I am a tutor attempting to help a student with a probability problem for a card game they invented, but I am afraid the probability may be too complex (the student is only in pre-calculus).

Here are the rules of the game:
1. There are four players in a circle. A standard 52-card deck is distributed among these four players so they each have a deck of 13 cards. They may not look at the cards that they are given.
2. The order for which the players will take their turns is that the shortest person goes first (I doubt this has to be included in the probability).
3. Player 1 flips a coin.
a. If the result is heads, each player (including player 1,) puts two of their cards forward face down (they do not know which cards these are), so there are a total of 8 cards in the middle.
b. If the result is tails, each player (including player 1), puts one of their cards forward face down(they do not know which cards these are), so there are a total of 4 cards in the middle.
4. Player 1 chooses two of the cards that are in the middle to be revealed.
a. If one of the cards is a multiple of 3 (Queen included), Player 1 gets 1 point.
b. If both cards are multiples of 3 (Queen included), Player 1 gets 3 points.
c. If neither card is a multiple of 3 (nor a Queen), Player 1 gets no points.
5. The cards are redistributed to the owners of those cards (they are returned to the deck from where they came), and each player shuffles their deck of 13 cards.
6. The next player (counter-clockwise) takes their turn (repeats steps 3-5).
7. The first player to 6 points wins the game.

What is the probability of Player 1 winning the game?

Now, I think calculating the probability of this game is more complicated than it first appears. Here is how I first went about it:

• In the end, what's really happening is that Player 1 is choosing 2 cards from the 52 card deck.
• There are 16 multiples of 3 (3, 6, 9, Q in each suite) in the 52 card deck.
• Therefore, the probability of choosing two multiples of three is 16/52 * 15/51 (0.0905). The probability of choosing one multiple of three is 16*36 / 52C2 (0.4344). The probability of getting no multiples of three is 1 - P(one multiple of 3) - P(two multiples of 3) = 0.4751

From here, I think this is the direction to go in:
• Assume all players have the same skill level so this does not affect the probability
• How many ways are there in which player 1 could win? Lets call getting two multiple of threes B, and one multiple A. Player one could win if they get: BB, BAAA, BAB, BAAB, ABB, ABAB, ABAA, AABB, AABA, AAAB, AAAAB, AAAAAB.
• HOWEVER, this does not include two factors: the possibility that Player 1 does not get any multiples of 3, and therefore no points, and the probability of another player winning the game. This makes the probability more difficult to calculate because there could be up to an infinite number of C's (no multiples of three), so the number of events is unknown.
• I think there is another problem if the cards are reinserted into the same deck, as then there is the consideration that each player knows that there may be, or may not be, multiples of 3 that exist in that persons deck. This affects the probability because there is some given information about someone's deck and you can choose whether to avoid or choose that deck based on that knowledge. However, I'm not sure if this is something that can be ignored since this is just supposed to be a two-week probability assignment and it shouldn't be that difficult for a precalculus class? I am not entirely sure.

My suggestions to eliminate these factors:
• Change rule 4a so a player gets 2 points for one multiple of three.
• Change rule 4c so each player, besides the player who chose the card, gets 1 point.
• Change rule 3/5, and instead put all the cards in the center and shuffle them. So that the player choosing the cards does not know which deck they came from AND the cards are randomly returned.

This change is for still making the game fun and does not make the probability easier like the other changes:
• Change rule 4, so that if the coin was heads, the player chooses two cards, and if the coin is tails, the player chooses one card.

My main questions are, with my proposed changes:
• Will my proposed changes make the probability of player 1 winning the game easier to calculate?
• Do the following really have any effect on the probability of winning:
• separating the deck into 13-card decks (I don't think that this will have an effect).
• the order in which the player goes (I think that this will, first player has the greatest chance, but I'm not sure)
• only allowing the player to choose from 4/8 cards in the middle (I don't think this will have an effect).

• Is the game at least kind of fun? Are there other changes I could make that will make more fun and/or, more importantly, easier to calculate?

Any advice would be greatly appreciated!! Thank you!

Originally Posted by jordanvhn
I am a tutor attempting to help a student with a probability problem for a card game they invented, but I am afraid the probability may be too complex (the student is only in pre-calculus).

Here are the rules of the game:
1. There are four players in a circle. A standard 52-card deck is distributed among these four players so they each have a deck of 13 cards. They may not look at the cards that they are given.
2. The order for which the players will take their turns is that the shortest person goes first (I doubt this has to be included in the probability).
3. Player 1 flips a coin.
a. If the result is heads, each player (including player 1,) puts two of their cards forward face down (they do not know which cards these are), so there are a total of 8 cards in the middle.
b. If the result is tails, each player (including player 1), puts one of their cards forward face down(they do not know which cards these are), so there are a total of 4 cards in the middle.
4. Player 1 chooses two of the cards that are in the middle to be revealed.
a. If one of the cards is a multiple of 3 (Queen included), Player 1 gets 1 point.
b. If both cards are multiples of 3 (Queen included), Player 1 gets 3 points.
c. If neither card is a multiple of 3 (nor a Queen), Player 1 gets no points.
5. The cards are redistributed to the owners of those cards (they are returned to the deck from where they came), and each player shuffles their deck of 13 cards.
6. The next player (counter-clockwise) takes their turn (repeats steps 3-5).
7. The first player to 6 points wins the game.

What is the probability of Player 1 winning the game?
Why not just cut to the chase?
Have player one flip a coin. If a head appears, then deal that player from a well shuffled deck two cards. Otherwise deal the player one card. If one card is multiple of three then the player gets one point. If both cards are multiple of three then the player gets three points. This game is identical to the one you described above. As hard as it is to believe that it is a well known fact. You say that you tutor. Then you should have access to a reasonably good mathematics library. In Jim Pitman's book PROBABILITY there is an entire section on Symmetry. It is 3.5 in the edition that I have.Some editions of Larsen & Marx also deal with this material.

If one is dealt two cards, what are the probabilities of exactly one multiple of three; of two multiples of three?
Note that at a minimum, the first time player I can win is on game five.
Player I can win on game five, nine, thirteen, etc...

Even with the simplification given by symmetry this is still a beast of a question.
In order for player I to win on the fifth game, that player must have gotten three points on the first game & on the fifth game.
For that player to win on the ninth game, no one else has won on any of the first eight games.
You must list all possible situations that the player can win on the ninth game.

Using conditioning, given that none has won on the first eight games, what is the probability that player I wins?