# Probability of winning a score based game

#### luscra

##### New member
I have a question about the probability of winning a score based game.

It's a game of dice and cards, the rules are:

• There are 2 to 17 players (including you).
• Each person rolls two 6 faced dice.
• Each person extract 3 cards from a full standard 4x13 cards deck
For each die:

• If the die results in an odd number add a³ points to the player, where a is the number on the die face.
• If it's even, add a² - 1 points.
For each card:

• Add b² if b ∈ {2, 3, 4, 5, 6, 7, 8, 9, 10}, where b is the value of the card.
• If the card is an A, add 0.
• If the card is J, add 1.
• And for K, add 3.
The player with the highest score wins. The question is: How many ways can someone win the game?

I would like to know how to start developing this problem. Where can I learn more about the subject? How do I check for all possible games whe you would win.

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#### BigBeachBanana

##### Senior Member
Are you drawing the cards with or without replacement?

#### Steven G

##### Elite Member
Welcome to the forum. This is a math help forum where we help students solve their problems. What we never do here is solve a problem for a student. We prefer to give hints as to how to go about the problem.
What have you tried, where are you stuck? If you had read and followed the forum's posting guidelines you would have received help by now.

Can you explain what happens in the event of a tie after the 1st round?

#### luscra

##### New member
Are you drawing the cards with or without replacement?
Without replacement.
Welcome to the forum. This is a math help forum where we help students solve their problems. What we never do here is solve a problem for a student. We prefer to give hints as to how to go about the problem.
What have you tried, where are you stuck? If you had read and followed the forum's posting guidelines you would have received help by now.

Can you explain what happens in the event of a tie after the 1st round?
There is only 1 round per game, the player with the highest score wins. A tie happens when two or more players achieve the highest score.

What I got thus far is:
$\text{amount of possible victories}=\frac{\text{amount of possible games - amount of possible ties}}{\text{number of players}}$
Which is:
$\frac{6^{2\times J}\times \frac{52!}{(3\times J)!(52-(3\times J))!}-\text{ties}}{J}$
J is the number of players.

Now, I just need to know how to find the amount of possible ties based on the number of players. I don't know where to start from here.

#### Cubist

##### Senior Member
$\text{amount of possible victories}=\frac{\text{amount of possible games - amount of possible ties}}{\text{number of players}}$

I think your expression for the "amount of possible games" is good except the suit of card doesn't matter. Taking this into account is possible but more tricky. Alternatively you just consider the suits as being different when you calculate the number of possible ties (and accept this over-count in the results)

However, calculating the number of ties seems horrendous. You'd need to add:-
the ways that 2 players can tie and (J-2) players score less
+ the ways that 3 players can tie and (J-3) players score less
...
+ the ways that (J-1) players can tie and 1 player scores less
+ the ways that J players can tie

Did you make this problem up yourself? If so, then I'd recommend that you learn some programming skills and implement a Monte Carlo method to obtain a reasonable estimate of the probability for one player winning (for all J=2 to 17). I guess that the results will be close to 1 / J since a tie seems unlikely.

#### BigBeachBanana

##### Senior Member
1) All possible games - possible ties = Possible victories or losses. Not just victories.
2) Drawing cards without replacement follows a hypergeometric distribution and not binomial distribution as you have.
3) Not sure why you're dividing by the number of players when you're calculating counts?
4) I agree with @Cubist that this problem is horrendous. Calculating the ties is one of the few things on the laundry list.

#### Cubist

##### Senior Member
1) All possible games - possible ties = Possible victories or losses. Not just victories.
...
3) Not sure why you're dividing by the number of players when you're calculating counts?
I think @luscra is calculating the number of ways that any player wins, not just the player we're interested in. And then this is divided by the number of players (because each player has the same number of ways of winning).

2) Drawing cards without replacement follows a hypergeometric distribution and not binomial distribution as you have.
That part of the calculation @luscra provided is working out the number of ways to roll 2*J dice multiplied by the ways to choose 3*J cards from the deck of 52.

#### luscra

##### New member
Did you make this problem up yourself? If so, then I'd recommend that you learn some programming skills and implement a Monte Carlo method to obtain a reasonable estimate of the probability for one player winning (for all J=2 to 17). I guess that the results will be close to 1 / J since a tie seems unlikely.
It's a college assignment. Honestly I don't really think it was meant to be solved, it's just a way for our professor to get a good laugh. He does this question for most semesters and no one can solve it, even close to the end of their majors. We showed this question to another professor and they just laughed, so...

About the Monte Carlo method, I did not know there was a name for that, but it was using this method that I got to the games-ties/players answer.
I developed an algorithm in python that runs the game 100000 times and outputs a graph.

It's in portuguese, but basically the y-axis is the number of games, the x-axis the players. In green you have the games a player loses, in blue wins and orange ties.
I guess that the results will be close to 1 / J since a tie seems unlikely.
The ties are indedeed really unlikely, for 2 players there's a 0.36% chance of a tie, and that number just decreases.

I think I'm going to drop trying to get the absolute number and just do an approximation using what I have. If my theory is right, he doesn't really expects us to solve anything, which means I'm already above expectations lol.

Is it possible to mathematically prove (or at least try to) an argument using the Monte Carlo method, even though it will never be accurate enough? Or the best I can do is explain what I did, how I did, and what I got?

#### Cubist

##### Senior Member
It's a college assignment. Honestly I don't really think it was meant to be solved, it's just a way for our professor to get a good laugh. He does this question for most semesters and no one can solve it, even close to the end of their majors. We showed this question to another professor and they just laughed, so...
I like this professor

I developed an algorithm in python that runs the game 100000 times and outputs a graph.
Nice work!

The ties are indedeed really unlikely, for 2 players there's a 0.36% chance of a tie, and that number just decreases.
That's interesting. This isn't like the problem of there being two or more people with the same birthday in a group of N. It's more like Texas Hold Em' poker, if there are more people playing then it's more likely that someone will have a really good hand (or in your game a high winning score becomes more likely).

Is it possible to mathematically prove (or at least try to) an argument using the Monte Carlo method, even though it will never be accurate enough? Or the best I can do is explain what I did, how I did, and what I got?
Sorry, the Monte Carlo method can't provide an exact answer.

I usually run the code several times so that I can examine the variance in each result. By doing this you can obtain an idea of the accuracy. (Does the result change wildly on each run, or stay near to the mean average.)

You might be able to "brute force" the exact answer, on a computer, for J=2 but I expect that the runtime will be considerable. There are several ways you could improve the runtime over a basic "brute force all scenarios" algorithm but it's probably not worth the time and effort.