Dumb-assed stats question

Technology, computers, sciences, mysteries and phenomena of all kinds, etc., etc. all here at The Loresraat!!

Moderator: Vraith

User avatar
Fist and Faith
Magister Vitae
Posts: 23438
Joined: Sun Dec 01, 2002 8:14 pm
Has thanked: 6 times
Been thanked: 30 times

Post by Fist and Faith »

Ah! Not caring which of the three cards your partner gets does not mean the multiple possibilities don't have to be taken into account when figuring the odds!

I am but an egg.
All lies and jest
Still a man hears what he wants to hear
And disregards the rest
-Paul Simon
User avatar
wayfriend
.
Posts: 20957
Joined: Wed Apr 21, 2004 12:34 am
Has thanked: 2 times
Been thanked: 4 times

Post by wayfriend »

I should explain...

There are 3 players who could get the first card.
3 the second.
3 the third.

These are independent events*. So there should be 3x3x3 permutations.

When you use values that are X/27 you will get answers that correlate with the others.

__________
* Or are they?
.
User avatar
Hashi Lebwohl
The Gap Into Spam
Posts: 19576
Joined: Mon Jul 06, 2009 7:38 pm

Post by Hashi Lebwohl »

I never saw this question until just now. I will have to address it later, though--time for a quick nap before the second half-job.
The Tank is gone and now so am I.
User avatar
Rigel
The Gap Into Spam
Posts: 2096
Joined: Wed Sep 26, 2007 10:42 pm
Location: Albuquerque

Post by Rigel »

Fist and Faith wrote: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???
But the odds are not equal. You already know one card in your hand, meaning you only have 12 remaining cards, where the other 3 all have 13.
"You make me think Hell is run like a corporation."
"It's the other way around, but yes."
Obaki, Too Much Information
User avatar
Hashi Lebwohl
The Gap Into Spam
Posts: 19576
Joined: Mon Jul 06, 2009 7:38 pm

Post by Hashi Lebwohl »

Whoops--I forgot to revist this question.

It will be eaiser to calculate the probability that your partner does not have any of those cards, then just 1-p it.

You have the jack and 12 other cards, so we can reduce the deck to 39; the other 12 cards in your hand are irrelevant. Having either the queen, the king, or the ace is 3/39, which we can reduce to 1/13. From here, it is a straightforward binomial. Since none of their cards are the ones you want, each card is a 12/13 probability to have and they have 13 cards in hand.

p (not having queen, king, or ace) = (12/13)^13 = 0.353258498

1 - p = 0.646741502 = the probability that they have at least one spade higher than the jack you hold.
The Tank is gone and now so am I.
User avatar
Fist and Faith
Magister Vitae
Posts: 23438
Joined: Sun Dec 01, 2002 8:14 pm
Has thanked: 6 times
Been thanked: 30 times

Post by Fist and Faith »

Why didn't you stop at "Having either the queen, the king, or the ace is 3/39, which we can reduce to 1/13."? That seems to answer the question.

Although it certainly doesn't seem right.

But you came up with .65? That seems right, considering my above.
6 of 10 scenarios have at least one of those cards.
Last edited by Fist and Faith on Wed May 19, 2021 3:01 am, edited 1 time in total.
All lies and jest
Still a man hears what he wants to hear
And disregards the rest
-Paul Simon
User avatar
Skyweir
Lord of Light
Posts: 25188
Joined: Sat Mar 16, 2002 6:27 am
Location: Australia
Has thanked: 2 times
Been thanked: 18 times

Post by Skyweir »

You are all such smartypantelones .. geez :8O
ImageImageImageImage
keep smiling 😊 :D 😊

'Smoke me a kipper .. I'll be back for breakfast!'
Image

EZBoard SURVIVOR
User avatar
SoulBiter
The Gap Into Spam
Posts: 9190
Joined: Wed Jun 02, 2004 2:02 am
Has thanked: 74 times
Been thanked: 12 times

Post by SoulBiter »

Just remember, when you try to put this into play, the random nature of cards. Assuming that at the end of the first round, you have the suits mostly together, you have a better possibility of the cards being more evenly distributed if you dealt them right then. Depending on how the cards are shuffled and how many times before dealing will alter those odds somewhat. Example: If the cards had all the suits together (like a brand new set), if you shuffled 8 times the same way in a normal riffle shuffle, you are more likely to have put them very close to where they were when you started. In fact the worse the shuffler, the more random the shuffle will be and each time they shuffle, the cards will be even more random.

But that had nothing to do with the question and it looks like it was answered.
We miss you Tracie but your Spirit will always shine brightly on the Watch Image
Post Reply

Return to “The Loresraat”