Math problem (easy)

perceval · #1

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

Bonus Question (just thought of it ):
if you remove the 3 cards limitation to match the sum , is there a hand to be sure to finish next turn ? which hand ?

entropi · #2

Is there a clever trick to solve, or do we have to think of dice outcome probability histogram and find the sequence of cards covering the biggest range?

perceval · #3

that s not a book problem but a problem i thought of while playing, so i don't know of a pleasant trick.
i jus tried to find the hand with the max proba manually.

Magicwand · #4

entropi wrote:

Is there a clever trick to solve, or do we have to think of dice outcome probability histogram and find the sequence of cards covering the biggest range?

two dice... #6 is the most common out come.
so i solved half of the problem

perceval · #5

Magicwand wrote:

entropi wrote:

Is there a clever trick to solve, or do we have to think of dice outcome probability histogram and find the sequence of cards covering the biggest range?

two dice... #6 is the most common out come.
so i solved half of the problem

actually with 2 dice 7 is the most common outcome
proba for each dice sum outcome hidden:

Tryss · #6

Quote:

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 ?

Probably 1,2 and 6, with this hand you win with : 3, 5, 7 and 9

it represent 44.44% of winning

Quote:

Bonus Question (just thought of it ):
if you remove the 3 cards limitation to match the sum , what is the minimal hand to be sure to finish next turn ?

Is the number of card limited?

If it's not limited, then you can't be sure to finish next turn. If you have only even card and the dice is odd, you've lost

entropi · #7

Magicwand wrote:

entropi wrote:

Is there a clever trick to solve, or do we have to think of dice outcome probability histogram and find the sequence of cards covering the biggest range?

two dice... #6 is the most common out come.
so i solved half of the problem

For 4, you need 1-3; 3-1; 2-2
For 5, you need 1-4; 4-1; 2-3; 3-2
For 6, you need 1-5; 5-1; 2-4; 4-2; 3-3
For 7, you need 1-6; 6-1; 2-5; 5-2; 3-4; 4-3
For 8, you need 2-6; 6-2; 3-5; 5-3; 4-4
For 9, you need 3-6; 6-3

So 7 is more probable than 6

It should be such a card sequence that covers 6, 7, 8

So, if you have two cards as 1 and 7, you cover 6 and 8 (but miss 7 itself

). Anyway, with 1 and 7 the probability you finish next round is 10/36. Not bad

hyperpape · #8

entropi · #9

hyperpape wrote:

Wildclaw · **#10**

perceval wrote:

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 ?

The two first cards are 1 and 2. The final card can be either 6 7 or 8. All three give you the same 16/36 chance on winning.

perceval wrote:

Bonus Question (just thought of it ):
if you remove the 3 cards limitation to match the sum , is there a hand to be sure to finish next turn ? which hand ?

No. If a hand contains an even number of even cards it will only be able to produce even results. If a hand contains an odd number of even cards it will only be able to produce odd results.

daniel_the_smith · **#11**

Three way tie.

Edit: added more detail

extra: for 2x 10 sided dice, the best you can do is:

for 3x 6 sided dice, it's:

hyperpape · **#12**

Wildclaw wrote:

No. If a hand contains an even number of even cards it will only be able to produce even results. If a hand contains an odd number of even cards it will only be able to produce odd results.

I think you're slightly off: [1,2,3] can produce 2,4,6. [1,2,3,4] can produce 2,4,6,8,10.

shapenaji · **#13**

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.

Wildclaw · **#14**

hyperpape wrote:

I think you're slightly off: [1,2,3] can produce 2,4,6. [1,2,3,4] can produce 2,4,6,8,10.

Ah yeah. I meant an even number of odd numbers makes the sum even.

perceval · **#15**

I think Tryss gave a good solution first, wildclaw was the first to give all 3.

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

)

but brute force is useful: i didnt even try with even numbers and actually its a tie with odd numbers

a proof without listing all cases ?

perceval · **#16**

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

daniel_the_smith · **#17**

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...

cyclops · **#18**

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?

Harleqin · **#19**

daniel_the_smith wrote:

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

daniel_the_smith · **#20**

Harleqin wrote:

daniel_the_smith wrote:

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:

Math problem (easy)

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