Conquer Club

Does anybody know?

I'm too lazy to do the math

- As sometimes happens when you can cash mid-turn after an elimination.

What about doubling with 7?

Sometimes, clear thought is better than doing the maths. With 6 spoils, you are guaranteed to have a set. So you are left with 3 spoils, wishing for them to mix. That is 1/3
PS this logic would not apply if there was a case that selecting the "correct" set from the 6, would offer you a second set, but a "wrong" choice would not. Such a case does not exist though

yeah you are right. i think i have chosen the wrong way to cash with 7 before. so the odds are .33 for cashing with 6.

Is there a possible way to "cash wrong" with 7 spoils? Let me think about it a little

Yes, there is. 3 reds, 3 greens and 1 blue for example. You must not cash in the rainbow.
So, the calculations for 7 spoils does not grant equal result with that of 4 spoils, although I doubt it should be too difficult to calculate by hand.

I think the odds are pretty good of having 2 sets with 7 spoils,

all 7 of the same colour- 2 sets

6 of the same colour, 1 of a different colour- 2 sets

5 of the same colour, one of a second colour, one of a third colour- two sets
5 of the same colour two of a second colour one set

4 of the same colour- 3 of a second colour, two sets
4 of the same colour-two of a second colour, one of a third colour, two sets

if you drop to 3 of the same colour
3 of a second colour, one of a third colour- two sets
2 of a second colour, two of a third colour, two sets

so the only combo that would not give you two sets would be 5 of the same colour two of a second colour
something like that

so that would be
1* [(1/3)^5] *2* [(1/3)^2] *3

to get 5 same 2 different. 0.27 %, so 2 sets with 7 is 99.7%?

j1mathman wrote:so that would be
1* [(1/3)^5] *2* [(1/3)^2] *3

to get 5 same 2 different. 0.27 %, so 2 sets with 7 is 99.7%?

Of course my post doesn't really address the question asked odds of double cashing with 6 cards got the math on that? ?

Fewnix wrote:

j1mathman wrote:so that would be
1* [(1/3)^5] *2* [(1/3)^2] *3

to get 5 same 2 different. 0.27 %, so 2 sets with 7 is 99.7%?

Of course my post doesn't really address the question asked odds of double cashing with 6 cards got the math on that? ?

I was just doing the calculation based off the events you gave me. So it helped me, if I did the calculation right.

This is an ideal problem for a trinomial expansion!

There are 3^6 possibilities. Each digit in the trinomial expansion of 3^6 represents a different color partitioning.

1
6 6
15 30 15
20 60 60 20
15 60 90 60 15
6 30 60 60 30 6
1 6 15 20 15 6 1

Sum those numbers, and you should get 3^6, or 729 different possibilities.

Because of color symmetry and without loss of generality, we can consider 1/3 of the solutions out of 1/3 of the possibilities.

1
6 6
15 30 15
20 60 60
30 (One third of the 90 instances where there are two of each color)

So which outcomes represent two sets? Our proof is by exhaustion:

1: All Colors the same. Yes!
6 + 6: 5 of one color, 1 of another. No!
15 + 15: 4 of one color, two of another. No!
30: 4 of one color, one of each other color. Yes!
20: 3 of one color, 3 of another. Yes!
60 + 60: 3 of one color, 2 of another, 1 of a third color. No!
30: 2 of each color. Yes!

Summing the Yeses, we get 1 + 30 + 20 + 30 = 81. Out of 729 / 3 = 243 possibilities.

The odds are 81/243 or one in three.

TheFlashPoint wrote:This is an ideal problem for a trinomial expansion!

There are 3^6 possibilities. Each digit in the trinomial expansion of 3^6 represents a different color partitioning.

1
6 6
15 30 15
20 60 60 20
15 60 90 60 15
6 30 60 60 30 6
1 6 15 20 15 6 1

Sum those numbers, and you should get 3^6, or 729 different possibilities.

Because of color symmetry and without loss of generality, we can consider 1/3 of the solutions out of 1/3 of the possibilities.

1
6 6
15 30 15
20 60 60
30 (One third of the 90 instances where there are two of each color)

So which outcomes represent two sets? Our proof is by exhaustion:

1: All Colors the same. Yes!
6 + 6: 5 of one color, 1 of another. No!
15 + 15: 4 of one color, two of another. No!
30: 4 of one color, one of each other color. Yes!
20: 3 of one color, 3 of another. Yes!
60 + 60: 3 of one color, 2 of another, 1 of a third color. No!
30: 2 of each color. Yes!

Summing the Yeses, we get 1 + 30 + 20 + 30 = 81. Out of 729 / 3 = 243 possibilities.

The odds are 81/243 or one in three.

can you do odds of double cashing with 7 cards? I calculated it as being above 99%, but I'm not sure that is right.

j1mathman wrote:can you do odds of double cashing with 7 cards? I calculated it as being above 99%, but I'm not sure that is right.

Sure!

Using the same method:

1
7 7
21 42 21
35 105 105 35
35 140 210 140 35
21 105 210 210 105 21
7 42 105 140 105 42 7
1 7 21 35 35 21 7 1

Should sum to 3^7 or 2187.

We consider one third of the possibilities (which should sum to 729).

1
7 7
21 42 21
35 105 105 35
140 210

All same color. Yes! (1)
6 of one color, 1 of another. Yes! (7 + 7)
5 of one color, 2 of another. No! (21 + 21)
5 of one color, 1 of another, 1 of the last color. Yes! (42)
4 of one color, 3 of another. Yes! (35 + 35)
4 of one color, 2 of another, 1 of the last. Yes! (105 + 105)
3 of one color, 3 of another, 1 of the last. Yes! (140)
3 of one color, 2 of another, 2 of the last. Yes! (210)

As Fewnix noted above, only the 5/2 doesn't translate to two sets. Rather than summing the yesses, we can subtract the no's from the total possibilities.

(729 - 42) / 729 = 687 / 729 = 94.2%

Im not exactly sure if there's a more straightforward way to do this without using combinatorics.

5 out of 7 things, 1 out of 3 colors = nCr(7,5)*nCr(3,1) = 21 * 3
2 out of 2 things, 1 out of 2 colors = nCr(2,2)*nCr(2,1) = 1 * 2

21*3*2 = 126

out of 3^7 possibilities

These are my approximations:
Odds of cashing twice with 6 cards: 1/3
Odds of cashing twice with 7 cards: 2/3
Odds of cashing twice with 8 cards: 3/3

we calculated it for 7 and it was actually about 95%.

j1mathman wrote:we calculated it for 7 and it was actually about 95%.

I am famously not very good at maths or even counting, but I understand the truth lies between your two estimates. In fact, as per a post higher up, assuming you don't make a mistake cashing your first set and cash the wrong colour cads, you have the same chance of getting 2 sets with 7 cards as you do getting one set with 4 cards. The odds on that is a bit higher than 2/3 (which is the odds of getting a set with 4 cards if you don't have a set with 3 cards) but a lot lower than 94% or 97%.

Hope for it, don't stake your run on it.

Sometimes I've gotten a double-cash option but choose to hold back on the second cash anyway. Say if the next player to take out in the cascade only has 2 cards to win; if you cashed both you would wind up with only 2 cards and end of run; hold back and you have 5 cards and a sure option for more troops.

laughingcavalier wrote:

j1mathman wrote:we calculated it for 7 and it was actually about 95%.

I am famously not very good at maths or even counting, but I understand the truth lies between your two estimates. In fact, as per a post higher up, assuming you don't make a mistake cashing your first set and cash the wrong colour cads, you have the same chance of getting 2 sets with 7 cards as you do getting one set with 4 cards. The odds on that is a bit higher than 2/3 (which is the odds of getting a set with 4 cards if you don't have a set with 3 cards) but a lot lower than 94% or 97%.

Im confident in the odds I posted.

If the number of cards is divisible by three than the odds of cashing all the cards is exactly one third. All of the interesting math revolves around having a number of cards that is one more than a multiple of three.

The heuristic argument: In order to fail to cash in the last four cards, they have to be two pair. As you increase the number of cards, it becomes harder to be forcibly left with two pair.

The math:
1 set in 4 cards: 77.8%: 1 - 3 * (4 choose 2 ) / 3 to the 4th
2 in 7: 94.2%: 1 - 6 * (7 choose 2 ) / 3 to the 7th
3 in 10: 97.0%: 1 - 6 * (10 choose 2 + 10 choose 5 ) / 3 to the 10th
4 in 13: 99.5%: 1 - 6 * (13 choose 2 + 13 choose 5 ) / 3 to the 13th
5 in 16: 99.8%: 1 - 6 * (16 choose 2 + 16 choose 5 + 16 choose 8 ) / 3 to the 16th
6 in 19: 99.95%:1 - 6 * (19 choose 2 + 19 choose 5 + 19 choose 8 ) / 3 to the 19th

Another great RISK odds thread. Someone should compile all these.