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charliepatrick
charliepatrick
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Gialmere
June 17th, 2021 at 9:53:36 AM permalink
Each set is 0 1 2 4.

The way to show it is to consider the binary for the numbers 1 thru 6 as 001 010 011 100 101 and 110, some which use two digits and others which use one. So picking two from 1 2 4 covers the two digit ones while 0+ another cover the others. Thus each set create the numbers 1 thru 6, so are equivalent to a regular die.
Last edited by: charliepatrick on Jun 17, 2021
Wizard
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Wizard
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June 17th, 2021 at 11:29:41 AM permalink
I agree with Charlie on four cards. My answer is by trial and error.


I think this is a trick question. I suspect it is impossible if you stick with the constraints of using integers and adding the two cards.

I've tried using negative numbers and adding, but can't find a solution.

I've tried multiplying, instead of adding, but still can't find a solution.

I suspect there is some other function of the two cards I must take.

That's about where I am now.


Practical Question

In the land of Calizonia, casinos may not use dice in craps, but they may use cards. They must draw two cards, without replacement, from a seven card deck and add the results. The probabilities must match those with two dice. There are no regulations on what can be on the cards.

What should the Calizonia casino do?
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
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charliepatrickGialmere
June 17th, 2021 at 11:42:19 AM permalink
Quote: Wizard

I agree with Charlie on four cards. My answer is by trial and error.


I think this is a trick question. I suspect it is impossible if you stick with the constraints of using integers and adding the two cards.

I've tried using negative numbers and adding, but can't find a solution.

I've tried multiplying, instead of adding, but still can't find a solution.

I suspect there is some other function of the two cards I must take.

That's about where I am now.



Who said anything about the numbers having to be integers? Try having all of the numbers be {integer} + 1/2.

For example, the lowest two can be 1/2 and 3/2, so you can "roll" a 2.

charliepatrick
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Gialmere
June 17th, 2021 at 11:52:23 AM permalink
Quote: ThatDonGuy

...


Who said anything about the numbers having to be integers? Try having all of the numbers be {integer} + 1/2.

For example, the lowest two can be 1/2 and 3/2, so you can "roll" a 2.

That helps as it had to be symmetric and having the middle one being 3.5 seemed to work. Working from the top the first two have to be .5 and 1.5 (to get one 2) then two 2.5 to (get two 3's). Similarly 6.5 5.5 4.5 4.5. The rest falls out.
Wizard
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June 17th, 2021 at 11:55:35 AM permalink
Quote: ThatDonGuy



Who said anything about the numbers having to be integers? Try having all of the numbers be {integer} + 1/2.

For example, the lowest two can be 1/2 and 3/2, so you can "roll" a 2.



I tried that too and couldn't find an answer, although I guess I didn't try hard enough.
It's not whether you win or lose; it's whether or not you had a good bet.
charliepatrick
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June 17th, 2021 at 12:03:10 PM permalink
Quote: Wizard

...
Practical Question

In the land of Calizonia, casinos may not use dice in craps, but they may use cards. They must draw two cards, without replacement, from a seven card deck and add the results. The probabilities must match those with two dice. There are no regulations on what can be on the cards.

What should the Calizonia casino do?

Since if you only looked at the two cards there would be 21 permutations you must use the order. Then you have 42 combinations, so some of these must be "to try again".

The following method works.

The cards are labeled 1 2 3 4 5 6 "same". The first card drawn determine the roll of the first die. However if the first card drawn is "same" then you start over again. The second card drawn determines the roll of the second die, except if it's the "same" then you've thrown a repeat of the first roll. Using the "same" to replace the first roll means the second roll still has an equal chance for each outcome.
Wizard
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Wizard
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June 17th, 2021 at 12:23:18 PM permalink
Quote: charliepatrick

Since if you only looked at the two cards there would be 21 permutations you must use the order. Then you have 42 combinations, so some of these must be "to try again".

The following method works.

The cards are labeled 1 2 3 4 5 6 "same". The first card drawn determine the roll of the first die. However if the first card drawn is "same" then you start over again. The second card drawn determines the roll of the second die, except if it's the "same" then you've thrown a repeat of the first roll. Using the "same" to replace the first roll means the second roll still has an equal chance for each outcome.



I agree! One of the California casinos uses this method, I forget which one.
It's not whether you win or lose; it's whether or not you had a good bet.
Gialmere
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June 17th, 2021 at 4:57:33 PM permalink
Quote: charliepatrick

Each set is 0 1 2 4.

The way to show it is to consider the binary for the numbers 1 thru 6 as 001 010 011 100 101 and 110, some which use two digits and others which use one. So picking two from 1 2 4 covers the two digit ones while 0+ another cover the others. Thus each set create the numbers 1 thru 6, so are equivalent to a regular die.


Quote: charliepatrick

Quote: ThatDonGuy

...


Who said anything about the numbers having to be integers? Try having all of the numbers be {integer} + 1/2.

For example, the lowest two can be 1/2 and 3/2, so you can "roll" a 2.

That helps as it had to be symmetric and having the middle one being 3.5 seemed to work. Working from the top the first two have to be .5 and 1.5 (to get one 2) then two 2.5 to (get two 3's). Similarly 6.5 5.5 4.5 4.5. The rest falls out.


Correct!!

Yes, the problem only specifies that the result must be an integer.

-------------------------------------------------

My wife accused me of never achieving anything because of my addiction to board games.

She must have forgotten that time I won second prize in a beauty contest.
Have you tried 22 tonight? I said 22.
Ace2
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June 17th, 2021 at 8:15:42 PM permalink
Quote: Gialmere

But wait! You recount and find there are actually nine cards available. Hmm. It occurs to you that you could number the nine cards in such a way that randomly drawing two cards from the pile of nine cards also gives an integer from 2 to 12 with the same frequency of occurrence as rolling that sum on two standard dice.



Number the cards 1,1,2,3,4,5,6,6, Wild

Draw two cards and sum them normally with the following caveats:

2W, 3W, 4W and 5W means pair. So, for example, 4W means 44 or 8

1W is 5, W1 is 6, 6W is 8, W6 is 9. These are the only 4 cases when the order of the cards matters

I think this works but I may be wrong

Itís all about making that GTA
Ace2
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Gialmere
June 18th, 2021 at 8:52:23 AM permalink
Quote: Gialmere

There's only eight cards. Hmm. A ha! You can still divide the cards into two sets (piles) and number them in such a way that by drawing two cards from each pile, and adding the four cards together, you get an integer from 2 to 12 with the same frequency of occurrence as rolling that sum on two standard dice.



Number the cards in each pile 0,1,2,3

For each pile, draw two cards and value them by their sum as follows:

01 is 1
02 is 2
12 is 3
13 is 4
23 is 5
30 is 6 (this is the only exception to the sum rule)

So, for instance, a draw of 23 and 02 is valued at 52 or 7
Itís all about making that GTA

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