# #!@*?# Math

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Carlos
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Joined: 08/01/2009

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!

sedjtroll Offline
Joined: 07/21/2008
#!@*?# Math

I'm not sure about the math for the answer (in haste I have a guess, below), but if it turns out you do (or even if you don't), you could vary the values for each suit. Like for suit A you could have one 3, two 2's and five 1's or something.

Here's a guess...

So there are 15 total points in each color (4+3+2+2+1+1+1+1), and 7 colors, so there are 15*7=105 points total.

In your opening draw you will maybe have an average of 10/56 of those points in hand, or 18 to 19 points.

Since the cards are evenly distributed over the suits, then on the average you should have say 19/7=2 or 3 points per color.

Now, that's an average, so how often will you have significantly more than that in one color and less in another, and how oftn will that color be suit A? I don't know.

Finally, you could add a rule that says players dealt 5 or more points in suit A must declare as much, reveal their hand, and re-draw a new hand. It's wonky, but it might work if it only happens once in a while.

Emphyrio Offline
Joined: 02/10/2010
#!@*?# Math

I'm not a probability expert, but here's how I'd analyze the problem:

You have a 10 in 56 chance of drawing A4, and if you do, there is a 9 in 55 chance of drawing A3. So your chance of drawing A4 + A3 = (10*9)/(56*55) = about 2.9%.

Some are a bit trickier. For example, A4 + A2 + A1 you have 10/56 chance to draw A4, but since there are two A2 cards, your chance of drawing at least one is 1 - chance of not drawing either, which is 1 - (53/55 * 52/54 * 51/53 * ... * 45/47) = 30.3%. I'm sure there's an easier way to compute this. Then your chance of drawing at least one A1 is 1 - (50/54 * 49/53 * 48/52 * ... * 43/47) = 48.4%. So your total chance of drawing A4 + A2 + A1 = 10/56 * .303 * .484 = 2.6%.

Finally, you need to combine all these separate probabilities. These are not independent probabilities, so strictly speaking you can't just add them up. However, they're small enough that you can get an upper bound on the chances by adding them. Since you have two columns with 8 spaces, the upper bound is actually about twice as much. Then you also have to figure the chances you might get cards adding up to 9 for columns B and F, and so on... By this time your upper bound is becoming less and less useful.

Another alternative would be to write a program to enumerate all 3.5 billion possible hands (the binomial coefficient of 56, 10) and count how many of them fulfill the criteria. If I get a chance I'll try it.

Carlos
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Joined: 08/01/2009
#!@*?# Math

Emphyrio wrote:

Another alternative would be to write a program to enumerate all 3.5 billion possible hands (the binomial coefficient of 56, 10) and count how many of them fulfill the criteria. If I get a chance I'll try it.

I love you.

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

Thanks to for the responses so far.

Sebastian
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Joined: 07/27/2008
Re: #!@*?# Math

Carlos wrote:
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?

There is a 6% chance that this will occur. That is, with properly shuffled cards, it will come up every fiveteenth or sixteenth game.

Epigone
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Joined: 12/31/1969
#!@*?# Math

Emphyrio wrote:
Since you have two columns with 8 spaces, the upper bound is actually about twice as much. Then you also have to figure the chances you might get cards adding up to 9 for columns B and F, and so on... By this time your upper bound is becoming less and less useful.

Another alternative would be to write a program to enumerate all 3.5 billion possible hands (the binomial coefficient of 56, 10) and count how many of them fulfill the criteria. If I get a chance I'll try it.
It looks like you have information we don't... 9 for columns B and F, etc.?

For just column A you don't need to enumerate all of those hands, just... 60. And then multiply by the probabilities of getting those hands. I get the same result Sebastian got, 2279313410/35607051480 = 6.4%.

LSJ
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Joined: 12/31/1969
#!@*?# Math

If you want to analyze it for any color (possibly more than one at once), rather than just a single color, the solution is about 3 of every 10 hands.

Analysis follows:

Success means getting a single-color sum of 7 or more in a hand of 10.

Probability = #successful hands / # hands

# hands is the easy part: C(56,10) = 35607051480
That's number of ways to choose 10 things from 56 things.
C(n,k) = n!/(k! * (n-k)!)

# successful hands is the tricky part, especially since a hand that fails in getting to 7 in A may get to 7 in B.

First, calculate the prob of NOT getting a match given a set # of cards of a single color. For instance, if you get 3 cards of color A, what is the prob of not getting a 7 sum in A?

FailColor(N) = prob of missing 7 with N cards of a single color.

FC(8) = 0
FC(7) = 0
FC(6) = 0
FC(5) = 2 / C(8,5) = 2/56 = 1/28
FC(4) = (#1111 + #2111 + #2211 + #3111) / C(8,4)
= (1 + 8 + 6 + 4) / 70 = 19/70
FC(3) = 1 - ((#431 + #432 + #421 + #422) / C(8,3))
= 1 - ((4 + 2 + 8 + 1) / 56) = 41/56
FC(2) = 1 - (#43 / C(8,2)) = 1 - (1/28) = 27/28
FC(1) = 1
FC(0) = 1

Find out how likely it is to get exactly a particular pattern of color (sorting each hand by largest group of color to smallest group). For instance, to get a hand that looks like AAAAAAAABB, you have to pick a color to be A (7 choices) and a color to be B (6 choices). For A, you have to pick all 8 cards of that color (1 way to do that). For B, you have to pick 2 of the 8 = C(8,2) = 28 ways. So 7*6*28 = 1176 hands that match the pattern.

With this hand, you have FC(8)*FC(2) = 0*(27/28) = 0 probability of missing, or a prob 1 of success. Which means all 1176 hands of AAAAAAAABB are successful.

Shortcutting, we notice that all AAAAAAAAXX, AAAAAAAXXX, and AAAAAAXXXX hands are successful (with X being any color but A, to avoid overcounting), from the table of FC(N) values. The number of such hands is 7*C(8,8)*C(48,4) + 7*C(8,7)*C(48,3) + 7*C(8,6)*C(48,2) = 7*1*194580 + 7*8*17296 + 7*28*1128 = 2551724 (which includes the 1176 above).

#AAAAABBBBB = C(7,2)*C(8,5)*C(8,5) = 65856 (Note we use C(7,2) and not 7*6, since it A=red,B=green is the same hand as A=green,B=red).
Of these, FC(5)*FC(5) fail, so 1-FC(5)*FC(5) = 1-(1/28)(1/28) = 783/784 succeed. 65856 * 783/784 = 65772 successful hands

For AAAAABBBBC, we have 7*C(8,5)*6*C(8,4)*5*C(8,1) * (1-FC(5)*FC(4)*FC(1)) = 6521760 successful hands.

Successful hands in patterns:
AAAAAA??XX= 2551724
AAAAABBBBB = 65772
AAAAABBBBC = 6521760
AAAAABBBCC = 17974740
AAAAABBBCD = 82091520
AAAAABBCCD = 142618560
AAAAABBCDE = 325570560
AAAAABCDEF = 61931520
AAAABBBBCC = 13382565
AAAABBBBCD = 82355616
AAAABBBCCC = 19696005
AAAABBBCCD = 596245440
AAAABBBCDE = 675440640
AAAABBCCDD = 162772820
AAAABBCCDE = 206786160
AAAABBCDEF = 1244651520
AAAABCDEFG = 93585408
AAABBBCCCD = 119498400
AAABBBCCDD = 258966750
AAABBBCCDE = 213801840
AAABBBCDEF = 625766400
AAABBCCDDE = 945972160
AAABBCCDEF = 3013785600
AAABBCDEFG = 634454016
AABBCCDDEE = 60090681
AABBCCDDEF = 559204800
AABBCCDEFG = 325283840

Total successful hands: 10491066817

Odds of getting such a hand: 10491066817/35607051480 = 29.46%

Carlos
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Joined: 08/01/2009
#!@*?# Math

Thanks for helping me on this everyone.

Epigone (btw, that's a sad moniker! you are not an epigonist, but a creator of original games! :) ) and LSJ, what Emphyrio was alluding to in his post about varying lengths of columns was this less-than-lucid part of my post:

Carlos wrote:
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.

So to put in more clearly, let's assign each of the seven rows a letter, A-G; I'll put the length of each row next to each letter. Remember, the length of each includes the starting square:

A=8
B=10
C=12
D=14
E=12
F=10
G=8

So LSJ, the reason why I limited my question to column A was because the other columns (except G) are much less likely to allow a winning condition in the first hand--or so I thought. After seeing your analysis, I'm wondering if the possibility of getting an immediate winning hand in the other columns, especially B and F, might warrant their own statistical investigation.

But there are other factors in play. Every card has a total of four points on it, represented by the mix of symbols for the seven suits. So, for instance, there is an A A A A card, but there is also a B D E G card. And so the more cards that are used toward the conquering of one column, the less that are available to defend against your opponent's assaults. That's why I was initially more concerned with the outer columns and the 4-point and 3-point cards: there's little you can do with those cards except play them on the appropriate column. For cards with a mix of symbols, though, you have to choose which column you should play it, and choosing one will leave you more vulnerable on another.

After reading Sebastian's and Epigone's post that puts the chance at getting 7 or more points for the A column in the first column in the first draw of the game at 6%, I think I need to shrink the hand size. That just seems too lop-sidedly high.

Once more, thanks to all of you math wizzes for helping me and my math-addled brain out!

--Carlos

Epigone
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Joined: 12/31/1969
#!@*?# Math

Carlos wrote:
Epigone (btw, that's a sad moniker! you are not an epigonist, but a creator of original games! :) )

After reading Sebastian's and Epigone's post that puts the chance at getting 7 or more points for the A column in the first column in the first draw of the game at 6%, I think I need to shrink the hand size. That just seems too lop-sidedly high.
I use it in the sense of "standing on the shoulders of giants". It's good enough for Newton. :)

For reference, the probs for different hand sizes are:
2: 0.06%
3: 0.23%
4: 0.53%
5: 0.99%
6: 1.63%
7: 2.48%
8: 3.55%
9: 4.86%
10: 6.4%
11: 8.18%

Could the 4x cards just be too overpowered? They're easily several times as good as the variety-pack cards, since you should be able to get four times the effect from them for only a small versatility hit. Imagine a game where one player gets 5+ of the 4x cards and 5+ of the 3x cards throughout the game... this should happen uncomfortably often, does it lead to a pretty much guaranteed win?

Sebastian
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Joined: 07/27/2008
#!@*?# Math

Carlos wrote:

So LSJ, the reason why I limited my question to column A was because the other columns (except G) are much less likely to allow a winning condition in the first hand--or so I thought. After seeing your analysis, I'm wondering if the possibility of getting an immediate winning hand in the other columns, especially B and F, might warrant their own statistical investigation.

B has a probability of 1.28% of being instantly winable. Once every 78 draws or so.

Carlos
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Joined: 08/01/2009
#!@*?# Math

Thanks again guys; this is exactly the info. I needed. Much appreciated. 