## Poll

19 votes (46.34%) | |||

14 votes (34.14%) | |||

6 votes (14.63%) | |||

2 votes (4.87%) | |||

12 votes (29.26%) | |||

3 votes (7.31%) | |||

6 votes (14.63%) | |||

5 votes (12.19%) | |||

11 votes (26.82%) | |||

9 votes (21.95%) |

**41 members have voted**

November 30th, 2022 at 8:02:38 AM
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It’s the same number with different bases. Since 30 is divisible by 3 the number must be. One of these is decimal, since 18 isn’t there, it must be 15. This gives 120 and 1111 as the last two and F as a giveaway first!

November 30th, 2022 at 9:54:18 AM
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Quote:Gialmere

Complete this series:

10, 11, 12, 13, 14, 15, 16, 17, 21, 23, 30, 33, …

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The sequence is 15 in base 15-n

The remainder of the series is 120, 1111, 111111111111111

The remainder of the series is 120, 1111, 111111111111111

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

November 30th, 2022 at 4:39:12 PM
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Quote:GialmereComplete this series:

10, 11, 12, 13, 14, 15, 16, 17, 21, 23, 30, 33, …

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120, and 1111

The original list is 15 in base 15, 14, 13, ..., 4; 120 in base 3 and 1111 in base 2 = 15

If you want to get pedantic, it's not technically complete; although bases 1, 0, and -1 are undefined, 15 in base -2 = 10011, since (-2)^4 + (-2)^1 + (-2)^0 = 16 - 2 + 1 = 15.

November 30th, 2022 at 5:35:28 PM
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Quote:charliepatrickIt’s the same number with different bases. Since 30 is divisible by 3 the number must be. One of these is decimal, since 18 isn’t there, it must be 15. This gives 120 and 1111 as the last two and F as a giveaway first!

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Quote:gordonm888The sequence is 15 in base 15-n

The remainder of the series is 120, 1111, 111111111111111

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Quote:ThatDonGuy

120, and 1111

The original list is 15 in base 15, 14, 13, ..., 4; 120 in base 3 and 1111 in base 2 = 15

If you want to get pedantic, it's not technically complete; although bases 1, 0, and -1 are undefined, 15 in base -2 = 10011, since (-2)^4 + (-2)^1 + (-2)^0 = 16 - 2 + 1 = 15.

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Correct!!

Well done.

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Because Oct 31 = Dec 25

Have you tried 22 tonight? I said 22.

November 30th, 2022 at 8:18:38 PM
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How many unique five-card stud hands are there in a five-deck shoe?

Revised wording: How many distinct combinations of five cards can be made from a five-deck shoe of cards?

Revised wording: How many distinct combinations of five cards can be made from a five-deck shoe of cards?

Last edited by: Wizard on Dec 1, 2022

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

November 30th, 2022 at 9:17:20 PM
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Quote:WizardHow many unique five-card stud hands are there in a five-deck shoe?

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I think the wording is unclear.

Is this: how many distinct 5 card stud hands can be created from 5 decks of cards?

Or:

Is this: given five decks of cards are shuffled and arranged in a shoe, how many unique 5 card stud hands exist as five consecutive cards in the stack?

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

November 30th, 2022 at 9:20:19 PM
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I think there is no such a game in the world using a 5-deck shoe for 5-card stud games.

December 1st, 2022 at 3:07:45 AM
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Quote:gordonm888how many distinct 5 card stud hands can be created from 5 decks of cards?

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That wording is good.

Revised reply: Please do not think about poker. Rather refer to this revised wording, "How many distinct combinations of five cards can be made from a five-deck shoe of cards?"

Last edited by: Wizard on Dec 1, 2022

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

December 1st, 2022 at 6:42:03 AM
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As I approached this problem, I wondered why this was in a thread titled Easy Math Problems. Then, as I worked the first hand categories I found its simplicity.

5 singletons (includes straights, straight flushes and unpaired flushes)

c(13,5) *4

1 pair + 3 singletons (includes flushes with 1 pair)

c(13,4)* 4

2 pair + 1 singleton (includes flushes with 2 pair)

c(13,3)* 4

3oak +2 singletons (includes flushes with 3oak)

c(13,3)* 4

3oak + 2oak (includes flushes with full houses)

c(13,2)* 4

4oak + 1 singleton (includes flushes with quads)

c(13,2)* 4

5oak (includes flushes with 5oak)

c(13,1)* 4

Summing up the above numbers gives: 2,808,832

5 singletons (includes straights, straight flushes and unpaired flushes)

c(13,5) *4

^{5}=1,317,888

1 pair + 3 singletons (includes flushes with 1 pair)

c(13,4)* 4

^{5}=732,160

2 pair + 1 singleton (includes flushes with 2 pair)

c(13,3)* 4

^{5}= 292,864

3oak +2 singletons (includes flushes with 3oak)

c(13,3)* 4

^{5}= 292,864

3oak + 2oak (includes flushes with full houses)

c(13,2)* 4

^{5}= 79,872

4oak + 1 singleton (includes flushes with quads)

c(13,2)* 4

^{5}= 79,872

5oak (includes flushes with 5oak)

c(13,1)* 4

^{5}= 13,312

Summing up the above numbers gives: 2,808,832

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.