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**19 members have voted**

I would think that's a bad sign.Quote:RSYou sure? Doesn't show up on his recent posts list. But it does say he logged in today. So hopefully that's a good sign.

Can I get action on that? ;-)Quote:PokerGrinder...One was so disgusting that I can now say I will never eat one again.

Wonder how many nuggets she can down in one night if you know what I mean...

Quote:EvenBobMission's humiliation hits a new high. You

need to vindicate yourself, get back in

the game.

Why would I be humiliated? I literally choked, it had nothing to do with my stomach. There are quite a few who can attest that I slammed down beer after beer for hours almost immediately after my attempt, and then after that, I tore up the food court at...maybe Fremont?

Quote:WizardI

#2: Mission146: This one also ended in mid 80's for reasons debated to this day. As I understand it, a chunk went down the wrong pipe and Mission choked, literally, but never vomited. However, he did this over a trashcan so it looked like he was. However, he never ate more so nobody disputes he fell short of the 100 goal.

I swear on my life I didn't, but I thought that I might from the choking, that's why I went over the trashcan to be safe. My stomach felt just fine.

Quote:TomGFor this past week I've been telling myself I will never eat another large meal again. But I'll have nothing to do this summer and lots of buffet comps.

So, what do you demolish at a buffet? 150 oysters?

Quote:Wizardofnothinghttp://nypost.com/2017/04/30/woman-agreed-to-trade-sex-for-mcnuggets-cops/

Wow. What I'm confused about sort of, is, why would the undercover cop haggle her on the price? Perhaps I don't understand how undercover anti-prostitution cops or w/e work. But if it were a drug bust with a dude wearing a wire, I'm pretty sure he wouldn't be like, "Okay, you're offering $50k for 10 pounds of meth.....let's make it $5k for 20 pounds." then the drug dude being like, "Hmm....how about 15 pounds for $35k?" then they haggle back and forth. If anything, the undercover dude would be like, "You're offering $50k for 10 pounds, sure, let's do it" and not try to be a cheap-a** and haggle the price down.

Then again, maybe the undercover dude's natural instincts came in and he was thinking, "Nah, no way, this b**** nasty as all hell.....I'll drop the price from $100 to $25...and I'll throw in some chicken nuggets, she'll be all over that."

Why were you confident that you could beat the precious record by 50%? Have you done competitive eating before?

Then again, he didn't strike me as the kinda person that'd do something like that.

Also, since it was known he ate the 50 pancakes or whatever at Dupars (place at golden gate) with a relative breeze, likely would'a been difficult to get himself some good action. Then again, (theoretically) if someone thinks betting that a German doesn't know the capitol of Germany is a good bet, I guess anything's possible.

Added angle: playing gets anybody unbanned.

Quote:RigondeauxWait, the answer is so obvious.

Math puzzle time!

Assuming a revolver, with six chambers and one bullet, rotating the chamber before each pull, what is the probability the one to go first dies?

The way you wrote it, the answer is 100%.Quote:WizardMath puzzle time!

Assuming a revolver, holding six bullets is used, rotating the chamber before each pull, what is the probability the one to go first dies?

I assume you're talking about a typical six-shooter with only one of the six chambers loaded, and the barrel is spun with what looks like a random result.

The knee-jerk answer would be, duh, 1 in 6.

Of course, since Mike is asking, the answer obviously isn't 1 in 6.

So, recalling that Mike recently posted about biased coin flips, I'll bite that he read about this somewhere. In short:

Someone playing would load the top most chamber, so that when the barrel is closed, the bullet is in position to fire.

I seem to think that a barrel needs to be turned while loading. If so, that adds to the probability that the single bullet is usually loaded into a specific position. Possibly the firing position.

Now on to the 'random' spin.

A roulette wheel, or big six wheel, is designed to have minimal friction and a truly random spin duration. A gun barrel? Not so much.

I'm betting that the article Mike read said there is a huge bias in a gun barrel spin for a non-whole number of rotations, making the odds of dying on the first try much lower than 1 in 6.

It's all about the law and intent. They do this for a bunch of reasons... One being they want it to be as real as possible and of course people try to save money and haggle over anything, even drugs. Two, they might let the purchase happen and take the dealer down later and try to save the undercover identity, again, needing to make it as real as possible. Three, perhaps it's known in the community that they intentionally overbid and only cops say "yeah sure okay" to the first high ball price. Four, it shows further and further thought and intent in to exactly what they're doing. There is no insanity plea when you show cognitive responses and thought processes to haggle exact prices and amounts over and over rather than a one and done discussion.Quote:RSWow. What I'm confused about sort of, is, why would the undercover cop haggle her on the price? Perhaps I don't understand how undercover anti-prostitution cops or w/e work. But if it were a drug bust with a dude wearing a wire, I'm pretty sure he wouldn't be like, "Okay, you're offering $50k for 10 pounds of meth.....let's make it $5k for 20 pounds." then the drug dude being like, "Hmm....how about 15 pounds for $35k?" then they haggle back and forth. If anything, the undercover dude would be like, "You're offering $50k for 10 pounds, sure, let's do it" and not try to be a cheap-a** and haggle the price down.

Then again, maybe the undercover dude's natural instincts came in and he was thinking, "Nah, no way, this b**** nasty as all hell.....I'll drop the price from $100 to $25...and I'll throw in some chicken nuggets, she'll be all over that."

If you ever watched To Catch a Predator with Chris Hansen on MSNBC they'd always ask the guys to bring specific items when coming over. They said this furthered their case that it was a well thought out plan and that it showed exact intent because sometimes they'd show up and say "I was just here to check on her... not have sex!" but then they found the requested hand cuffs and Mikes Hard Lemonade in the car, and that spoke louder about their intent than their words.

I think DJT answered this correctly - The way you worded it is ambiguous. You didn't say he'd die from the game nor a gunshot and it is inevitable that we will all die one day, so 100%.Quote:WizardMath puzzle time!

Assuming a revolver, holding six bullets is used, rotating the chamber before each pull, what is the probability the one to go first dies?

Also didn't state how many bullets were in the 6 shooter... if 6, 100%, if 0, 0%... and if 1, 1/6 = .1667%... same as a 7-out in craps. So if you have "gun influence" you can shave your odds and get a great +EV situation.

Quote:DJTeddyBearThe knee-jerk answer would be, duh, 1 in 6.

Yes, I meant one bullet and six chambers. I thought it was understood but two players take turns until one loses.

Oh! That's different. My response was regarding the first SHOT. I wasn't thinking about if the gun gets returned.Quote:Wizard.... two players take turns until one loses.

I assume you mean a single bullet in a six shooter, with two 'players' going back and forth until somebody dies. Well, the first player is more likely to die simply because he was first.

So, the first player has a slightly more than 50% chance of eventually dying. The question is, how much more.

Now we're talking about a straight math problem, and assumes the barrel spin is truly random.

OK. I tried to wrap my head around it, but I'm failing.

Oh, sure, I can tell you that a single person playing alone has less than 50% chance of surviving 4 shots, less than 10% chance of surviving 13 shots, and less than 1% chance for 26 shots. But which of two people dies when passing back and forth? You got me!

"Gun influence" aside:Quote:WizardYes, I meant one bullet and six chambers. I thought it was understood but two players take turns until one loses.

If we go first:

1/6 chance to lose, if not 5/6 times we'll be safe.

Other guy will have 1/5 chance to lose, 4/5 chance to pass it back to us.

Then we have a 1/4 chance to lose, 3/4 chance to pack back...

Other guy has 1/3 chance to lose, 2/3 chance to pass back.

We have 1/2 chance to lose and 1/2 chance to pass back.

Other guy dies.

It all depends if the other guy finds the bullet or not:

- If he finds the bullet on his first attempt, we only had 1/6 chance of dying going first.

- If he finds the bullet on his 2nd attempt, we had 1/6 + 1/4 chance of dying going first.

- If we don't find the bullet on our 3rd attempt, he's doomed and we have a 0% chance of dying at this point.

What about us "don't" betters? Would be just be the opposite?Quote:AyecarumbaHow about a "Craps Golden Arm Challenge"? The winner will be the shooter who totals the most rolls in three turns. The tie-breakers could be longest single shoot, most points, most hardways, etc. Easy to watch, and easy to side bet. Not easy to record video though.

I'm thinking each player spins the barrel every time.

Ah, so with replacement essentially...Quote:DJTeddyBearRomes -

I'm thinking each player spins the barrel every time.

Each turn would be a 1/6, but we'll always be a 1/6 ahead of our opponent for going first. So it's going to be some fancy calculation of the probability over number of trials... + 1/6

Quote:RomesWhat about us "don't" betters? Would be just be the opposite?

Hmm, I don't think that would be as fun, but maybe one of the side bets could be who makes more at the end of three turns, the Do's or the Don'ts? Of course, everyone would have to start with same buy-in.

Quote:DJTeddyBearRomes -

I'm thinking each player spins the barrel every time.

If the other chambers are empty, the weight of the cartridge should influence the loaded chamber to end up at the bottom position. I don't think it is random.

How you hold the gun while spinning the chamber becomes very important. If you hold it sideways ("gangsta style"), allowing the cylinder to spin below the frame, then swinging it back into the locked position, I think there is a better chance of the cartridge being in the firing position, than if you were to spin the cylinder holding the frame in upright in a shooting position. The chamber with the bullet should tend to the bottom position, furthest from the firing position.

I think this only applies to swing-out cylinders. There are "break frame" models out there too, where the cylinder doesn't swing out the side.

As to Russian Roulette, with 1 bullet in a 6 chamber revolver, I think your odds would be slightly better going first, because spinning the chamber should put the bullet towards the bottom of the chamber to start. That assumes the chamber would not be re-spun before each shot. But either way, I wouldn't want to bet my life on it.

Quote:Rigondeaux

I say we just make it a cocaine challenge.

I've a resting heart rate of 40. Bring the noise.

Quote:Ayecarumba

How you hold the gun while spinning the chamber becomes very important. If you hold it sideways ("gangsta style"), allowing the cylinder to spin below the frame, then swinging it back into the locked position, I think there is a better chance of the cartridge being in the firing position, than if you were to spin the cylinder holding the frame in upright in a shooting position. The chamber with the bullet should tend to the bottom position, furthest from the firing position.

But what dominance is he?

A righty's natural wrist turn would be anti-clockwise, as is a revolver's (or at least all of mine). This would make any weight influence to favor the post fire position, aka further from the chamber. It would only put it closer if the gangsta was a lefty

Cocaine and guns. Thanks y'all for making me feel relevant again =)

Quote:AyeI think this only applies to swing-out cylinders. There are "break frame" models out there too, where the cylinder doesn't swing out the side.

Don't forget your cowboy shooters (single action). Those don't break OR swing, and can't be spun =)

Add on: In a quick test of a double action S&W 442 loaded with one round of +P, given a random spin from random locations, the lone round rested on the bottom most half of the wheel 22 out of 30 times.

In yet other words -- exactly like how they played in the Dear Hunter.

Please put answers in spoiler tags.

Quote:WizardMath puzzle time!

Assuming a revolver, with six chambers and one bullet, rotating the chamber before each pull, what is the probability the one to go first dies?

(1/6) + ((5/6) * (4/5) * (1/4)) + ((5/6) * (4/5) * (3/4) * (2/3) * (1/2))

Quote:WizardMath puzzle time!

Assuming a revolver, with six chambers and one bullet, rotating the chamber before each pull, what is the probability the one to go first dies?

One of three things will happen:

1. The first player dies; there is a 1/6 = 6/36 chance of this happening.

2. The first player lives, and the second player dies; there is a 5/6 x 1/6 = 5/36 chance of this happening.

3. Both live, in which case, we are back at the beginning.

The probability of the first player dying is 6 / (6 + 5) = 6/11.

Quote:ThatDonGuy

One of three things will happen:

1. The first player dies; there is a 1/6 = 6/36 chance of this happening.

2. The first player lives, and the second player dies; there is a 5/6 x 1/6 = 5/36 chance of this happening.

3. Both live, in which case, we are back at the beginning.

The probability of the first player dying is 6 / (6 + 5) = 6/11.

Thank you! That is correct. Good explanation too. Add one to the very large number of beers I owe you.

Quote:WizardQuote:ThatDonGuy

One of three things will happen:

1. The first player dies; there is a 1/6 = 6/36 chance of this happening.

2. The first player lives, and the second player dies; there is a 5/6 x 1/6 = 5/36 chance of this happening.

3. Both live, in which case, we are back at the beginning.

The probability of the first player dying is 6 / (6 + 5) = 6/11.

Thank you! That is correct. Good explanation too. Add one to the very large number of beers I owe you.

If it didn't rotate every round....would my answer have been correct?

Quote:AyecarumbaWill it change in subsequent rounds if: 1) the game is played until someone dies, or 2) the game is played for a maximum of 10 rounds?

What would happen after ten rounds?

Quote:RSOh. I missed the "rotates every round" part.

If it didn't rotate every round....would my answer have been correct?

It looks correct under the assumption of not rotating the barrel. For those who don't know how revolvers work, it automatically advances one chamber after a trigger pull.

Quote:WizardQuote:AyecarumbaWill it change in subsequent rounds if: 1) the game is played until someone dies, or 2) the game is played for a maximum of 10 rounds?

What would happen after ten rounds?

Survivors are released. The basis of the question is the odds have to favor someone getting the bullet the more rounds that are played. Does that mean the player least likely to die in round 1 is even less likely to die the longer the game goes on?

I already addressed "Gun Influence" ;-).Quote:billryanAn AP would control the spin so it always landed back on the same spot it was previously on.

Quote:RomesI already addressed "Gun Influence" ;-).

You would like to influence another man's "gun", wouldn't you?

Quote:AyecarumbaWill it change in subsequent rounds if: 1) the game is played until someone dies, or 2) the game is played for a maximum of 10 rounds?

"The quick answer" is:

(5/6)

^{10}= about 16.15% of the time, nobody would die (assuming "10 rounds" means each person pulls the trigger five times).

The remaining 83.85% of the time, the first player would die 6/11 of the time, and the second would die 5/11 of the time, as already shown.

The overall probabilities are:

First player dies = 6/11 of 83.85% = 45.74%

Second player dies = 5/11 of 83.85% = 38.11%

Both players live = 16.15%

Quote:ThatDonGuyQuote:Ayecarumba

"The quick answer" is:

(5/6)^{10}= about 16.15% of the time, nobody would die (assuming "10 rounds" means each person pulls the trigger five times).

The remaining 83.85% of the time, the first player would die 6/11 of the time, and the second would die 5/11 of the time, as already shown.

The overall probabilities are:

First player dies = 6/11 of 83.85% = 45.74%

Second player dies = 5/11 of 83.85% = 38.11%

Both players live = 16.15%

Thank you.

Player 1 dies = 6/11

Player 2 dies = 5/11

Both players live = 0

But the state where both players live will never reach exactly 0, since there is an infinitesimal possibility that both players will avoid the bullet no matter how many rounds are played. Therefore, is the answer ~ 6/11 more accurate than 6/11 for Player 1 catching "the lead express"?

Quote:AyecarumbaBut the state where both players live will never reach exactly 0, since there is an infinitesimal possibility that both players will avoid the bullet no matter how many rounds are played. Therefore, is the answer ~ 6/11 more accurate than 6/11 for Player 1 catching "the lead express"?[/spoiler]

No, the answer is a solid 6/11. We can say that for the same reason we can say the house edge in craps is 7/495, even though, in theory, the dice could be rolled zillions of times without the bet being resolved.

Quote:mipletJust over an hour ago, someone attempted to eat the most chicken nuggets in 3 minutes. facebook live recording

So the new world record is 646 grams in 3 minutes. Each nugget is approximately 17.5 g so that's only 37 in 3 minutes.

I think we've seen three or four WoV competitors who could do this.