How do you get 68% chance??? If you have 22 cards 2 fog and 3 horn there is no way it's 68% to pull one fog AND one horn?
68% for a fog only. I can reply something like fine I can add another fog to my deck as it won't change much, but instead let's turn the tables a little. If you insist on getting combined probabilities, note that now as you've pulled 15 cards, you get an honest, even if hardly 'guaranteed', 50% of getting the dragon. However, you must multiply it by the chance of having at least one decoy (89%) and at least one medic (76%). That's already just 34% - and your post-frost anti-monsters strategy relies on decoying and res'ing the dragon. Here's the kicker though - with a 34 cards deck, your chances of getting 2 spies are just 36%. I'm assuming neither of us has Avallachs here, since even though it'll give you (still unreliable) 55% chance of getting 2+ spies, I'll have a reasonable chance of getting it as well. Sure, monsters risk drawing musters with spies, but it's greatly reduced if a couple of musters are already used up.
Wait a sec. Your numbers are off. My total deck is 22 base plus 12 special (3 frost, 3 decoy, 3 horn and 3 scorch) for a total of 34 cards. I pull 10 have 2 changes then get 4 more on spy cards for 16 cards or half the deck. And you are saying I don't have a frost and one more frost or scorch? Remember you already ruled that my other two spy cards, and 2 reavers weren't picked. It is indeed very likely that I have 1 of each of the specials. Please show the math where I only have a 43% chance to have a frost/scorch or frost/frost.
Well your starting hand had 10 cards with no scorches (not sure why you chose it to be this way). Now that you've pulled 4+1 more, you chances of getting at least one scorch went up to 47%. Better, but still not 'most likely'. They are quite a bit lower for another frost or decoy. As for the reavers, I actually
expected you to have a pair, that's why I took the fogs.
I compute chances like this. If you have a remaining deck of N cards, M of which are the ones you need, your chances of
not getting it after C pulls are p0 = (N-M)/N * (N-M-1)/(N-1) ... (N-M-(C-1))/(N-(C-1)). Correspondingly, your chances of getting at least 1 are 1-p0. Your chances of getting
exactly one in C pulls are a bit trickier. You'd have to sum the probabilities of getting at an i[SUP][SUB]th[/SUB][/SUP] draw, multiplied by chances of
not getting it before and after (I can post a code fragment if you like). This p1 can be used to get the chances of having more than one such card as 1-p0-p1.
You have a 22 deck plus 5 specials (3 horn, 2 fog) and were able to get a horn and fog. Not to mention I already gave you the unlikely situation of getting a vampire, crone and anarachs card without having a duplicate in your hand. That is one pretty lucky pull but I wasn't complaining.
That's not
so lucky actually. I played quite a few monsters games, and very rarely I wasn't able to correct the initial set with 2 re-draws.
I play a frost card which means your point total is now (6 siege, 10 hero, 10 close combat which are 5 cards plus horn plus frost) 26. I play a 8 point siege and then Foltest on the siege row. I now have (10 hero, 2 points range which are 2 cards and fog and 16 siege 8 point with horn) 28 points.
I win the round 28 -26.
You have 6 cards in your hand and I have 8. I had 9 played 2 plus Foltest and got one for NR faction. BTW my odds of getting another reaver or that scorch that I might not have had has increased.
It is your turn to start round two.
I'm pretty sure the player who won a round starts the next one, actually. Just checked it in game.
What was the power of your leader?
Why, the one that lets me play one of my discarded cards, of course.
