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Topic: Black Jack (Read 703 times) |
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ThudnBlunder
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A well-shuffled pack of 52 playing cards is placed face-down on a table. Then one-by-one the cards are turned over. If you had to choose the position of the first black Jack to be revealed, what position would you pick and why?
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ima1trkpny
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Re: Black Jack
« Reply #1 on: Jun 5th, 2007, 5:41pm » |
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The top (within the top 10 cards) because assuming you cut the deck around the middle region the initial pattern was black suit-red suit-black suit-red suit so when you cut and shuffle the deck you mix the two black suits in the bottom half of the deck and then when you cut and shuffle again they come out in the top and remain there?
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| « Last Edit: Jun 5th, 2007, 5:43pm by ima1trkpny » |
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ThudnBlunder
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Re: Black Jack
« Reply #2 on: Jun 5th, 2007, 6:09pm » |
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on Jun 5th, 2007, 5:41pm, ima1trkpny wrote:| ...because assuming you cut the deck around the middle region the initial pattern was black suit-red suit-black suit-red suit... |
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No, the deck is randomly cut, randomly shuffled, etc, etc. So the initial arrangement is irrelevant.
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Grimbal
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Re: Black Jack
« Reply #3 on: Jun 6th, 2007, 2:14am » |
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Intuitively, I would say the first black jack would appear around 1/3 of the pack. But that is if you average the position.
But the most likely single position for the first black jack is the first card.
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| « Last Edit: Jun 6th, 2007, 3:29am by Grimbal » |
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towr
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Re: Black Jack
« Reply #4 on: Jun 6th, 2007, 3:27am » |
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The probability of the Kth card out of N being the first black Jack seems to be (N-K)/[N*(N-1)]
So the least K, i.e. 1, is always the most probable
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Miles
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Re: Black Jack
« Reply #5 on: Jun 6th, 2007, 7:20am » |
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on Jun 6th, 2007, 3:27am, towr wrote:The probability of the Kth card out of N being the first black Jack seems to be (N-K)/[N*(N-1)]
So the least K, i.e. 1, is always the most probable |
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You need to double your formula, but obviously that does not change the conclusion.
If our success is measured by how small the difference is between the position we predicted and the actual position, what position would you choose to minimise the expected difference?
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towr
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Re: Black Jack
« Reply #6 on: Jun 6th, 2007, 9:16am » |
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on Jun 6th, 2007, 7:20am, Miles wrote:| You need to double your formula, but obviously that does not change the conclusion. |
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Ah yes, two interchangable black Jacks; I concentrated on one specific one being first..
Quote:| If our success is measured by how small the difference is between the position we predicted and the actual position, what position would you choose to minimise the expected difference? |
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I suppose the easiest way is to try all 52 cases
At the moment no simpler expression comes to mind minargk  Ni=1 P(i) |k-i|
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| « Last Edit: Jun 6th, 2007, 9:17am by towr » |
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rmsgrey
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Re: Black Jack
« Reply #7 on: Jun 6th, 2007, 9:45am » |
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on Jun 6th, 2007, 2:14am, Grimbal wrote:Intuitively, I would say the first black jack would appear around 1/3 of the pack. But that is if you average the position.
But the most likely single position for the first black jack is the first card.
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Intuitively, I'd say each card has the same probability of being a black jack, so it's the probability of having an earlier black jack that controls it - and that's obviously minimised when there are no previous cards...
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ThudnBlunder
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Re: Black Jack
« Reply #8 on: Jun 6th, 2007, 10:05am » |
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on Jun 6th, 2007, 9:16am, towr wrote:
At the moment no simpler expression comes to mind minargk Ni=1 P(i) |k-i| |
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Speaking of moments, I would say it is analogous to the continuous case of balancing a beam whose weight varies uniformly from 1 to 51 per unit length, although it is not obvious that this gives the minimum.
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| « Last Edit: Jun 7th, 2007, 1:15am by ThudnBlunder » |
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Grimbal
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Re: Black Jack
« Reply #9 on: Jun 6th, 2007, 3:50pm » |
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If you'd like to minimize the square of the difference, it is easy, it would be the mean value, which is at 1/3 of the pack, at 51/3 = 17 (zero-based).
To minimize |i-j|, you need the median, which is around 51*(1-1/sqrt(2)) = 14.94. The best guess is card 15 (zero-based).
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towr
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Re: Black Jack
« Reply #10 on: Jun 7th, 2007, 1:13am » |
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on Jun 6th, 2007, 3:50pm, Grimbal wrote:| To minimize |i-j|, you need the median, which is around 51*(1-1/sqrt(2)) = 14.94. The best guess is card 15 (zero-based). |
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16 is slightly better.
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ThudnBlunder
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Re: Black Jack
« Reply #11 on: Jun 7th, 2007, 1:15am » |
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What is the most probable position for the 2nd black Jack?
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Grimbal
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Re: Black Jack
« Reply #12 on: Jun 7th, 2007, 1:51am » |
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on Jun 7th, 2007, 1:13am, towr wrote:
zero-based?
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towr
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Re: Black Jack
« Reply #13 on: Jun 7th, 2007, 3:46am » |
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on Jun 7th, 2007, 1:51am, Grimbal wrote:hmm, I missed that. If that means you count the first card as zeroth, then I concur..
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towr
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Re: Black Jack
« Reply #14 on: Jun 7th, 2007, 3:48am » |
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on Jun 7th, 2007, 1:15am, ThudanBlunder wrote:| What is the most probable position for the 2nd black Jack? |
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Well, the second black jack from the front, is the first from the back. So I would say the last position.
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Grimbal
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Re: Black Jack
« Reply #15 on: Jun 7th, 2007, 4:25am » |
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on Jun 7th, 2007, 3:46am, towr wrote:
hmm, I missed that. If that means you count the first card as zeroth, then I concur.. |
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It was a bit silly to give a zero-based result. Except to give away the fact that I did a simulation in C first.
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SWF
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Re: Black Jack
« Reply #16 on: Jun 7th, 2007, 10:23pm » |
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My first thought was the most likely position is the 2nd card, but upon re-reading the question I see it says "black Jack" and not "blackjack".
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Random Lack of Squiggily Lines
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Re: Black Jack
« Reply #17 on: Jun 8th, 2007, 9:13am » |
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on Jun 7th, 2007, 1:15am, ThudanBlunder wrote:| What is the most probable position for the 2nd black Jack? |
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20?
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ThudnBlunder
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Re: Black Jack
« Reply #18 on: Jun 8th, 2007, 9:20am » |
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on Jun 8th, 2007, 9:13am, tiber13 wrote:
No, if 1 is the most probable position for the first black Jack, then by symmetry 52 is the most probable position for the second black Jack (as towr claimed).
on Jun 7th, 2007, 3:48am, towr wrote:
Well, the second black jack from the front, is the first from the back. So I would say the last position.
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Grimbal
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Re: Black Jack
« Reply #19 on: Jun 9th, 2007, 8:18am » |
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If the most probable position for the 1st BJ is the first card and the most probable position for the 2nd BJ is the last card, certainly the most probably position for the pair is the first and last card.
Or isn't it?
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| « Last Edit: Jun 9th, 2007, 10:20am by Grimbal » |
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rmsgrey
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Re: Black Jack
« Reply #20 on: Jun 9th, 2007, 8:59am » |
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on Jun 9th, 2007, 8:18am, Grimbal wrote:If the most probable position for the 1st BJ is the first card and the most probable position for the 2nd BJ is the last card, certainly the most probably position for the pair is the first and last card.
Or isn't it? |
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Of course it isn't - all positions for the pair are equally probable, it's just that the second card, say, is first black jack for some of them, and last for some - it's only the first card which is only ever the first black jack, and similarly for the last...
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