From: Germany Posts: 4 Thanks: 0 Thanks for your efforts @JeffM1 I am not sure about your way of calculating the probability, because even though your control looks pretty convincing, the result of 17.17% is not correct. The expected value is 17.96%. I assume I misexplained the example or there is something else I don't understand. Maybe it's related to the ordering (ignoring/ not ignoring) of the binomial coefficients? I would appreciate some feedback, so I know where my mistake is. It looks a bit weird because I don't know how to insert an equation with MathJax. My solution is as follows: 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))) Last edited by skipjack; November 7th, 2018 at 08:33 PM. November 12th, 2018, 05:10 PM   #6
Senior Member

Joined: May 2016
From: USA

Posts: 1,310
Thanks: 551

Quote:
 Originally Posted by M2theK Thanks for your efforts @JeffM1 I am not sure about your way of calculating the probability, because even though your control looks pretty convincing, the result of 17.17% is not correct. The expected value is 17.96%. I assume I misexplained the example or there is something else I don't understand. Maybe it's related to the ordering (ignoring/ not ignoring) of the binomial coefficients? I would appreciate some feedback, so I know where my mistake is. It looks a bit weird because I don't know how to insert an equation with MathJax. My solution is as follows: 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
That is correct. Let's calculate that.

$\dfrac{\dbinom{45}{3}}{\dbinom{48}{3}} = \dfrac{\dfrac{45!}{3! * 42!}}{\dfrac{48!}{3! * 45!}} = \dfrac{45 * 44 * 43}{3!} * \dfrac{3!}{48 * 47 * 46} \approx 82.04\%.$

Notice that I calculated that in my first post as the probability of no aces.

Notice as well that I calculated the probabilities of drawing exactly one ace or of exactly two aces or of exactly three aces. If you add those up, you get the probability of drawing one or more aces. Another way to say the same thing is the probability of getting at least one ace.

$17.17\% + 0.78\% + 0.01\% = 17.96\%.$

And that is equal to $100\% - 82.04\% = 17.96\%.$

In other words, the probability of drawing at least one more ace is 17.96%. But the probability of drawing exactly one more ace is 17.17%.

Quote:
 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)))
That is the probability of drawing at least one more ace. BUT THAT IS NOT WHAT YOU ASKED in your first post.

Last edited by JeffM1; November 12th, 2018 at 05:14 PM. Tags card, correctly, denote, draw, equation, probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post cbenson4 Probability and Statistics 1 March 5th, 2015 11:56 AM jimbobob Probability and Statistics 3 October 22nd, 2014 08:37 AM xtyz Probability and Statistics 1 October 16th, 2012 12:34 AM Erimess Probability and Statistics 4 May 10th, 2011 09:14 PM Niko Bellic Probability and Statistics 1 March 18th, 2010 01:15 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      