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 Replies: 678 / Views: 16,136 |
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Valued Member
Canada
287 Posts |
My denominator is incorrect. It should not be every combination possible but every possible combination that includes a Jack. The numerator should be every combination that includes a Jack and a Six.
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Pillar of the Community
United States
2669 Posts |
heh.. gave number of 3 number possibilities here https://goccf.com/t/70139&whichpage=41#591510 and in 2 or 10 others.. Every combination that includes a jack should be total * 4/52 = 10200. Problem with the 800 is it only accounts for Jack being first. Needs to be doubled for reverse order. If you figure 8/52 of 132600 permutations will start with the first number, and 4/52 of that will end with the 2nd number, you get 1569, instead of 1600, though. Which makes no sense to me. Errand time, you guys better solve this by the time I get back!  |
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Valued Member
Canada
287 Posts |
Quote:
you get 1569, instead of 1600, though. Which makes no sense to me.
Wow, one more thing we can agree on. Makes no sense to me either. |
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Valued Member
Canada
287 Posts |
I would agree that every 3 card combination that starts with a Jack would be 4*51*50 which is 10200. It would only make sense that the number of 3 card combination's that end with a Jack would be the same amount. I on;y wish I cold remember why we needed that. I need to go back and regroup my thoughts.
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Valued Member
Canada
287 Posts |
The 800 does account for left handed matches (6 being first).
So, 800/20400 = .039 or 4%.
Ah, do I get a second final answer on this question?
So this would show that there is a 4% chance of a 6 being within 2 cards of a Jack within a deck.
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Valued Member
United States
297 Posts |
wow, I looked at some of the puzzles I missed, these are hard! Did you make them your self? If its hard to solve how hard is it to make!?
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Pillar of the Community
 United States
4000 Posts |
LOL! They were very hard at first. They are still a little hard to create, but getting easier.
The latest isn't a puzzle I created. It is a card trick that I have been trying to figure out for a few years now.
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Pillar of the Community
United States
4000 Posts |
It seems like all the formulas I throw at it say the expected win % should be much lower than reality.
I just ran twenty 100-hand experiments on a card generator/testing program that rikcando wrote:
19 times of the 20, it found a match 90 or more times. One round had 88 hits out of 100.
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Pillar of the Community
United States
4000 Posts |
The player chooses one of 156 combinations.
Let's say they choose A-2.
Let's call A our anchor card.
Somewhere in the deck, the 4 Aces will be found and we need to know if our second card is either in front of it or behind it within two places:
xxAxx
xxAxx
xxAxx
xxAxx
So, we are given 16 chances to find 1 of 12 remaining values. That should indicate high probability there, right? 16 chances to find a 2 through K? Or 16/48. Normal distribution would suggest that we will find almost an entire run of 2-K with some duplicates.
Or not quite a complete run, which is where you run the risk of losing, but also means other values (and in a lot of those cases yours) were duplicated.
Am I getting off track?
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Pillar of the Community
United States
4000 Posts |
No, not 16/48. Some of those x's would overlap.
16/48 would be the extreme amount of chances, but not likely.
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Valued Member
Canada
287 Posts |
I have to apologize Scooby. I just realized that my program looks at a flat deck and does not check the beginning against the end, or vice versa. It should only take a few minutes to fix that tonight.
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Pillar of the Community
United States
2669 Posts |
Quote:
Let's say they choose A-2.
Let's call A our anchor card.
Somewhere in the deck, the 4 Aces will be found and we need to know if our second card is either in front of it or behind it within two places:
xxAxx
xxAxx
xxAxx
xxAxx
So, we are given 16 chances to find 1 of 12 remaining values. The first part simplifies things a bit, but I don't understand the 2nd part. Wouldn't we have 16 chances to find 1 of 4, not 12? We're only looking for the 2nd card. |
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Pillar of the Community
United States
4000 Posts |
I meant 1 of any 12 that they may have chosen, or 4/48.
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Pillar of the Community
United States
4000 Posts |
Quote:
I have to apologize Scooby. I just realized that my program looks at a flat deck and does not check the beginning against the end, or vice versa. It should only take a few minutes to fix that tonight. Then some of the potential losses may have turned into wins. I thought the stats were sick enough already! |
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Pillar of the Community
United States
8904 Posts |
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Replies: 678 / Views: 16,136 |
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Posted 10/26/2010 10:57 am
