It was two years ago today...

[a2a3dseddie]

In order for the answer to be 1/11, the die that originally shows a 2 would have to rotate to show other positions.

Here is the original problem again:

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?


Let's say you are the peeker Alan. We are using a pair of regulation identical looking craps dice. In a previous post you gave an example where you say you see:

Dice 1=6
Dice 2=2

This is 1 of the 11 possible combinations you could be looking at.

Dice 1=2
Dice 2=6

Here is the other possible combination you could be looking at.

They appear identical, but are not. You play a lot of craps, you know why there are 6 combinations for a 7.

So, 2 of the 11 possible combinations where at least one dice is a 2 are represented here. Other people have tried to demonstrate this using coloured dice. i.e. mickeycrimm

No dice need to rotate or spin.

[jbjb]

The dice rotate or spin when they're rolled. When will these blockheads realize that "other" doesn't mean "certain"?

[redietz]

Next up: How many angels can fit on the head of a pin?

Much more interesting, and has more real-world applications.

[Alan Mendelson]

guys take two dice. set one of the dice on the 2. that die with the 2 cannot change. it is set on 2.

now ask yourself what are the chances that both dice will show 2-2?

there is only one answer: 1/6. Why? Because the die showing a 2 cannot change.

Why doesn't it change? Because after two dice are shaken up and slammed on the table they don't move again... at least not in the real world.

Here's the problem for the 1/11 crowd: they are not visualizing or actually using two real, physical dice. They keep referring to charts and graphs showing the various combinations of two dice. If you used real, physical dice you would see that the solution can only be 1/6... hopefully.

But... I doubt it. You still want to be looking at your charts and graphs of all the combinations of two dice, and all of the combinations of dice with at least one two.

This is hopeless.

About the angels on the head of pin: there are 18, just like the number of successive rolls of a yo on the craps table at Caesars. LOL

[Rob.Singer]

Let's assume you're the so-called peeker, Eddie. You look under the cup and see a 2 but do not look at the other die. You report "at least one of the dice is a 2". You also will answer your own question that the probability of there being 2-2 is 1 in 6--not 1 in 11, because you know which of the two dice shows a 2 at the moment.

Alan, OTOH, also "knows" it's 1 in 6 because in his mind he has taken one of the dice completely out of the picture, leaving only one die. But what does a WoV armchair "AP" have to say about this? Why....1 in 11 of course! And this is only because they are assuming nobody but the peeker knows what he has seen.

Day-in/day-out the prevailing answer of 1 in 6 will be correct, except in the flawed world of theorists, who will never admit that the original question was incompetently formulated. The OP outsmarted himself, because he never bounded the question with necessary absolutes. And for a forum loaded with supposed "mensas" how surprising it is to see one of them fail so extraordinarily.

[a2a3dseddie]

guys take two dice. set one of the dice on the 2. that die with the 2 cannot change. it is set on 2.

now ask yourself what are the chances that both dice will show 2-2?

there is only one answer: 1/6. Why? Because the die showing a 2 cannot change.

Why doesn't it change? Because after two dice are shaken up and slammed on the table they don't move again... at least not in the real world.

Here's the problem for the 1/11 crowd: they are not visualizing or actually using two real, physical dice. They keep referring to charts and graphs showing the various combinations of two dice. If you used real, physical dice you would see that the solution can only be 1/6... hopefully.

But... I doubt it. You still want to be looking at your charts and graphs of all the combinations of two dice, and all of the combinations of dice with at least one two.

This is hopeless.

About the angels on the head of pin: there are 18, just like the number of successive rolls of a yo on the craps table at Caesars. LOL

Don't take it from me Alan...

simply telling you that one of the dice shows a 2 does not remove it from its numerical relationship with the die with the unknown number showing. Whereas rolling 2-2 is a 36-1 possibility, knowing half the outcome, ie. one of the dice is showing a 2, reduces the odds to 11-1 since it is no longer a single die event. You now have to consider all combinations of the two dice where either one of them shows a 2.

[a2a3dseddie]

Rob, it's spelled out. The only question is 9-for-1 or 9-to-1 and the Wizard says it doesn't matter to him.

Gone are the "old rules". This is a new bet and quite frankly I like it.

Let me answer your questions specifically and I am sure this will be the understanding:

1. Two dice are in a cup or other device, shaken and in the cup placed on the table.
2. A witness will peek. If a 2 is shown the bet is on.
3. If a 2 is not shown, there is no betting.
4. In both cases, the cup will be removed and the dice can be viewed. This will prevent the original dice from showing 2-2 and the witness lying.
5. With one deuce the bet is on (#2) and if there is not a second bet the "player" will lose their bet.
6. With one deuce the bet is on (#2) and if there is a second deuce the "bank" will pay either 9-to-1 or 9-for-1 (the Wiz doesn't care.)

My own personal thought: I can't imagine why the Wizard agreed to this? There is a 1/6 chance that when one die shows a 2 that the other will also be a 2 yet he is willing to pay 9-for-1 or 9-to-1.

Is it possible he misread this bet the same way I suspect he misread the original question? Somebody wake me up from this (bad) dream.

Reminding everyone from post 178

On the Wizard of Vegas site, here's what someone posted as a bet:

1) shake dice in the cup and then turn upside down on the table
2) reveal both dice
3) pay 8:1 if dice show 2-2
4) get paid 1:1 if dice show 2-1, 2-3, 2-4, 2-5, or 2-6
5) no action if dice show any other combination

Are these terms agreeable to you regnis or Alan?

I just booked my flight to Las Vegas and will be there from June 21 to June 28. I would be willing to meet either of you or anyone else believing it's 1/6.

Alan suggested betting stakes be lunch. It doesn't matter to me.

I suggest we each start with $25 in quarters. We play until one person has no quarters left and we use the $50 to get lunch somewhere.

These are the basic details of the bet, which are essentially the same as what Alan posted. Any takers? We can negotiate the minimum number of rolls, stakes, venue etc. Like I said I'll be in Las Vegas June 21 to 28.

[coach belly]

The only thing funnier is coach actually thinks we can win. Of course, coach was chomping at the bit to get at that hundred play a couple of months ago. Coach, you haven't posted your results on that. How'd you come out?

I can win if I roll 2-2 before rolling 2-x 7 times.

But I read this is a one-roll exercise, so I'm willing to take jbjb's bet, and Alan can witness...heck he can even roll the dice.

I can be in LV first thing Wednesday morning...what's up jbjb?...are we on?

As far as my million hands of 7-5 BP...I did play them on a 100-play machine....and I was up after a million hands.

I used a Singer-type strategy, dropping down in denomination after I was dealt a FH...assuring myself a profit...even after a million -EV hands.

That's exactly what I would have done to take your 100-1 bet down, but you wormmed your way out of it.

No go find somebody else to troll...you fucking bum.

[a2a3dseddie]

It is in fact the same (in my opinion) of rolling one die after the first die shows a 2.

Instead, ask jbjb if he would accept this bet:

Set a die on the number 2 and roll a second die. If the second die also shows a 2 you will get paid $800. If the second die shows any other number you will pay him $100.

THAT would be 1/6. But it's not even close to the original question.

Alan, I asked you this earlier but you either missed or ignored it.

Let's say during a crap game, you throw the dice, one lands on 2 and the other bounces off the table. Do the dealers let you just re-roll ONE dice, add it to the 2 and resolve all the bets on the layout?

No they don't! Why?

[jbjb]

These morons have a comprehension problem. The puzzle clearly didn't say set a die to a number and roll the other.

They're just trolling looking for attention. I suggest just ignoring them from now on and don't give them the attention.

[mickeycrimm]

The only thing funnier is coach actually thinks we can win. Of course, coach was chomping at the bit to get at that hundred play a couple of months ago. Coach, you haven't posted your results on that. How'd you come out?

I can win if I roll 2-2 before rolling 2-x 7 times.

But I read this is a one-roll exercise, so I'm willing to take jbjb's bet, and Alan can witness...heck he can even roll the dice.

I can be in LV first thing Wednesday morning...what's up jbjb?...are we on?

As far as my million hands of 7-5 BP...I did play them on a 100-play machine....and I was up after a million hands.

I used a Singer-type strategy, dropping down in denomination after I was dealt a FH...assuring myself a profit...even after a million -EV hands.

That's exactly what I would have done to take your 100-1 bet down, but you wormmed your way out of it.

No go find somebody else to troll...you fucking bum.

However you think you can beat negative expectation gambling is fine with me. Why don't you quit your job and go to it full time?

[jbjb]

They're scared wimps, Mickey.

[coach belly]

However you think you can beat negative expectation gambling is fine with me. Why don't you quit your job and go to it full time?

Machine play is a solitary exercise...spending full time looking for money in casino machines only appeals to misfits like you.

[coach belly]

They're scared wimps, Mickey.

I'm ready to take money from you gambling on a -EV play.

What kind of AP are you to turn down such a strong play?...a scared wimp?

[Biloxi Bill]

However you think you can beat negative expectation gambling is fine with me. Why don't you quit your job and go to it full time?

Machine play is a solitary exercise...spending full time looking for money in casino machines only appeals to misfits like you.

Maybe like looking for stray men in queer bars eh Cooch?

[Rob.Singer]

The only thing funnier is coach actually thinks we can win. Of course, coach was chomping at the bit to get at that hundred play a couple of months ago. Coach, you haven't posted your results on that. How'd you come out?

I can win if I roll 2-2 before rolling 2-x 7 times.

But I read this is a one-roll exercise, so I'm willing to take jbjb's bet, and Alan can witness...heck he can even roll the dice.

I can be in LV first thing Wednesday morning...what's up jbjb?...are we on?

As far as my million hands of 7-5 BP...I did play them on a 100-play machine....and I was up after a million hands.

I used a Singer-type strategy, dropping down in denomination after I was dealt a FH...assuring myself a profit...even after a million -EV hands.

That's exactly what I would have done to take your 100-1 bet down, but you wormmed your way out of it.

No go find somebody else to troll...you fucking bum.

However you think you can beat negative expectation gambling is fine with me. Why don't you quit your job and go to it full time?

Hey mickey that's exactly what I've been doing every year since 1999, and on a scale that would bring all AP's to their knees, you know, like I brought the entire HP crowd--and really, EVERY singer-hating AP in Vegas--to their knees with my published bet ten years ago. In fact, I even made bigger sissies out of the wizard's bunch as they ran for the hills as soon as a mere $25k bet took on some teeth. They ALL safely chose the ability to keep lying about and disbelieving me rather than facing the music. Just like the anonymous cowards do here....and just as any of you pocket-change specialists would do with the coach. Similar in much the same way as in how the libtards got their asses handed to them by Trump, so all they now have left is to make up lie after lie about him and pretend it's all real.

[RS__]

These morons have a comprehension problem. The puzzle clearly didn't say set a die to a number and roll the other.

They're just trolling looking for attention. I suggest just ignoring them from now on and don't give them the attention.

+1

[mickeycrimm]

Alan, I've got another dice game for you. It's played with ONE DIE and a cup.

Player A is betting he can roll a 6 within 4 rolls of the die.

Player B is betting against Player A rolling a 6 within 4 rolls.

The chance of rolling a 6 is 1 in 6. That's 5 to 1 against.

Which side of the bet would you take, Alan?

Everyone please give Alan a chance to respond to this before you put up any information.

Alan has had ample time to respond to this. He has avoided it because he really doesn't know which player holds the advantage. This is one of the oldest dice hustles there is. It goes back to the 17th century.

In mid 17th century France a dice hustler went around propositioning other dice gamblers that he could roll a 6 in four rolls of the die. He got lots of action at first. It's pretty easy to figure out why. Most people would look at the odds of a single roll, see that it is 5 to 1 against rolling a 6 so anyone betting they could do it in four rolls is a fool. Or so they thought. But this dice hustler kept beating them with the proposition. He beat them to the point that they quit playing him. Why was he winning?

To calculate his chance of success you have to multiply out the failure rate of each roll. Since the chance is 1 in 6, the failure rate is 5 in 6.

5/6 = .8333333%

.8333333 X .8333333 X .8333333 X .8333333 = .4822530%

The shooter has a 48.22% chance of rolling the die four times and a six won't appear. And that means the shooter has a 51.78% chance of rolling a six in four rolls. A great hustle by this French dice gambler.

And the two dice puzzle is a great hustle too with people like coach and Alan, they are not the only ones, believing what they do. If they play the game long enough to see that they just keep slowly losing their money they will quit the game.

[mickeycrimm]

While Alan is ready to film any gambling action over the two dice puzzle he doesn't seem to want to throw down any money on it himself. When you think the true odds on an event are 5 to 1 against but you are getting an 8 to 1 payoff when it occurs you have a massive advantage. At $100 a pop, per every six decisions you would make $600 in action and get an $800 return. That's a 25% edge. Why would you turn down a game like this but have no problem betting $25 a hand at 99% video poker?

[Alan Mendelson]

Here is the original problem again:

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?


Let's say you are the peeker Alan. We are using a pair of regulation identical looking craps dice. In a previous post you gave an example where you say you see:

Dice 1=6
Dice 2=2

This is 1 of the 11 possible combinations you could be looking at.

Dice 1=2
Dice 2=6

Here is the other possible combination you could be looking at.

They appear identical, but are not. You play a lot of craps, you know why there are 6 combinations for a 7.

So, 2 of the 11 possible combinations where at least one dice is a 2 are represented here. Other people have tried to demonstrate this using coloured dice. i.e. mickeycrimm

No dice need to rotate or spin.

Let me say this again:

When die #1 is a 2, die #2 has six faces to choose from.
When die #2 is a 2, die #1 has six faces to choose from.
When the green die is a 2, the red die has six faces to choose from.
When the red die is a 2, the green die has six faces to choose from.
When the top die is a 2, the bottom die has six faces to choose from.
When the bottom die is a 2, the top die has six faces to choose from.

The original problem tells you the value of at least one die is known. It doesn't matter which.

The only way to get eleven faces to choose from is to rotate the die showing a 2. Get yourself a pair of dice and see for yourself. But don't change the die that lands on a 2.