I suck at Maths

For those that don't - what's the probability of this happening. I have a crypt of 13 minions, 9x Imbued and 4x a Vampire. I don't get the Vampire in my opening crypt, I pay 1 to see 1 until I get the Vampire, but find that the Vampire is the final 4 cards of the crypt. Yes, this happened to me the other night. I would like to know how unlucky I was. It was a bad night for me. I also played a Parity Shift deck packing 8 Shifts in 75 cards. I went through close to 30 cards and still didn't find a single Shift, and effectively died to my own First Trad. --> J grail_pbem "at" hotmail.com

[ quoted text not captured ] I suck too, but I have this cool link here for you and me and all the other "suckers" ;-)) http://irafay.com/combo.php "A combo card drawing simulation" Cheers, Alf - can't do nothing about your luck though

> I have a crypt of 13 minions, 9x Imbued and 4x a Vampire. > > I don't get the Vampire in my opening crypt, I pay 1 to see 1 until I > get the Vampire, but find that the Vampire is the final 4 cards of the > crypt. The math for this one is pretty simple. The number of possible draws (including multiple permutations where different copies of the same card show up in different places) is 13!, or 13x12x11...x2x1. With 4 copies of the vampire in question, there are 4! ways to arrange the different copies, and there are 9! ways to arrange the cards above them in the crypt. So the probability comes out to p = 9! x 4! / 13! = 1 / 715 or .0014 > Yes, this happened to me the other night. I would like to know how > unlucky I was. Very. > It was a bad night for me. I also played a Parity Shift deck packing > 8 Shifts in 75 cards. I went through close to 30 cards and still > didn't find a single Shift, and effectively died to my own First Trad. The math on this one is harder, since you have to account for all the different ways you could draw 30 cards without seeing a Parity Shift. I'm not inclined to do it right now. I am however inclined to say that given both of these unlikely circumstances, there is probably a common variable involved other than pure randomness. In other words, you should probably shuffle your decks better.

CthuluKitty wrote: > >> It was a bad night for me. I also played a Parity Shift deck packing >> 8 Shifts in 75 cards. I went through close to 30 cards and still >> didn't find a single Shift, and effectively died to my own First Trad. > > The math on this one is harder, since you have to account for all the > different ways you could draw 30 cards without seeing a Parity Shift. I make the probability of drawing 30 cards without seeing any of 8 Parity Shifts in a 75 card deck (67! 45!) / (75! 37!) or roughly a 1.3% chance. Appolonius.

> I'm not inclined to do it right now. I am however inclined to say > that given both of these unlikely circumstances, there is probably a > common variable involved other than pure randomness. In other words, > you should probably shuffle your decks better. Oh, they were shuffled properly.... i shuffled the Parity Shift deck for about 15 minutes before the start of the game. [ quoted text not captured ]

> > Oh, they were shuffled properly.... i shuffled the Parity Shift deck > for about 15 minutes before the start of the game. > > --> J > grail_pbem "at" hotmail.com assuming they were, you had: 67/75 chances to have a card different than a Parity shift as the first card you draw 66/74 chances to have a card different than a Parity shift as the second card you draw 65/73 chances to have a card different than a Parity shift as the third card you draw ... so if you didn't geta Parity shift in your top 30 cards, that means: 67! 46! / 38! 75! = 46*45*44*43*42*41*40*39 / 75*74*73*72*71*70*69*68 = 1,55% good luck, Jim ^^

> I don't get the Vampire in my opening crypt, I pay 1 to see 1 until I > get the Vampire, but find that the Vampire is the final 4 cards of the > crypt. > > Yes, this happened to me the other night. I would like to know how > unlucky I was. Number of combinations possible with 13 crypt cards: 13! = 6,227,020,800 Number of combinations that have the 4 vampire cards as bottom four cards: 4!*9! = 8,709,120 Odds of this happening: 8,709,120 / 6,227,020,800 = 0,001389, or about 0,14%. Another way to look at it: Number of ways to put 4 vampire cards in a stack of 13: 13nCr4 Numbers of ways to have them as the four bottom ones: 1 Odds of it happening: 1 / 13nCr4 = 1 / 715 which also is 0,001389 or 0,14%. In short: you were "rather unlucky" :-)

J <gra...@hotmail.com> wrote: > For those that don't - what's the probability of this happening. > I have a crypt of 13 minions, 9x Imbued and 4x a Vampire. > > I don't get the Vampire in my opening crypt, I pay 1 to see 1 until I > get the Vampire, but find that the Vampire is the final 4 cards of the > crypt. > > Yes, this happened to me the other night. I would like to know how > unlucky I was. You weren't unlucky, and your problem has nothing to do with luck. You just have bad shuffling skill. :) Kevin M., Prince of Las Vegas "Know your enemy, and know yourself; in one-thousand battles you shall never be in peril." -- Sun Tzu, *The Art of War* "Contentment... Complacency... Catastrophe!" -- Joseph Chevalier

J <gra...@hotmail.com> wrote: > i shuffled the Parity Shift deck for about 15 minutes > before the start of the game. You are shuffling too much. Pile shuffle in any way you like, then riffle-shuffle a few times. Then STOP. :) [ quoted text not captured ]

> You are shuffling too much. Pile shuffle in any way you like, then > riffle-shuffle a few times. Then STOP. :) I might have to do that... I shuffled for so long because I was waiting for the next game to start after getting ousted in the game prior. And there's only so much death match Red Faction you can play whilst waiting. [ quoted text not captured ]

I agree with Kevin. You are a bad shuffler. Dont you ever tough my decks!!! :) greetz, Oortje

[ quoted text not captured ] Is toughing a deck like thugging it? :) Seriously though, the optimum randomization of a deck occurs at 3 shuffles, ie riflle shuffle, mix shuffle, repeat 2 more times, STOP!

Two more math questions. 1) If my crypt is 4x copies each of three different guys, what is the chance that all three of them will show up in my opening crypt draw? 2a) If my crypt is 3x copies each of four different guys, what is the chance that all four of them will show up in my opening crypt draw? b) What is the chance that at least three of them will show up in my opening crypt draw? Thanks! John Eno

[ quoted text not captured ] 1) 58,1% 2) a) 16,36% b) 28.181%

Kushiel wrote: > Two more math questions. > > 1) If my crypt is 4x copies each of three different guys, what is the > chance that all three of them will show up in my opening crypt draw? Choose 1 to double-up on, then draw two of those: C(3,1) * C(4,2) Draw 1 of each of the others: * C(4,1) * C(4,1) Divide by number of possible hands: / (12*11*10*9 / 4!) 3 * 6 * 4 * 4 * 24 / 12*11*10*9 = 58.1% > 2a) If my crypt is 3x copies each of four different guys, what is the > chance that all four of them will show up in my opening crypt draw? 3/12 * 3/11 * 3/10 * 3/9 * 4! = 16.4% > b) What is the chance that at least three of them will show up in my > opening crypt draw? First compute the chance of getting exactly 3: (= choose 1 to double-up on, choose 2 to get singles, then draw them) 4 * 3 * 3 * 3 * 3 * 4! / 12*11*10*9 = 65.5% So the odds of getting at least three = 65.5% + 16.4% = 81.9% (or compute the odds of seeing only two: C(4,2) * C(6,4) * 4! / 12*11*10*9 = 18.1 and subtract from 100%)

On Apr 20, 2:41 pm, LSJ <vtes...@white-wolf.com> wrote: > Kushiel wrote: > > Two more math questions. > > > 1) If my crypt is 4x copies each of three different guys, what is the > > chance that all three of them will show up in my opening crypt draw? <snip> > First compute the chance of getting exactly 3: > (= choose 1 to double-up on, choose 2 to get singles, then draw them) > > 4 * 3 * 3 * 3 * 3 * 4! / 12*11*10*9 = 65.5% > > So the odds of getting at least three = 65.5% + 16.4% = 81.9% > > (or compute the odds of seeing only two: C(4,2) * C(6,4) * 4! / 12*11*10*9 = > 18.1 and subtract from 100%) -on 2b) I've forgot to multiply by 4 and LSJ added extra 3. This is exact: 4 * 3 * 3 * 3 * 4! / 12*11*10*9 = 28.181% 28.181% + 16.4% = 44,581%

[ quoted text not captured ] Um, there was no extra three in my formula (as you can verify by the computation of seeing only two, as shown above, and comparing the answers). Choose 1 to double and take 2 of those: C(4,1) * C(3,2) = 4 * 3 Then choose two to get singles of, and choose 1 of each: C(3,2) * C(3,1) * C(3,1) = 3 * 3 * 3 That makes (4 * 3) * (3 * 3 * 3) Then divide by the number of different draws of 4 (12*11*10*9 / 4!) That makes 4 * 3 * 3 * 3 * 3 * 24 / 12*11*10*9 = 0.6545454... = 65 and 5/11 percent.

[ quoted text not captured ] You have just chosen three instead of two. It should be: C(3,1) * C(3,1) * C(3,0) = 3 * 3 * 1 C(3,0) means that you have chosen none of the ramaining 3 copies of 4th vamp. > > That makes (4 * 3) * (3 * 3 * 3) > > Then divide by the number of different draws of 4 (12*11*10*9 / 4!) > > That makes > > 4 * 3 * 3 * 3 * 3 * 24 / 12*11*10*9 = 0.6545454... = 65 and 5/11 percent.- Hide quoted text - > > - Show quoted text - This is how i did ti: Eg. we have vamps 3xA, 3XB, 3xC, 3xD in crypt Uncontroled region posibilities for EXACTLY tree diferent vamps (order inside row is irelevant): 1. AABC 2. ABBC 3. ABCC 4. BCDD 4 * C(3,2) * C(3,1) * (3,1) * C(3,0) / C(12,4) = 4 * 3 * 3 * 3 * 1 *4! * 8! / 12! = 0,21818182 C(3, 2) - chose 2 of 3 copies of vamp (eg. AA) C(3,1) - cohose 1of 3 copies of vamp (eg. BC) C(3, 0) - cohse none of 3 copies of vamp (eg. no D) C(12, 4) - chose 4 of 12 copies of crypt cards 4 - four ways to chose vamp which will show up in 2 copies (we can chose eather A,B,C or D to be in 2 copies)

ResurrectioN wrote: >> Um, there was no extra three in my formula (as you can verify by the computation >> of seeing only two, as shown above, and comparing the answers). >> >> Choose 1 to double and take 2 of those: >> C(4,1) * C(3,2) = 4 * 3 >> >> Then choose two to get singles of, and choose 1 of each: >> C(3,2) * C(3,1) * C(3,1) = 3 * 3 * 3 > > You have just chosen three instead of two. No. Choose two to get singled. You have three to choose from, since you cannot choose the one you've already chosen to double. So that makes: C(3,2) Say you choose Bob and Rob. Then choose 1 of the Bobs. C(3,1) Then choose 1 of the Robs. C(3,1) > It should be: > C(3,1) * C(3,1) * C(3,0) = 3 * 3 * 1 > > C(3,0) means that you have chosen none of the ramaining 3 copies of > 4th vamp. But you've neglected to choose which of the vampires is the fourth. If you like, choose 1 to get doubled and choose 1 to omit. A. C(4,1) <- choose who gets doubled B. C(3,2) <- choose two copies of that vampire C. C(3,1) <- choose who gets omitted D. C(3,1) <- choose 1 of the other vampires (neither omitted nor doubled) E. C(3,1) <- choose 1 of the remaining vampires (not chosen in A,C, or D above). And, again, this can be verified by computing the probability of seeing exactly 2 vampires. >> That makes (4 * 3) * (3 * 3 * 3) >> >> Then divide by the number of different draws of 4 (12*11*10*9 / 4!) >> >> That makes >> >> 4 * 3 * 3 * 3 * 3 * 24 / 12*11*10*9 = 0.6545454... = 65 and 5/11 percent.- Hide quoted text - >> >> - Show quoted text - > > > This is how i did ti: > > Eg. we have vamps 3xA, 3XB, 3xC, 3xD in crypt > Uncontroled region posibilities for EXACTLY tree diferent vamps (order > inside row is irelevant): > 1. AABC > 2. ABBC > 3. ABCC > 4. BCDD > What about AABD AACD ABBD BBCD ACCD BCCD ABDD ACDD > 4 * C(3,2) * C(3,1) * (3,1) * C(3,0) / C(12,4) = 4 * 3 * 3 * 3 * 1 > *4! * 8! / 12! = 0,21818182 > > C(3, 2) - chose 2 of 3 copies of vamp (eg. AA) > C(3,1) - cohose 1of 3 copies of vamp (eg. BC) > C(3, 0) - cohse none of 3 copies of vamp (eg. no D) > C(12, 4) - chose 4 of 12 copies of crypt cards > 4 - four ways to chose vamp which will show up in 2 copies (we can > chose eather A,B,C or D to be in 2 copies) and still needed: 3 -- three ways to choose vamp which will not show up at all (we can choose any of the three not already chosen to be the double-up vampire).

Freaks the lot of you... :P [ quoted text not captured ]