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TheFallen Master of Innominate Surquedry
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Posted: Wed Jan 20, 2021 11:33 am Post subject: Dumbassed stats question 


As may be known, I am absolutely no mathematician.
Simple question. Say I am playing a four player trickbased game of cards, using a standard 52 card deck. I am playing with a partner sat opposite me against two opponents, also sat opposite each other.
Immediately after the deal, I see the jack of spades as the highest spade card in my 13 card hand.
Statistically speaking, what are the odds of my partner holding any spade card higher than that (i.e. at least one of either the queen, king or ace)?
And how is this calculated? _________________ Newsflash: the word "irony" doesn't mean "a bit like iron"
Shockingly, some people have claimed that I'm egocentric... but hey, enough about them
"If you strike me down, I shall become far stronger than you can possibly imagine."
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* I occasionally post things here because I am a fan of SRD.
This should not be considered as any condoning of the hypocritical, snowflake, bleeding heart, white guiltridden, compulsively virtuesignalling and totalitarian prog Left woke zealotry which some try to force upon this forum. 

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wayfriend whilom witling
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Posted: Wed Jan 20, 2021 1:18 pm Post subject: 


In a loose sense, it's the odds of a person having any 1 of 3 cards GIVEN that you don't have any of those cards. (That is, the odds of one divided by the odds of the other.)
[I thought long and hard about it, and I don't think we have to consider the other players. Any cards that they have can be considered to have been left in the deck for the purposes of this calculation. And I don't think we have to consider that you have a Jack  we only need to take into account that you don't have any of those 3.]
The first is calculated by finding the odds of not having any of the three cards, and then inverting by subtracting from 1.0.
The second is also calculated as the odds of not having any of the three cards.
[ So if those odds are P, the answer is (1P)/P. ]
That much I am certain. The rest is more complicated.
The odds of not drawing one of three cards in seven draws is
49/52 x 48/51 x 47/50 ..... x 43/46.
or (49! x 45!) / (52! x 42!).
Some stuff should cancel out, leaving (45x44x43)/(52x51x50)
which is 0.642. or 64%.
So the odds of someone not having Q/K/A of Spades is about 64%.
The odds of having at least one them are thus 36%.
That seems good to me.
The odds of your partner having them, given that you don't have them, or (1P)/P, is 0.558. or 56%.
Seems wrong. Maybe someone can spot my error. _________________ * I occasionally post things on KevinsWatch because I am a fan of Stephen R. Donaldson; this should not be considered as condonation of the white nationalist propaganda which is posted far too frequently on this website. 

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I'm Murrin Aren't you?
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Posted: Wed Jan 20, 2021 6:10 pm Post subject: 


You can safely eliminate your own entire hand from the calculation, so you'd only be looking at determining the odds of a combination of 13 cards from a 39 card set containing any of those three cards. _________________ Inspiration Struck.
Mod of the General Fantasy/SciFi Forum 

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Rigel Tenno Awoken
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Posted: Wed Jan 20, 2021 6:58 pm Post subject: 


I started to write out the work and got some errors.
I'll have to dig into my math later. _________________ "You make me think Hell is run like a corporation."
"It's the other way around, but yes."
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TheFallen Master of Innominate Surquedry
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Posted: Wed Jan 20, 2021 11:33 pm Post subject: 


Guys I am genuinely grateful for your efforts so far... as well as being mildly encouraged that the problem/calculation isn't ludicrously simple (something which may in part at least excuse my own abject helplessness in the face of this).
I'm sure that Murrin is correct when he mentions that the calculation only has to involve odds relating to 3 in 13 of the remaining 39 cards, because my own hand is already known and thus surely must be excluded in its entirety from any workings?
I'll be interested to see more thoughts.
By the way, my own moron maths answer came out at a 70.3% probability that my partner is holding at least one of those three cards, but I am next to certain that my assumptions are wrong, because the equation I came up with to assess the odds in this particular scenario seems way too simple... hence this heartfelt enquiry. _________________ Newsflash: the word "irony" doesn't mean "a bit like iron"
Shockingly, some people have claimed that I'm egocentric... but hey, enough about them
"If you strike me down, I shall become far stronger than you can possibly imagine."
_______________________________________________
* I occasionally post things here because I am a fan of SRD.
This should not be considered as any condoning of the hypocritical, snowflake, bleeding heart, white guiltridden, compulsively virtuesignalling and totalitarian prog Left woke zealotry which some try to force upon this forum. 

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wayfriend whilom witling
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Posted: Thu Jan 21, 2021 2:40 pm Post subject: 


I'm Murrin wrote:  You can safely eliminate your own entire hand from the calculation, so you'd only be looking at determining the odds of a combination of 13 cards from a 39 card set containing any of those three cards. 
I disagree.
The cards in your hand are not truly random. It is a priori that they do not have the A, K, or Q of Spades. (If they were truly random, there would be a chance that they might have those cards.)
This means that the odds of the partner drawing the Ace of Spades as his first card is 1 in 45, not 1 in 52. Etc. etc.
Quote:  By the way, my own moron maths answer came out at a 70.3% probability that my partner is holding at least one of those three cards 
What made me suspect that this is wrong is that one of your opponents would also have a 70.3% chance. And so would your other opponent.
But now that I think about it  there are three cards, and your partner having one does not preclude an opponent from having one, and your other opponent from having one. So the outcomes are not preclusive  the odds do not have to add up to 1.00.
So maybe this is correct. _________________ * I occasionally post things on KevinsWatch because I am a fan of Stephen R. Donaldson; this should not be considered as condonation of the white nationalist propaganda which is posted far too frequently on this website. 

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TheFallen Master of Innominate Surquedry
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Posted: Thu Jan 21, 2021 3:23 pm Post subject: 


I'll shamelessly share the childish modelling I came up with on this (which may actually be valid  but probably isn't).
I thought this way.
I'm interested in where three specific cards are. Out of 39, because I have 13 cards, and none of those three cards are with me.
There are three potential recipients for each of those cards, namely my partner and my two opponents.
So after all 52 cards have been dealt out, what then are the odds of my partner not having been dealt the Ace of spades? 2 in 3 would seem the obvious answer.
And what then are the odds of my partner not having been dealt the King of spades? Also 2 in 3.
And what then are the odds of my partner not having been dealt the Queen of spades? Again 2 in 3.
So to my maths dunce mind, the odds of my partner not having been dealt any of those three cards are 2/3 * 2/3 * 2/3 = 8/27.
That's a 29.6% probability of my partner not having been dealt any of those three cards, and so a 70.4% probability of him having been dealt at least one of them.
(And as you undoubtedly correctly point out, also a 70.4% probability of each of my opponents having been dealt at least one of them... but does that matter? Maybe not?)
I'm fairly sure this is a fallacious piece of modelling, but if so, would love to know how and where.
Added edit: what concerns me here is that, if there's a 29.6% possibility of my partner not having been dealt any of the three named cards, then there's an equal 29.6% possibility that each of my opponents have not been dealt any of the three named cards.
29.6 * 3 does not equal 100. I can't help feeling it should, as I think you were also considering.
I'm actually staggered that what seems like a very simple question is potentially rather complex. Either that or I am thick. _________________ Newsflash: the word "irony" doesn't mean "a bit like iron"
Shockingly, some people have claimed that I'm egocentric... but hey, enough about them
"If you strike me down, I shall become far stronger than you can possibly imagine."
_______________________________________________
* I occasionally post things here because I am a fan of SRD.
This should not be considered as any condoning of the hypocritical, snowflake, bleeding heart, white guiltridden, compulsively virtuesignalling and totalitarian prog Left woke zealotry which some try to force upon this forum. 

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wayfriend whilom witling
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Posted: Thu Jan 21, 2021 11:13 pm Post subject: 


13 card hand?
I missed that. I assumed 7.
TheFallen wrote:  29.6 * 3 does not equal 100. I can't help feeling it should, as I think you were also considering. 
That alludes to what I was saying earlier. There's no reason that two or even three people could have been dealt one of the three cards. Similarly, there's no reason why one or two people might not have gotten any of them. The possibilities are not exclusive of each other. So the math is harder. _________________ * I occasionally post things on KevinsWatch because I am a fan of Stephen R. Donaldson; this should not be considered as condonation of the white nationalist propaganda which is posted far too frequently on this website. 

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Fist and Faith Magister Vitae
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Posted: Fri Jan 22, 2021 12:20 am Post subject: 


Not looking at anybody's calculations, I'm thinking:
One of three people has the Ace. The odds of it being your partner are 1 in 3.
One of three people has the King. The odds of it being your partner are 1 in 3.
One of three people has the Queen. The odds of it being your partner are 1 in 3.
The odds of your partner (or anybody else) having two of them are 1 in 6. The odds of your partner (or anybody else) having all three are 1 in 9. _________________ We are not required to save the world. We are required to stand up as truly as we can for what we love. SRD
All lies and jest
Still a man hears what he wants to hear
And disregards the rest Paul Simon 

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TheFallen Master of Innominate Surquedry
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Posted: Fri Jan 22, 2021 7:52 am Post subject: 


Aah but Fist... that was not the question asked.
It was "What are the odds of my partner (or anyone else) having at least one of those three cards?"
Plus I think you'll find that the odds of my partner (or anyone else) having all three of the specified cards is actually 1 in 27, not 1 in 9. That piece of maths is a lot simpler to work out.
wf, with your 13 card hand full deal realisation, do you think my 70.4% answer is accurate? _________________ Newsflash: the word "irony" doesn't mean "a bit like iron"
Shockingly, some people have claimed that I'm egocentric... but hey, enough about them
"If you strike me down, I shall become far stronger than you can possibly imagine."
_______________________________________________
* I occasionally post things here because I am a fan of SRD.
This should not be considered as any condoning of the hypocritical, snowflake, bleeding heart, white guiltridden, compulsively virtuesignalling and totalitarian prog Left woke zealotry which some try to force upon this forum. 

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Avatar Immanentizing The Eschaton
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Fist and Faith Magister Vitae
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Posted: Fri Jan 22, 2021 11:15 am Post subject: 


TF, I'm saying the odds of your partner having at least one of those three cards is 1 in 3.
I wasn't sure about the 1 in 6 for two and 1 in 9 for three. I suspect 1 in 9 and 1 in 27 is correct.
Now, what are the odds of having two specific of those three. I'm not even going to try. _________________ We are not required to save the world. We are required to stand up as truly as we can for what we love. SRD
All lies and jest
Still a man hears what he wants to hear
And disregards the rest Paul Simon 

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wayfriend whilom witling
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Posted: Fri Jan 22, 2021 12:47 pm Post subject: 


Fist and Faith wrote:  Not looking at anybody's calculations, I'm thinking:
One of three people has the Ace. The odds of it being your partner are 1 in 3.
One of three people has the King. The odds of it being your partner are 1 in 3.
One of three people has the Queen. The odds of it being your partner are 1 in 3. 
Brilliant. I didn't think of looking at it that way. (Because I was thinking 7.)
Fist and Faith wrote:  The odds of your partner (or anybody else) having two of them are 1 in 6. The odds of your partner (or anybody else) having all three are 1 in 9. 
Ah, no.
I think we can assume that these are orthogonal events. So P(both) = P(one) x P(other).
The odds of having any SPECIFIC two are 1 in 9.
The odds of having all three are 1 in 27.
The odds of having none are 2/3 x 2/3 x 2/3 or 8/27 or 30%.
So the odds of having at least one of them is 70%.
(TF earns the ding.)
The odds of having at at least two (any two) are the odds of having Q&K (a SPECIFIC two) plus Q&A plus K&A, or 1/9 x 3, or 33%.
So:
30% your partner has none.
70% your partner has at least one.
33% your partner has at least two.
04% your partner has all three.
And no, it doesn't add up to 100%, as these events are not all mutually exclusive. E.g. if you have at least two, then you also have at least one.
Now for Double Jeopardy.
33% your partner has the A.
22% your partner has the K but an opponent has the A.
14% your partner has the Q but an opponent has the K and the A. _________________ * I occasionally post things on KevinsWatch because I am a fan of Stephen R. Donaldson; this should not be considered as condonation of the white nationalist propaganda which is posted far too frequently on this website. 

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Damelon Lord
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Posted: Fri Jan 22, 2021 2:07 pm Post subject: 


I've been following this for the last couple days and have spent too much time on it. I'm not good at math, but like Av, I think 70% is too high.
My figuring is:
1/3 = The chance to get 1 of the 3
1/9 = The chance to get 2 of the 3 (1/3 x 1/3)
1/27 = The chance your partner has all 3. (1/3 x 1/3 x1/3)
For the odds then, add 1/3 + 1/9 + 1/27
9/27 + 3/27 + 1/27 = 13/27 or 48.1%
But I am no math whiz, so I'd be happy to be proved wrong. _________________ But the fact that some geniuses were laughed at does not imply that all who are laughed at are geniuses. They laughed at Columbus, they laughed at Fulton, they laughed at the Wright brothers. But they also laughed at Bozo the Clown. ~ Carl Sagan


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TheFallen Master of Innominate Surquedry
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Posted: Fri Jan 22, 2021 4:19 pm Post subject: 


wayfriend wrote:  Fist and Faith wrote:  Not looking at anybody's calculations, I'm thinking:
One of three people has the Ace. The odds of it being your partner are 1 in 3.
One of three people has the King. The odds of it being your partner are 1 in 3.
One of three people has the Queen. The odds of it being your partner are 1 in 3. 
Brilliant. I didn't think of looking at it that way. (Because I was thinking 7.)  Sheesh. Although FnF didn't look at anyone's workings, he just reversed what I'd previously said (namely the odds of my partner NOT having each of the three specified cards were 2/3 in each case)
wayfriend wrote: 
I think we can assume that these are orthogonal events. So P(both) = P(one) x P(other).
The odds of having any SPECIFIC two are 1 in 9.
The odds of having all three are 1 in 27.
The odds of having none are 2/3 x 2/3 x 2/3 or 8/27 or 30%.
So the odds of having at least one of them is 70%.
(TF earns the ding.) 
I'm frankly staggered I got it right. And Av and Damelon, I'm now supremely confident  solely courtesy of wf confirming (because he is dead good at this sort of thing... FFS he even managed to explain the two envelope paradox to a maths dunce like me in a way that made sense)  that the odds are exactly 70.37% (or 19/27, if you want to be pedantic) of my partner having at least one of the three specified cards.
Having said that, I am massively impressed at the more complex stats displayed by wf here to evaluate the odds of other outcomes. For example, I wouldn't have been able to figure out the model that shows there is a 33% chance of my partner having any 2 of the three specified cards. _________________ Newsflash: the word "irony" doesn't mean "a bit like iron"
Shockingly, some people have claimed that I'm egocentric... but hey, enough about them
"If you strike me down, I shall become far stronger than you can possibly imagine."
_______________________________________________
* I occasionally post things here because I am a fan of SRD.
This should not be considered as any condoning of the hypocritical, snowflake, bleeding heart, white guiltridden, compulsively virtuesignalling and totalitarian prog Left woke zealotry which some try to force upon this forum. 

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Fist and Faith Magister Vitae
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Posted: Thu Feb 11, 2021 2:54 am Post subject: 


I was thinking about this some more. Without considering specific cards, we have a fairly small number of possibilities, and can actually count them. So there are three other players: A, B, C. Each can have 0, 1, 2, or all 3 of those cards. So the possibilities are:
A B C
3 0 0
2 1 0
2 0 1
1 1 1
1 2 0
1 0 2
0 3 0
0 0 3
0 2 1
0 1 2
Ten possibilities. The chances of your partner having:
all three is 1/10 or 10%
two of three is 2/10 or 20%
one of three is 3/10 or 30%
none of three is 4/10 or 40%
Where did I go wrong??? _________________ We are not required to save the world. We are required to stand up as truly as we can for what we love. SRD
All lies and jest
Still a man hears what he wants to hear
And disregards the rest Paul Simon 

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Fist and Faith Magister Vitae
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Posted: Mon Mar 08, 2021 3:49 am Post subject: 


Hey, come on, guys! _________________ We are not required to save the world. We are required to stand up as truly as we can for what we love. SRD
All lies and jest
Still a man hears what he wants to hear
And disregards the rest Paul Simon 

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sgt.null jack of odd trades; master of fun
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Posted: Mon Mar 08, 2021 4:47 am Post subject: 


You went wrong by attempting maths. _________________ life's not a paragraph
And death i think is no parenthesis”
― E.E. Cummings


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Fist and Faith Magister Vitae
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Posted: Tue Mar 09, 2021 12:21 am Post subject: 


So it seems. My first line of thinking was too simplistic. (My thanks anyway, wf.) I only considered the number of cards and the number of players. Which gave a 33 1/3% chance. But that's not the same as all the possibilities. Which give a 30% chance. And that's only if TF's original questions is "What are the odds of my partner holding one, and only one, spade card higher than that?" But that's not his originial question. So we have to add the two scenarios where your partner has two of the cards, and the one scenario where your partner holds all three.
So the odds of your partner holding any spade card higher than the Jack are 60%. _________________ We are not required to save the world. We are required to stand up as truly as we can for what we love. SRD
All lies and jest
Still a man hears what he wants to hear
And disregards the rest Paul Simon 

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wayfriend whilom witling
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Posted: Tue Mar 09, 2021 2:11 pm Post subject: 


I think (not sure) that the problem with the "ten possibilities" approach is that you are assuming all 10 are equal probability.
There is only one way to have a player have 3 of the cards.
But there are three ways to have 1 of the cards.
And there are three ways to have 2 of the cards.
That is,
3 0 0 can be represented by
AKQ ... ...
2 1 0 can be represented by
AK. Q.. ...
AQ. K.. ...
KQ. A.. ...
And so one is 3 times more likely than the other.
If you weight the cases with a 2 or a 1 three times heavier than the ones with a 3, you get a total weight of 24.
Your partner having all three is 1/24.
Your partner having two is 6/24.
Your partner having one is 9/24.
Your partner having zero is 8/24 or 33%.
Your partner having some is 66%.
Which is still not correct!
Total weight should be 27. But I am at a loss where at the moment. _________________ * I occasionally post things on KevinsWatch because I am a fan of Stephen R. Donaldson; this should not be considered as condonation of the white nationalist propaganda which is posted far too frequently on this website. 

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