Best cards on hand ever ( crazy )

[NightWolf]

I was laughing up a storm today after I noticed the order, and names, of the cards I got for my Conquer man game I'm playing right now. Take a look, and see if you see something funny about my cards:



Now, would any interested mathematicians like to determine the odds of getting such a set of cards? this is killing me!

[Metsfanmax]

I count 151 territories. The chance of getting those cards in that order is then (1/151)*(1/150)*(1/149)*(1/148), or 2 * 10^-9 (i.e. the chances are one in 500 million). Of course, that ignores what the other players have, but it's a rough idea ;P

[Incandenza]

It would be cooler if you had the LI CK ME CC cards...

[AAFitz]

Thats pretty cool. Its also surprising you even picked up on it. I imagine most wouldnt have.

What are some other possibilities.

[AAFitz]

It would be cooler if you had the LI CK ME CC cards...

that has had to have happened at some point, has to be the best possibility...especially if its nuclear and you won every one of the spots.

AA cashed in LI CK ME and was eliminated from the game.

[Sir. Ricco]

I think the best ever would be DA MN BI CH. And yes, it is possible.

[karelpietertje]

I count 151 territories. The chance of getting those cards in that order is then (1/151)*(1/150)*(1/149)*(1/148), or 2 * 10^-9 (i.e. the chances are one in 500 million). Of course, that ignores what the other players have, but it's a rough idea ;P

It also ignores the amount of games that have been played with spoils on Conquer Man, and the amount of possible combinations that would have triggered a forum post.
imo, Lame Poem isn't that interesting a set of spoils.

[the.killing.44]

IM AF AG

but there is no IM

[Metsfanmax]

I count 151 territories. The chance of getting those cards in that order is then (1/151)*(1/150)*(1/149)*(1/148), or 2 * 10^-9 (i.e. the chances are one in 500 million). Of course, that ignores what the other players have, but it's a rough idea ;P

It also ignores the amount of games that have been played with spoils on Conquer Man, and the amount of possible combinations that would have triggered a forum post.
imo, Lame Poem isn't that interesting a set of spoils.

He only asked what the probability of getting those particular spoils in that order was, and my answer was right, except for what the opponents have (because that lowers the total number of cards left in the "deck"). That being said, the probability is the same for any particular set of four cards in a row.

[karelpietertje]

I count 151 territories. The chance of getting those cards in that order is then (1/151)*(1/150)*(1/149)*(1/148), or 2 * 10^-9 (i.e. the chances are one in 500 million). Of course, that ignores what the other players have, but it's a rough idea ;P

It also ignores the amount of games that have been played with spoils on Conquer Man, and the amount of possible combinations that would have triggered a forum post.
imo, Lame Poem isn't that interesting a set of spoils.

He only asked what the probability of getting those particular spoils in that order was, and my answer was right, except for what the opponents have (because that lowers the total number of cards left in the "deck"). That being said, the probability is the same for any particular set of four cards in a row.

False. From his post, I can only deduct he asked for the odds on such a set, not this particular set in this order.

Now, would any interested mathematicians like to determine the odds of getting such a set of cards?

The odds of getting a set that is funny in the same way remain not too small.

[Metsfanmax]

I count 151 territories. The chance of getting those cards in that order is then (1/151)*(1/150)*(1/149)*(1/148), or 2 * 10^-9 (i.e. the chances are one in 500 million). Of course, that ignores what the other players have, but it's a rough idea ;P

It also ignores the amount of games that have been played with spoils on Conquer Man, and the amount of possible combinations that would have triggered a forum post.
imo, Lame Poem isn't that interesting a set of spoils.

He only asked what the probability of getting those particular spoils in that order was, and my answer was right, except for what the opponents have (because that lowers the total number of cards left in the "deck"). That being said, the probability is the same for any particular set of four cards in a row.

False. From his post, I can only deduct he asked for the odds on such a set, not this particular set in this order.

Now, would any interested mathematicians like to determine the odds of getting such a set of cards?

The odds of getting a set that is funny in the same way remain not too small.

It's obvious that there are only a few sets that would be amusing in the same way, and the odds remain at under one in a million if there are less than 500 such ordered sets, which I presume is the case.

[pmchugh]

Haha brilliant

[AAFitz]

KI LL AA. HE AB AD PI CK. HE LA CK GG ON CC.

[Leehar]

HE AB AD PI CK. HE LA CK GG ON CC.

6 cards?

[AAFitz]

HE AB AD PI CK. HE LA CK GG ON CC.

6 cards?

I eliminated the other players.

[lordhaha]

OM FG KI LL ME !! 5 CARD SET

[NightWolf]

lol, I figured people might find this mildly interesting. To add to the mix, I would determine, by flying more numbers around, the odds of getting my unqiue set.

for this situation, 4 cards slots and 3 colors gives each color slot "12" states, which may hold 151 different cards, of which leaves us with 1,812 combos, which can be presented in 4 slots, leaving a final odds count of 10,780,347,822,336, or 10.78 trillion in one, presuming my math is correct. Giving the fact that I'm working with nested odds here, I've got a bit of room for error, since I have relatively little experience here. If I'm correct, that's a number that needs 44 bits for storage! lol...

[MrBenn]

I thought this was going to be a thread where I could share my game where I had an opportunity to cash in 3 rainbow sets (I started my turn with four cards and took out somebody with 5). TO OB AD

[tdans]

I thought this was going to be a thread where I could share my game where I had an opportunity to cash in 3 rainbow sets (I started my turn with four cards and took out somebody with 5). TO OB AD

nice one Mrbenn

[rdsrds2120]

I count 151 territories. The chance of getting those cards in that order is then (1/151)*(1/150)*(1/149)*(1/148), or 2 * 10^-9 (i.e. the chances are one in 500 million). Of course, that ignores what the other players have, but it's a rough idea ;P

It also ignores the amount of games that have been played with spoils on Conquer Man, and the amount of possible combinations that would have triggered a forum post.
imo, Lame Poem isn't that interesting a set of spoils.

The previous amount of games doesn't matter Getting that on your first game is the same odds of getting it on your 1243 game.

[stahrgazer]

He only asked what the probability of getting those particular spoils in that order was, and my answer was right, except for what the opponents have (because that lowers the total number of cards left in the "deck"). That being said, the probability is the same for any particular set of four cards in a row.

No... there was a thread where people got duplicates. From that thread, apparently each spoil pulls from a full deck, so your odds of getting any particular spoils is the same for every pull.

[Metsfanmax]

He only asked what the probability of getting those particular spoils in that order was, and my answer was right, except for what the opponents have (because that lowers the total number of cards left in the "deck"). That being said, the probability is the same for any particular set of four cards in a row.

No... there was a thread where people got duplicates. From that thread, apparently each spoil pulls from a full deck, so your odds of getting any particular spoils is the same for every pull.

Okay, so then my answer was correct from the start.

[ljex]

It would be cool if i got, IT IS I(i)J EX...too bad only one of those 4 territories exists

[Gilligan]

Does anyone else find it HUMORU(m)S that I have all of the China spoils?

Click image to enlarge.

[natty dread]

Slightly?