Okay boys and girls, I have an advanced degree in English, which means by definition I have no ability in Mathematics. I have a probability question that game up after a recent playtesting session that I really don't know how to answer. Hoping some of the fabulous folks here will look at this and say: "Pfft! That's easy." :)

My game is a boardgame that uses a deck of cards. The goal of the game is to conquer four of the seven columns that are on the board. For simplicity's sake, let's say you win a column by starting at one end and moving to the other, and the columns are of different length. The first column has 8 squares that must be traversed, the starting square plus seven others; the 2nd 10; the 3rd 12; the 4th 14; the 5th 12; the 6th 10; and the 7th 8. Here's what the first row looks like, with the "A" indicating the starting square:

A

1

2

3

4

5

6

7

So assume for purposes of this example that your goal for any given column is to start on the letter of a given column and move to the terminal number in that column (in the example above, 7 is the terminal number). The way that you get there is by playing cards, moving your token across the column in accordance with the value of all the column-appropriate cards you play.

Your hand size is ten cards. The deck for the game is a 56 card deck with seven suits, one suit corresponding to each column. You can only play cards of the suit that correspond to the column in play to advance your token. Each suit has

* One card with a value of 4

* One card with a value of 3

* Two cards with a value of 2

* Four cards with a value of 1

In a typical hand, you will get a mix of cards that are playable on several columns. BUT, finally, here is the question that came up after playtesting:

--Is there too great a probability that in YOUR FIRST HAND you will draw enough cards FOR COLUMN A that add up to a value of 7, thereby guaranteeing you a victory on that row?--

This is a strategic two player game with elements of luck, and I'm trying to make sure that luck does not govern more than strategy. But I don't know how to determine the probability of this.

Here's as far as I got: I know (or at least I think I know) the chance of drawing the A card with the value of 4, followed by the A card with a value of 3, to be 1 in 3,080. I got that by multiplying the probability of drawing A4 (1 in 56) by the chance of consecutively drawing the A3 card (1 in 55). But the hand size is 10. So do I then divide 3,080 by 10 to get the probability of drawing those two cards in the first round (1 in 308)? That doesn't sound right to me, but I don't know how to fix it.

And then it gets more complicated. Here are all the different ways you can get a value of seven (or more) A cards in your first draw:

A4 + A3

A4 + A2 +A2

A4 + A2 + A1

A4 + A1 + A1 + A1

A3 + A2 + A2

A3 + A2 + A1 + A1

A3 + A1 + A1 + A1 + A1

A2 + A2 + A1 + A1

A2 + A1 + A1 + A1 + A1

So, what are the chances that you will get any combination of cards in your first hand that will add up to a value of seven?

I know that Excel wizards could work up a wonderful spreadsheet that would do all of this for you, but, alas, I am not even an apprentice Excel person. Any help with this math would be much appreciated!

Emphyrio wrote:I love you.

I mean ... thanks! I'd be really grateful! :)

Thanks to for the responses so far.