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39 comments:
Good work as always, Red. The only request I have is please publish the last video(s) featuring Alexis and Ashley asap. Alexis is amazing.
Thanks.
SomeGuy.
Hi Red
I will buy absolutely anything with Candle in it. I'm her biggest fan and luckily she loses alot. Mallory is cute and I don't see how you can have a better body than Addie, but Candle is the ultimate. My ultimate game would be
Candle and Ashley with the ultimate forfeit. Anyway thanks and keep up the good work.
Hey Red, thanks for the pricing bud. I was kinda joking when I said that, but I appreciate it all the same.
Really enjoyed last weeks update. I didn't really care who won, and I think that would have been hot with any of those girls. I really liked the forfeit, and allowing the winning girl to choose who took off what.
I also really like the Candle, Addie and Malloy games, though I can't lie, the foot one wasn't all that enticing. And I even enjoyed 155, though I'm not sure I ever want to see that again.
Keep up the good work bud.
Random Guy
red i was just wondering you always seem to complain about the prices a C4S, why dont u start up ur own website where u can your own prices.
Thanks
Hi Red,
I bought the 2 clips from the double update weekend. Watched the game with the foot worship forfeit first. Loved the gameplay, but the forfeit wasn't up to my taste (I bought the clip out of curiosity. Wasn't sure what to expect from 'foot-worshipping'. But now I'm very sure it's absolutely not my cup of tea (and from the part of the forfeit I've watched, the 2 losers weren't too keen in doing the forfeit either).
Over to the 4-player game : this was a lot of fun. I sincerely believe not all players did understand the rules of the game and therefor messed up things completely during the game (making themselves lose pieces of clothing). Glad I didn't participate to the game myself, since I'm very sure I would end up completely naked too.
The winners had great fun during the forfeit. You have my permission to use this forfeit in other games ;-)
LOL, ofcourse you don't need my permission to do or don't do things.
@SomeGuy: Duly noted. I'll get that one into the queue, expect it within a month.
@Anon 10:43 (please sign your posts): Yeah, I know. That baby face, those shining eyes, and above all those dimples, she's really something else.
@RG: Oh, I knew you weren't seriously complaining, don't worry about it :)
@Anon 3:01 (please sign your posts): That's a very good question. The short answer is: convenience. See, lostbets.com isn't my primary gig. I've got a full-time job, which I love, and I started lostbets.com as a hobby. C4S handles billing, fulfillment, and customer service, and I used them because I just didn't have the time for that sort of crap. I still don't, and honestly, C4S isn't as bad as I paint them to be. They make a convenient target when I'm frustrated about something, though.
That being said, who's to say what the future holds? I've sworn off talking about what I'm planning to do (since I never seem to actually do it) but I do have plans and I am working on them.
@Myself: No, I don't need your permission, but it's nice to hear what people like, and it definitely influences what I do in the future. As for the foot thing: yeah, email I've received from folks who are, into feet confirmed what I'd feared: these girls are just too strong-willed and too unwilling to submit to dominance to have been good candidates for this forfeit. I'll try it again, although with a slightly different format. I'm pretty sure I can make it work.
Red,
I like the idea of the foot-fetish forfeit in principle... however you are right when you say that the girls were "too strong-willed and too unwilling to submit to dominance". What you need for that forfeit are some girls who understand that they've lost and who just get on with the licking, sucking & kissing, rather than whinging & giggling & time-wasting... but if they'd submitted properly it'd have been awesome.
But on the plus side - "nothing ventured nothing gained" and at least you now know how to improve so all in all it was still a good clip, and worth the money.
Red, sorry forgot to leave my "handle" with the last post I wrote:-
"I like the idea of the foot-fetish forfeit in principle... however you are right when you say that the girls were 'too strong-willed and too unwilling to submit to dominance'."
It's Simon Templar again
Red,
I have purchased a number of your harder videos. Love the site.
I most appreciate hardcore, couples group sex games. If I had my perfect game, it would be two teams of husband/wife or boyfriend/girlfriend with a participatory 'audience' of a couple extra males.
In my dream game, it is the guys who compete in a game of skill. Each round the winner removes a piece of clothing from the other guys’ girl.
Once a girl is fully naked, and her guy loses yet again, that team is the losing team.
Her guy has to get naked and watch as his girl is turned over to the rest of the group for whatever they want for as long as they want, until every guy (except her boyfriend/husband) is satisfied.
At the very end, after watching his girl get truly and completely fucked, the losing guy has to masturbate in front of everybody.
Call me a perv, but that is what I would like to see.
Thanks.
JW
Hi Red,
I would like to see a game between Ashley and Lilly with a high stakes forfeit because both players are very competetive. A good forfeit would be using the spreader bars and do the asshole tickling, etc.
JW - If Red filmed your idea, I would absolutely buy it.
Hi RED,
Any more games with Camilla, Taylor and Nicole. Would love to see more of them!
Cheers,
Hi,
the idea to tie up a girl with spreader bars is great (doggy style position would be great so you can see her pussy spread wide). The idea of spreading her cheeks and tickle her asshole with a feather or a brush is also very good the ass shaking of the tied up girl would be great.
It would be the best video ever if you could do this Red.
i like the hog-cuffed sex idea even better than that.
the winning girls get to move her into different positions while the lucky guy does what lucky guys do.
and yes, more camilla, taylor and nicole, if possible.
Shredder
More Taylor, definitely. Make that girl come on camera, please. :P
I prefer games where there's only one big loser. I am very curious about how Taylor reacts to being dominated / humiliated as sole loser.
JW, you're a perv, but then, aren't we all.
On the subject of 1 vs 2 girl forfeits, a suggestion: with 2 girls and dissimilar forfeits, once they've been determined, have them do a quick face off (high card, RPS, etc). Winner gets her choice of forfeits. For instance, someone once suggested a Bukkake forfeit; the winner could chose between being the "target" or a fluffer...
ta192
I know you probaly dont get asked this, would it be possible to have a POV where the player loses ?
BTW, I don't have to clarify what I mean by 'come' in my earlier post, do I ?
AV
Taylor, Camilla, Nicole...
let's subliminally suggest this to Red until he gets it
Old Guy
Yes Taylor and Camilla! sugestion +1.
Major Lurker
Taylor, Camilla, Nicole...
Shredder says so.
Sorry fellas,
Can't join you on this one. However hot Taylor, Camilla and Nicole are, (and they are all very cute) my favorites are Cora, Renna and Britney.
Bring them back!
JW
No problems with Taylor, Nicole and Camilla - but my favourites are Ashley, Ashton, Raven, Charlie and Julie. Would love to see one of them in a hardcore forfeit. But maybe that will never happen...
Ha! I'm still holding out for the return of Tristen.
I miss that amazon goddess.
-Piss Tron
I hope everything is well, Red. You're a patron of fine services and some of us at least hope you're doing ok. Been a while since we saw you around. Best wishes. -InternetHateMachine.
Game 195 is AWESOME!! First of all, the cleavage shots during the game are a great teaser, and the crawling over to get a card is a great visual tease too. Having the clothes get cut as a part of the game contributes to more ladies gettin' their clothes off - great touch Red. But that trib/scissors to orgasm showdown was incredible! Boobs bouncin', squeals and giggles from the players, and the tactic of pressing hard on the wand to get the other player to squeal and squirm was just great! Then finishing it off with the tickling, five minutes of boobs bouncing, women squirming, lotsa nudity in front of the camera, it's just a blast!!
I'll have to watch the clips4sale main page and see how high the standings on lost bets gets this week!
Gordon.
Phreeeow. I bet y'all thought I'd forgotten about you, didn't you?
Going back to December 1,
@SimonTemplar: Believe me, I have no regrets about shooting that one, even if it didn't turn out quite right and foot fans will have to wait awhile for a foot worship video worthy of the name. I did indeed learn from it, and I'm always glad to learn from experience when it makes my videos better.
@JW: Of course I'll call you a perv; it's a title you should wear proudly. I certainly do, which is why I say that your idea sounds hot as hell. For awhile, Amber and Mary have talked about doing a shoot with their husbands, but I don't know if they'd be willing to go quite that far. I'll certainly ask, and if not, I'll certainly look for a more daring couple of couples.
@Anon 12/3 12:29 (please sign your posts!): It'd be a good idea, if the two were ever in the same city at the same time. Lily (please do that classy lady the favor of spelling her name correctly) has retired from modeling, I believe, but I know she'd be happy to shoot with me. It's long past time she and Ashley met, don't you think?
@Anon 12/3 12:34 (please sign your posts!): As a matter of fact, I have done a game where the loser was put in the spreader bars and tickled everywhere,, including her asshole. Stay tuned.
@Anon 12/4 11:18 (yadda yadda): I don't think there's more of those three, but I may be wrong, (My copy of that footage is corrupeded, I have to go back to the archives to check. Hopefully, I'll find something.
@Shredder: As a matter of fact, I've used the hogtie sex forfeit, too. You may have to wait a bit for that one to be postproduced, though.
@ta192: That's a great idea. I'd love to do a bukkake forfeit (the Japanaese are pioneers in my genre and it's an honor to learn from them), but so far it's been a challenge finding players willing to risk it. I won't stop looking.
@Melcx: Believe it or not, you are far from the first who has asked. I'm considering it.
@AV: I've lost touch with Taylor. I'll redouble my efforts to find her again.
@a whole lot of people: and Camilla and Nicole as well. All I can do is try.
@InternetHateMachine: Seriously, man, that means a lot to me. Into each life some rain must fall, and when it falls into mine, knowing that there are folks like you out there helps me through. Cheers, mate.
@Gordon: So glad you liked 195. Hey, since 61% of you self-identify as geeks (a fact which makes me proud) maybe some of you will like this puzzle:
Four girls play a game. Each girl is assigned a color (red, green, blue, or yellow) and wears three articles of clothing. A stack of 24 cards contains two cards for each article of clothing worn by a player. Cards are drawn in essentially random order. The first time a specific card is drawn, that player must remove that article. The second time, that article is destroyed.
The game is played until two players are both naked and have had all their clothing destroyed -- i.e, cards of their color have been drawn six times. What are a player's odds of finishing the game wearing any clothing?
I'll give a free clip to whoever first posts the answer along with a reasonably-clear explanation of how it was derived. Hint: it's very, very low.
Ok here is my shot at the puzzle:
For one girl to be naked she would have to have her cards pulled only twice which makes the the statement P(x<3)=P(x=0)+(Px=1)+P(x=2). The probability that a persons card is pulled is 1/6 and therefore a persons probability that there card is not pulled is 5/6. The distribution will be binomial because of the fact that there are only two outcomes (card pulled or card not pulled). Since there is four different girls and the game is played until two girls are naked the minimum number of cards that would have to be pulled is 12. For a girl to end up with no clothes lost, P(x=0), there could be between 12-17 cards pulled. So by the binomial theorem the number of attempts will be between 12-17 which means the probability that a girl will remain fully clothed, P(x=0) is the sum of all of those probabilities. For a girl to end up losing only one article of clothing, P(x=1) means that there would be between 13-18 cards pulled and a girl with only one article of clothing left, P(x=2), means that there would be between 14-19 cards pulled. The probabilities of each of the attempts goes as fallowed:
Side note (12,0) means that out of the number of attempts that is the, 12 the number of articles of clothing lost were 0.
Side Note 2: the binomial probibilities has the fallowing set up: (attempts, aricltes lost)*(Probability of not losing an article of clothing)^(number of articles lost)*(probability of losing an article of clothing)^(attempts - articles lost)
Sorry it was so long it had to be on two posts
P(x=0), 12 attempts = (12,0)(5/6)^0(1/6)^12=4.59*10^-10
P(x=0), 13 attempts = (13,0)(5/6)^0(1/6)^13=7.6566*10^-11
P(x=0), 14 attempts = (14,0)(5/6)^0(1/6)^14=1.2761*10^-11
P(x=0), 15 attempts = (15,0)(5/6)^0(1/6)^15=2.127*10^-12
P(x=0), 16 attempts = (16,0)(5/6)^0(1/6)^16=3.54*10^-13
P(x=0), 17 attempts = (17,0)(5/6)^0(1/6)^17=5.9*10^-14
So the sum of all P(x=0) is 5.51261*10^-10 or
.000000000551261 This is the probability that a girl
will not lose an article of clothing
Now of the probabilities that a girl will lose only one article of clothing , P(X=1)
P(x=1), 13 attempts = (13,1)(5/6)^1(1/6)^12=4.593937*10^-9
P(x=1), 14 attempts = (14,1)(5/6)^1(1/6)^13=8.29461*10^-10
P(x=1), 15 attempts = (15,1)(5/6)^1(1/6)^14=1.48878*10^-10
P(x=1), 16 attempts = (16,1)(5/6)^1(1/6)^16=2.6585*10^-11
P(x=1), 17 attempts = (17,1)(5/6)^1(1/6)^16=4.726*10^-12
P(x=1), 18 attempts = (17,1)(5/6)^1(1/6)^17=8.37*10^-13
Ok here is my shot at the puzzle:
For one girl to be naked she would have to have her cards pulled only twice which makes the the statement P(x<3)=P(x=0)+(Px=1)+P(x=2). The probability that a persons card is pulled is 1/6 and therefore a persons probability that there card is not pulled is 5/6. The distribution will be binomial because of the fact that there are only two outcomes (card pulled or card not pulled). Since there is four different girls and the game is played until two girls are naked the minimum number of cards that would have to be pulled is 12. For a girl to end up with no clothes lost, P(x=0), there could be between 12-17 cards pulled. So by the binomial theorem the number of attempts will be between 12-17 which means the probability that a girl will remain fully clothed, P(x=0) is the sum of all of those probabilities. For a girl to end up losing only one article of clothing, P(x=1) means that there would be between 13-18 cards pulled and a girl with only one article of clothing left, P(x=2), means that there would be between 14-19 cards pulled. The probabilities of each of the attempts goes as fallowed:
Side note (12,0) means that out of the number of attempts that is the, 12 the number of articles of clothing lost were 0.
Side Note 2: the binomial probibilities has the fallowing set up: (attempts, aricltes lost)*(Probability of not losing an article of clothing)^(number of articles lost)*(probability of losing an article of clothing)^(attempts - articles lost)
P(x=0), 12 attempts = (12,0)(5/6)^0(1/6)^12=4.59*10^-10
P(x=0), 13 attempts = (13,0)(5/6)^0(1/6)^13=7.6566*10^-11
P(x=0), 14 attempts = (14,0)(5/6)^0(1/6)^14=1.2761*10^-11
P(x=0), 15 attempts = (15,0)(5/6)^0(1/6)^15=2.127*10^-12
P(x=0), 16 attempts = (16,0)(5/6)^0(1/6)^16=3.54*10^-13
P(x=0), 17 attempts = (17,0)(5/6)^0(1/6)^17=5.9*10^-14
So the sum of all P(x=1) is 5.604423*10^-10 or
.0000000005604423 This is the probability that a girl
will lose an article of clothing
Now of the probabilities that a girl will lose only one article of clothing , P(X=2)
P(x=1), 14 attempts = (14,2)(5/6)^2(1/6)^12=2.10555*10^-8
P(x=1), 15 attempts = (15,2)(5/6)^2(1/6)^13=3.509*10^-9
P(x=1), 16 attempts = (16,2)(5/6)^2(1/6)^14=5.84876*10^-10
P(x=1), 17 attempts = (17,2)(5/6)^2(1/6)^15=9.7479*10^-11
P(x=1), 14 attempts = (18,2)(5/6)^2(1/6)^16=1.6247*10^-11
P(x=1), 14 attempts = (19,2)(5/6)^2(1/6)^17=2.708*10^-12
So for the probability that a girl will only have one article of clothing left, P(x=2) is equal to 2.52566*10^-8 or .0000000252566
Now taking the values of P(x=0)+P(x=1)+P(x=2) which is equal to 3.1421794*10^-9 or 0.000000031421794 is the probability that one girl will remained clothed through the game.
sorry once again I had some pasting issues there but that is my attempt at the puzzle! Thanks red it was entertaining to work on!
Well, that was a nice try, but the worst-case odds of a girl having an article of clothing left is if the game ends with only 2 cards in play being the matching cards for her last remaining article of clothing.
The probability of that is 1/12 * 1/12, or about 0.7%. This is because there are two cards with that article, so the probability of one being in the last spot is 2/24 and the second-to-last spot is 2/24.
From there the chances get better because the game could end sooner, she could have more than a single article of clothing, etc. These are all additive to the probability, so in the end it's going to be greater than 0.7%.
Getting the numbers would take longer than any of the clips i haven't bought is worth, so i'll meet Red halfway. The probability is P(cards left = x) * P(her clothing left >= 1 given x cards left). For x from 2 to 12.
PN
Actually, x is from 12 to 3. It's impossible to have a game end with only 2 cards left. So it's actually a minimum possible probability of about 1.6%, with the caveat that once you derive the probabilities i posted above the chance will be higher.
PN
Well, the problem with posting anonymously is you can't delete your posts.
PN
Red,
I don't know the answer, but I agree GamacheRedSox is wrong.
The contest would be fairly simple if you were playing until three girls had all their clothes destroyed instead of only two. In that case each girl would have a 1.54% chance to still be wearing any clothes. (I can explain the math, if you are at all interested.)
Obviously, the chances of a girl keeping some clothes have to be larger if you are only playing until two girls have had all their clothes destroyed instead of three. So GRS's number is way too low. His math was way over my head, so I have no idea where he went wrong.
Dex
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