## Poll

16 votes (51.61%) | |||

12 votes (38.7%) | |||

5 votes (16.12%) | |||

2 votes (6.45%) | |||

9 votes (29.03%) | |||

3 votes (9.67%) | |||

5 votes (16.12%) | |||

5 votes (16.12%) | |||

10 votes (32.25%) | |||

7 votes (22.58%) |

**31 members have voted**

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.

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?

Quote:WizardI 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.

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.

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.

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?

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.

Quote:charliepatrickSince 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.

Quote:charliepatrickEach 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:charliepatrickQuote: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!!

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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.

Quote:GialmereBut 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.

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

Quote:GialmereThere'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.

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