Math problem (easy)

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perceval
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Re: Math problem (easy)

Post by perceval »

shapenaji wrote:This game sounds like fun,

I've thought of a variant which might be enjoyable too:

Roll 4 six-sided dice, now you may use any of the cards in your hand to add/subtract/multiply/divide to get the final outcome. (With the caviat that you may not multiply/divide by 1)

So, for example, say the dice come down 18, and you hold 1,4,7,3

You remove the 3-card limit, and 3*7-(4-1) = 18

You might need to have more cards in hand to play this way though, it would be quite easy to run out fast.


As to the original problem, I imagine backgammon has already solved this one.


sounds like fun BUT at start it was a game with my 7 year old. The level of difficulty with 2 dice is just right for him

Never thought of it as a backgammon problem
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daniel_the_smith
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Re: Math problem (easy)

Post by daniel_the_smith »

perceval wrote:but dts gave the brute force proof (you either did that by computer or you are crazy ;-) )


Yeah, I wrote a quick program. But that doesn't exclude the other option...
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Re: Math problem (easy)

Post by cyclops »

perceval wrote:I played a game with my 7 year old son this week end:

The players have cards on their hands, numbered from 1 to 9.

One of the player throws 2 dice.

Then all players must try to match the total of the dice by playing 1,2 or,3 cards, such that they can match the dice sum by substracting or adding the values of the cards played.

For example, if the total of the dice is 5:
you can play:
-1 card "5" OR
- 2 card "3" + "2" OR

- 2 cards "7" - "2" OR
- 3 cards "3" +"1" + "1" OR
- 3 cards "3" +"3" - "1" (you can mix substraction and additions)
etc....

If you cannot match the dice sum with the cards in hand, you must draw a card.
The winner if the first one to get rid of all his cards

Question:
what is the hand of cards that maximize the probability that you will finish next turn, ie play all your cards at once to match the dice sum ?


(you can play at most 3 cards to match the dice sum).
i have a proposal but i am not sure that it is the optimum



It took me to read the answers to get the intended interpretation of the problem.
But my initial interpretation might be fun also.
Given a number as the sum of the dice, you may lay down at the table from your hand as many correct combinations as you can. What is the hand of cards that maximizes the probability that you will finish next turn?
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Re: Math problem (easy)

Post by Harleqin »

daniel_the_smith wrote:
[1 2 9] [8 12 6 10] 0.30000000000000004



That's what you get when representing ratios with floating point numbers. :blackeye:
A good system naturally covers all corner cases without further effort.
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Re: Math problem (easy)

Post by daniel_the_smith »

Harleqin wrote:
daniel_the_smith wrote:
[1 2 9] [8 12 6 10] 0.30000000000000004



That's what you get when representing ratios with floating point numbers. :blackeye:


I agree, accuracy to 4 parts in 10^17 is completely unacceptable. :twisted:
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Re: Math problem (easy)

Post by hyperpape »

daniel_the_smith wrote:
Harleqin wrote:
daniel_the_smith wrote:
[1 2 9] [8 12 6 10] 0.30000000000000004



That's what you get when representing ratios with floating point numbers. :blackeye:


I agree, accuracy to 4 parts in 10^17 is completely unacceptable. :twisted:
The accuracy is good enough. It's the aesthetics of it that are garbage.

I'm waiting for the day when we use some sort of super high-level languages that automatically avoid floating points wherever possible.
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