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Posted: Tue Mar 09, 2021 2:29 pm
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.
Posted: Tue Mar 09, 2021 6:52 pm
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?
Posted: Tue Mar 09, 2021 10:11 pm
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.
Posted: Mon May 17, 2021 6:42 pm
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.
Posted: Mon May 17, 2021 9:04 pm
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.
Posted: Mon May 17, 2021 10:51 pm
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.
Posted: Tue May 18, 2021 5:38 am
by Skyweir
You are all such smartypantelones .. geez :8O
Posted: Tue May 25, 2021 8:11 pm
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.