Two Dice Puzzle: Part Deux

Quote: indignant99

I still disagree with you, OnceDear.
Regardless whether ONE die were viewed, or BOTH, an announcement is made. The announcement is "X." And now, FIVE possible pairs are ruled out. What has not been ruled out...

Left=X X=Right
Left=X non-X=Right
Left=non-X X=Right

I will surrender, IF you say that it's always and only the LEFT die viewed.

The peaker can pick either of the two dice. And it doesn't matter which -- after all, we only care if it's a double or a non-double. When the peaked picks a die, you are removing that die from being the unknown die.

If he reveals the left die, you remove "left = non-x" from the equation (there's be 5 non-X's). If he reveals the right die, you remove "right = non-x" from the equation (ie: 5 non-X's). In both cases, you are removing 5 from the total number of possibilities (11), ending with 1/(11-5) = 1/6.

I propose anyone to make a video like Alan did, only this time have at least 100 rolls and record only the single 2's to double 2's ratio. No editing and one continuous shot.

First quibble: PEEKER, not "peaker."

He can announce either die, correct? You don't know (nor care) whether it's the left or the right. Thus, you cannot eliminate any of the possibilities. Your "subtract 5" is an error.

Quote: Ibeatyouraces

I wonder how many times Alan retook this shot???

http://youtu.be/3jDKXRhgWyA

First comment on the video is 1/11.

Quote: Ibeatyouraces

I propose anyone to make a video like Alan did, only this time have at least 100 rolls and record only the single 2's to double 2's ratio. No editing and one continuous shot.

Already done. way over 200 rolls
earlier post

Quote: OnceDear

Already done. way over 200 rolls
earlier post

I want video of actual dice. Not that I don't believe a sim, but just as more evidence for the naysayers.

Quote: Ibeatyouraces

I wonder how many times Alan retook this shot???

http://youtu.be/3jDKXRhgWyA

As I posted on my forum, I did this out of boredom. You are looking at the first "try." Four out of ten times, when at least one die was a 2 there was a 2-2 showing. Granted, it's not scientific but done out of boredom.

By the way: The dice are new. I was given a new stick some years ago. I used two of the five to practice my throw. Three are still pristine.

EDITED TO ADD: I wonder if someone is going to say the dice are unbalanced, unfair or loaded? LOL

Quote: AlanMendelson

As I posted on my forum, I did this out of boredom. You are looking at the first "try." Four out of ten times, when at least one die was a 2 there was a 2-2 showing. Granted, it's not scientific but done out of boredom.

By the way: The dice are new. I was given a new stick some years ago. I used two of the five to practice my throw. Three are still pristine.

EDITED TO ADD: I wonder if someone is going to say the dice are unbalanced, unfair or loaded? LOL

Not questioning the integrity of the dice. Just fancy editing to me. Typical in the news business.

Again, Minimum 100 rolls. Unedited, continuous shot. Record are single two's to double two's. You don't have to do this. Anyone can. I don't have video equipment or I would.

I like the idea of a clear cup too.

Edit: 10 rolls mean nothing. Short term variance. A week and a half ago, out of the first 8 hands of video poker, 7 were a loser. The 8th was a royal flush. Doesn't tell me that royals are a 1/8 shot! Means nothing in the grand scheme of things.

Quote: indignant99

I still disagree with you, OnceDear.
Regardless whether ONE die were viewed, or BOTH, an announcement is made. The announcement is "X." And now, FIVE possible pairs are ruled out. What has not been ruled out...

Left=X X=Right
Left=X non-X=Right
Left=non-X X=Right

I will surrender, IF you say that it's always and only the LEFT die viewed.

This is indeed a far more interesting issue. Let's see if I can come up with a proof. I am still thinking this out, so rough draft....

We know that 'at least one of the dice is an x': Probability of that is 1

We know that both die were observed.: Probability of that is 1

I'll assume for now that we can refer to the dice spacially as 'Left Die' or 'Right Die', we could use alternatives such as dieA or dieB or by colour.

We know that it was either the left die that was announced, or the right die was announced:: Probability of that is 1

We DO NOT know whether left die was announced or right die was announced. and we have no reason to assume bias.

So probability that it was left die(PL)=probability that it was Right die(PR) = 1/2 = 0.5

We know that for any one die there are 6 equally likely outcomes.

Without anything called out and no constraints...

There are 36 ways the dice could land
There are 6 ways of rolling a pair
There are 30 ways of not rolling a pair

We know, or can commonly agree that the maths is done like so for equally probable events

Probability of throwing a pair and Announcer called the left die
= (count of ways of throwing a pair/count of ways of throwing two dice) x Probability of calling left die
= (6/36) x 0.5 = 3/36

Probability of throwing a pair and Announcer called the right die
= (count of ways of throwing a pair/count of ways of throwing two dice) x Probability of calling right die
= (6/36) x 0.5 = 3/36

We know announcer called left die or right die ( With equal probability)
Probability of throwing a pair and announcer called either of the dice is sum of the above
= (3/36) + (3/36) =6/36 = 1/6

ALL BECAUSE we have to assume that calling left die was equally likely to calling right die.


EDIT
Actually it just occurred to me that even if the announcer did have a bias towards left die or right die, the final answer would still be 1/6.
Lets assume some silly bias where he called left 13 times out of 36 and called right 23 times out of 36

Then it becomes
Probability of throwing a pair and Announcer called the left die
= (count of ways of throwing a pair/count of ways of throwing two dice) x Probability of calling left die
= (6/36) x (13/36)= 78/1296

Probability of throwing a pair and Announcer called the right die
= (count of ways of throwing a pair/count of ways of throwing two dice) x Probability of calling right die
= (6/36) x (23/36)= 138/1296

We know announcer called left die or right die ( With equal probability)
Probability of throwing a pair and announcer called either of the dice is sum of the above
= (78/1296) + (138/1296) =216/1296= 1/6

Really? You think it's fancy editing? To show something like two dice in a cup, shaken, dumped on a table and it shows 2-2 ?? Wow. I'll take this bet. Examine the video and if you can prove it's "fancy editing" in that video I will walk the Strip from Caesars to Paris naked. LOL

(Can't you accept that it happened? Of course not, it seems.)

Quote: AlanMendelson

Really? You think it's fancy editing? To show something like two dice in a cup, shaken, dumped on a table and it shows 2-2 ?? Wow. I'll take this bet. Examine the video and if you can prove it's "fancy editing" in that video I will walk the Strip from Caesars to Paris naked. LOL

(Can't you accept that it happened? Of course not, it seems.)

Why not show all of the rolls then???

Just be prepared to lose to Mike.

Guys, do me a favor. Put all the fancy math aside and simulate the actual question with two dice. It can be two dice from your Monopoly game, I don't care.

Take two dice and roll them until at least one of the two dice shows a 2. Take that die that shows a 2 and put it in your pocket and then look at the remaining die. If both dice showed a two, take one of the two dice showing a two and put it in your pocket.

What you now have is one die in front of you. Examine the remaining die carefully.

Now tell me as you look at this six sided die in front of you, what are the odds that this "other die" could also show a 2 ??

Each and every time I do this, I come up with 1/6. I do not come up with 1/11 for a six-sided die.

Very simple.

Quote: Ibeatyouraces

Not questioning the integrity of the dice. Just fancy editing to me. Typical in the news business.

Again, Minimum 100 rolls. Unedited, continuous shot. . . .

Edit: 10 rolls mean nothing. Short term variance.

OK. My experimernt used virtual dice from random.org It was way over 200 rolls and the very first roll was 2-2.
It was so unedited and so live and once only that I even posted to this forum in the midst of the exercise and all that is caught in the video. Multiple cockups were recorded in the video and a few times I gave player benefit of the doubt. Player had a few lucky streaks.

But guess who lost 25 units first :o)

Doing this with real dice would be so slow and boring. My video extends to 169 Meg. It's sat on my server. URL on request.

Quote: Ibeatyouraces

Why not show all of the rolls then???

Just be prepared to lose to Mike.

Because I did it holding my cell phone camera in one hand, and I wasn't trying to prove anything. I just did it to amuse myself. Even if I did show all of the videos what would it prove? Even I concede it proves nothing.

More importantly, do this:

Simulate the actual question with two dice. It can be two dice from your Monopoly game, I don't care.

Take two dice and roll them until at least one of the two dice shows a 2. Take that die that shows a 2 and put it in your pocket and then look at the remaining die. If both dice showed a two, take one of the two dice showing a two and put it in your pocket.

What you now have is one die in front of you. Examine the remaining die carefully.

Now tell me as you look at this six sided die in front of you, what are the odds that this "other die" could also show a 2 ??

Each and every time I do this, I come up with 1/6. I do not come up with 1/11 for a six-sided die.

Very simple.

Quote: OnceDear

OK. My experimernt used virtual dice from random.org It was way over 200 rolls and the very first roll was 2-2.
It was so unedited and so live and once only that I even posted to this forum in the midst of the exercise and all that is caught in the video. Multiple cockups were recorded in the video and a few times I gave player benefit of the doubt. Player had a few lucky streaks.

But guess who lost 25 units first :o)

Doing this with real dice would be so slow and boring. My video extends to 169 Meg. It's sat on my server. URL on request.

Watching some movies is slow and boring too. Some things just need to be done. :-)

Quote: Ibeatyouraces

Watching some movies is slow and boring too. Some things just need to be done. :-)

If your profile allowed Private Messages, you'd have received the sleepytime video url
$:o)

Quote: AlanMendelson

Because I did it holding my cell phone camera in one hand, and I wasn't trying to prove anything. I just did it to amuse myself. Even if I did show all of the videos what would it prove? Even I concede it proves nothing.

More importantly, do this:

Simulate the actual question with two dice. It can be two dice from your Monopoly game, I don't care.

Take two dice and roll them until at least one of the two dice shows a 2. Take that die that shows a 2 and put it in your pocket and then look at the remaining die. If both dice showed a two, take one of the two dice showing a two and put it in your pocket.

What you now have is one die in front of you. Examine the remaining die carefully.

Now tell me as you look at this six sided die in front of you, what are the odds that this "other die" could also show a 2 ??

Each and every time I do this, I come up with 1/6. I do not come up with 1/11 for a six-sided die.

Very simple.

Problem is, this is doing the experiment by looking at a single result, and trying to induce the answer

There's a 11 times from 36 (equally possible) results you can be picking up a die and putting it in your pocket. Only one of those ways is when there's a two on both dice.

You induction method misses the first step, 25 times out of 36 you'll be reshaking the dice. So, yeah, there's six sides to the remaining dice, but not all sides are equally likely when you get to the point examining the other dice. The two being on the remaining dice is twice as unlikely as any other side. 1, 3,4,5 or 6 will be there twice as often as a two.

That's what you are missing... it's not an eleven sided dice, and it's not random. You've made a decision (wait for at least one two, and pick up a two) that stops it being an evenly distributed, random event on the second dice.

But, it doesn't matter what anyone says, I guess.

(and yes, I have done this live a few times, but not enough to be statistically accurate... I'm convinced by those who have, my quick and easy excel sheet shows a double 2 is about 10 times less likely than any two).

I reported in the other thread that I rolled two dice repeatedly for 10 minutes.
102 times exactly one of the dice was a 2
7 times both dice were twos

That is closer to 1/11 than 1/6.

For the umpteenth time, the problem is this: 1/11 is not the answer to the question.

The question wants you to remove one die and look at the second of the two dice.

1/11 is indeed the combination of 2-2 when you have the 11 dice combinations showing at least a single 2. But 1/11 does not answer the question.

This reminds me of the phrase: you can't see the forest for the trees. Well, you can't see the question for your knowledge of math probability.

I really invite you all to revisit the original queston and ponder what the question is asking you to do. In no way does the question ask you to consider all of the possibilities of two dice with at least one die showing a 2.

Quote: AlanMendelson

...In no way does the question ask you to consider all of the possibilities of two dice with at least one die showing a 2.

Sure it does.

Quote: AlanMendelson

For the umpteenth time, the problem is this: 1/11 is not the answer to the question.

You may be good at marketing and TV videos, but you are wrong here.

Quote:

The question wants you to remove one die and look at the second of the two dice.

Yes.

Quote:

1/11 is indeed the combination of 2-2 when you have the 11 dice combinations showing at least a single 2. But 1/11 does not answer the question.

It does.

Quote:

This reminds me of the phrase: you can't see the forest for the trees. Well, you can't see the question for your knowledge of math probability.

And you can't see the question for your ignorance of math probability.

Quote:

I really invite you all to revisit the original queston and ponder what the question is asking you to do. In no way does the question ask you to consider all of the possibilities of two dice with at least one die showing a 2.

It asks you the probability of a 2 on the second dice, given certain conditions. All sides are NOT equally likely in the case of the set up, so the answer is NOT 1 in 6, despite their being only six sides to the dice. This is what you fail to see. The removal of one die into your jacket pocket means something quite important. There's only one case out of 11 where you can choose which dice you put into your jacket pocket and examine the other dice, consider it, and give it your best thoughts on the matter and find it to be a two.

Why I am bothering to convince you, I dunno. Go ahead, make bets, spend hours watching dice roll then buy the Wizard Lunch.

Quote: OnceDear

Bad proof. HORRIBLE.

There are only 6 Announcements possible: 1, 2, 3, 4, 5, or 6.
Half the time, it's the Left Die. Half the time, it's the Right Die.

When the Left Die is called:
1 of the roll-events, the Right Die matches. 5 of the roll-events, Right Die doesn't.

When the Right Die is called:
1 of the roll-events, the Left Die matches. 5 of the roll-events, Left Die doesn't.

Half the time (Left announcement) we get 1/6 of a pairing, 5/6 non-pairing.
Half the time (Right announcement) we get 1/6 of a pairing, 5/6 non-pairing.

Add them: 1/6 + 1/6 pairing. 5/6 + 5/6 non-pairing. 2 (sixths) versus 10 (sixths).
Now the nuance: Subtract ONE of the "1/6 pairing" because they're the exact identical same event. 1 versus 10. ONE versus TEN.

Quote: indignant99

Bad proof. HORRIBLE.

There are only 6 Announcements possible: 1, 2, 3, 4, 5, or 6.
Half the time, it's the Left Die. Half the time, it's the Right Die.

When the Left Die is called:
1 of the roll-events, the Right Die matches. 5 of the roll-events, Right Die doesn't.

When the Right Die is called:
1 of the roll-events, the Left Die matches. 5 of the roll-events, Left Die doesn't.

Half the time (Left announcement) we get 1/6 of a pairing, 5/6 non-pairing.
Half the time (Right announcement) we get 1/6 of a pairing, 5/6 non-pairing.

Add them: 1/6 + 1/6 pairing. 5/6 + 5/6 non-pairing. 2 (sixths) versus 10 (sixths).
Now the nuance: Subtract ONE of the "1/6 pairing" because they're the exact identical same event. 1 versus 10. ONE versus TEN.

Wanna make a bet?

Quote: RS

1 - Dealer/peeker only says something when a "2" is present, and says nothing when neither die is a 2.
#1 - there is a 1/11 chance that both dice are a 2 / the same.

This is correct.

Quote: RS

2 - Dealer/peeker says something every time, like "there is at least one ___ present" (could be 1, 2, 3, 4, 5, OR 6).
#2 - there is a 1/6 chance that both dice are the same.

This is *not* correct without more information, but this presents a good opportunity to explain why the answer to the original question is. The unstated assumption of your 2nd scenario, under which 1/6 is actually the correct answer, is that the peeker names one of the dice randomly with uniform probability. However, if the peeker's rules are different, the probability of the dice being equal may not be 1/6 after you have learned what one die is.[1]

Consider a scenario wherein the peeker's rules are "look at both dice and name the largest."

Under that rule, if the peeker says "there is at least one 1 present" then the probability of the other die being a 1 is 100%. Similarly, the probability of the other die being 2, 3, 4, 5, or 6 is zero. If the other die was greater than 1, the rules dictate that the larger number should have been disclosed, not the first 1. The only permutation of dice that satisfies this game, when the peeker says "there is at least one 1" is (1,1). For any other permutation, the peeker would not have named a 1.

But if the peeker says "there is at least one 2 present" the probability of the other die being a 2 is not 100%. It's not 1/6 either, in fact, it is 1/3. There are only three permutations of two dice that fulfill the requirement that the largest of the dice is 2: (1,2), (2,2), (2,1). Of those three, only 1 of them has 2 as the other die. The probability of the other die being a 1 is 2/3, and the probability of the other die being 3, 4, 5, or 6 is zero.

Continuing the pattern, the probability of the other die being equal to the first named die, conditional on that the first die named is the largest of the two, is:
1: 1/1
2: 1/3
3: 1/5
4: 1/7
5: 1/9
6: 1/11
(We've seen that 1/11 figure before.)

Now consider that if the peeker has just announced that one of the dice is a 6, the knowledge of the other die under the rule "name the largest of the two dice" is equivalent to the knowledge of the other die under the rule "if a 6 is present, name it." In other words, if the rule is "name the largest die" and the peeker names a 6, the probability that the other die is a 6 is 1/11. Equally, if the rule is "name a 6 if one is present" and the peeker names a 6, the probability that the other die is a 6 is also 1/11.

Finally, the rule "name a 6 if one is present" is equivalent to "name a 2 if one is present." Thus, referring back to the original problem, if the peeker's rule is "name a 2 if one is present" and the peeker does name a 2, then the probability of the other die also being 2 is 1/11. QED, I hope.


[1] Note that an assumed knowledge of the rules is also required to make the original problem 1/11. Echoing DJTeddyBear's comments about "Table 1" (page 1 of this thread), some rules will result in a conditional probability different than 1/11. Here's an example: "name an odd over an even number if both are present, otherwise name the highest odd or lowest even number." Under those rules, what are the chances of two 2s conditional upon a first 2 being announced? It's neither 1/11 nor 1/6...

Long time no see, ME. Nice to have you back. I hope you're able to educate some of the 1/6 believers. I think your odds are about 1 in 11.

Quote: Wizard

Long time no see, ME. Nice to have you back.

I second that.

Quote: Wizard

I hope you're able to educate some of the 1/6 believers. I think your odds are about 1 in 11.

Wanna bet? ;)

Quote: AlanMendelson

Put all the fancy math aside and...

Require a double payout on the two 2's.

Try this: https://www.youtube.com/watch?v=uCT-5Rye9bk

Quote: RS

Wanna make a bet?

Yeah. On this here game.....

I walk up to the craps table, and I'M BLIND.
I throw the dice, but cannot perceive the result.
[36 possible outcomes, 6 of which are doubles.]
Now the announcer (OnceDear) - a good little shill of the casino -
announces "Mr. Indignant, one of the dice shows a boing {1,2,3,4,5,6},
Would you like to bet that the complete result is a pair of boings?"

My answer according to this pay-table...

over 10-to-1	YES. Every time. What's table max?
exactly 10-to-1	NO. Pointless, no edge either way.
under 10-to-1	NO. This bet's a loser.

I don't care which die the announcer announced, as long as truthful.

deleted... some faulty stuff.

Quote: indignant99

Yeah. On this here game.....

I walk up to the craps table, and I'M BLIND.
I throw the dice, but cannot perceive the result.
[36 possible outcomes, 6 of which are doubles.]
Now the announcer (OnceDear) - a good little shill of the casino -
announces "Mr. Indignant, one of the dice shows a boing {1,2,3,4,5,6},
Would you like to bet that the complete result is a pair of boings?"

My answer according to this pay-table...

over 10-to-1	YES. Every time. What's table max?
exactly 10-to-1	NO. Pointless, no edge either way.
under 10-to-1	NO. This bet's a loser.

I don't think RS was offering to pay 10 to 1 on this wager where he and I agree that the probability is 1/6 for the pair. But if YOU want to bank the bet and YOU be the announcer and WE be blindfolded, then I reckon you'd have a full table at say 8 to 1.

I'd be really grateful if someone qualified would go through my earlier proof of the 1/6 and show me where I went wrong.( or right)

Quote: indignant99

Yeah. On this here game.....

I walk up to the craps table, and I'M BLIND.
I throw the dice, but cannot perceive the result.
[36 possible outcomes, 6 of which are doubles.]
Now the announcer (OnceDear) - a good little shill of the casino -
announces "Mr. Indignant, one of the dice shows a boing {1,2,3,4,5,6},
Would you like to bet that the complete result is a pair of boings?"

My answer according to this pay-table...

over 10-to-1	YES. Every time. What's table max?
exactly 10-to-1	NO. Pointless, no edge either way.
under 10-to-1	NO. This bet's a loser.

I don't care which die the announcer announced, as long as truthful.

deleted... some faulty stuff.

Would you like to be "the casino" and I'll be the patron? (ie: the rolls are flip flopped).

I bet $1 (or however much). If it's a double I win $9 from you. If it's not a double you win my $1. In other words, pays 9-to-1.

PS: The announcer does announce a number every roll, yes? If you or anyone else is interested in taking all my money with this bet, send me a PM.

Quote: MathExtremist

This is correct.


This is *not* correct without more information...

Excellent and very lucid explanation. I agree with every word. THANK YOU.

It was stated on the first page of the first thread, that the probability could not be known unless the rules of announcing were also known.

Long Long ago Mango Said...
Shortly afterwards I said...

Now, please consider this.
If the announcer had a set of rules such as 'Announce the biggest number seen' and we have a paytable as
1: 1/1
2: 1/3
3: 1/5
4: 1/7
5: 1/9
6: 1/11

Then consider the fact that we will see lots of rolls as we sit near the table. Some ploppies will bet every time because they're addicted and don't understand anything. Sometimes we will get mean payouts of 1:1 and sometimes they will whoop at their 11:1. What would be the average payout rate over a large number of rolls?

Welcome back, ME!

Quote: indignant99

Bad proof. HORRIBLE.

There are only 6 Announcements possible: 1, 2, 3, 4, 5, or 6.
Half the time, it's the Left Die. Half the time, it's the Right Die.

When the Left Die is called:
1 of the roll-events, the Right Die matches. 5 of the roll-events, Right Die doesn't.

So we do agree that when left die called it's 1/6?

Quote:

When the Right Die is called:
1 of the roll-events, the Right Die matches. 5 of the roll-events, Left Die doesn't.

So we do agree that when right die called it's 1/6?

Quote:

Half the time (Left announcement) we get 1/6 of a pairing, 5/6 non-pairing.
Half the time (Right announcement) we get 1/6 of a pairing, 5/6 non-pairing.

I think I agree with that.
So for half of games played, it's 1/6 and for the other half of games played it's 1/6. Which games have it as not 1/6?

That sort of completes my rebuttal of your rebuttal $;o)

Quote: AlanMendelson

Try this: https://www.youtube.com/watch?v=uCT-5Rye9bk

Nice! I think this is the first WoV "video response." Let me try to draw some more attention to it below.



Before anybody takes this post out of context, the probability IS 1 in 11.

Enjoy your Lunch, Wizard.

In his vid, Alan asks :-
"What are the odds of this same die also can show a two" (Bad grammar: Shame on you, Alan)

What he should ask, if being faithful to the original question is...

"What were the odds that this die IS NOW showing a two" (Answer absolutely = 1/11)('Were', because we have now revealed it as 6)

Those are two 100% totally different interpretations of the original question. Alan's interpretation is semantically flawed and shows poor comprehension of the original question as posed in the English language.

If the issue is not one of maths, but is one of English comprehension and/or logic, then Alan still fails dismally.

But of course, that was not the question posed, was it, because the tosser and peeker never did reveal which die was 'this same die'. Indeed, in the original question, no dice were ever revealed.

Quote: Dween, many years ago

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?

or as Ayacarumba later paraphrased

Quote: Ayacarumba at the start of this thread

The dealer places two six sided dice in a cup, shakes them, then slams the cup down on the table hiding the results. He then peeks under the cup, and states, "At least one of the dice is a two". Please place your bets. The only wagers available are "Total 4" and "Total not 4" ... Is the probability of "Total 4" ... 1-11?

(Please don't suspend me Face, I was responding to the Wizard's post $:o) )

Quote: RS

...Would you like to be "the casino"...
If it's a double I win $9 from you...
PS: The announcer does announce a number every roll, yes?

Yes, the announcer does announce a number every roll.
Yes, I will be the casino, paying 9-to-1 on a double of the announced number.

Announcement	Rolls that satisfied the announcement
Ace	11 . 12 . 13 . 14 . 15 . 16 . 21 . 31 . 41 . 51 . 61
Deuce	22 . 21 . 23 . 24 . 25 . 26 . 12 . 32 . 42 . 52 . 62
Three	33 . 31 . 32 . 34 . 35 . 36 . 13 . 23 . 43 . 53 . 63
Four	44 . 41 . 42 . 43 . 45 . 46 . 14 . 24 . 34 . 54 . 64
Five	55 . 51 . 52 . 53 . 54 . 56 . 15 . 25 . 35 . 45 . 65
Six	66 . 61 . 62 . 63 . 64 . 65 . 16 . 26 . 36 . 46 . 56

Quote: indignant99

Yes, the announcer does announce a number every roll.
Yes, I will be the casino, paying 9-to-1 on a double of the announced number.

Announcement	Rolls that satisfied the announcement
Ace	11 . 12 . 13 . 14 . 15 . 16 . 21 . 31 . 41 . 51 . 61
Deuce	22 . 21 . 23 . 24 . 25 . 26 . 12 . 32 . 42 . 52 . 62
Three	33 . 31 . 32 . 34 . 35 . 36 . 13 . 23 . 43 . 53 . 63
Four	44 . 41 . 42 . 43 . 45 . 46 . 14 . 24 . 34 . 54 . 64
Five	55 . 51 . 52 . 53 . 54 . 56 . 15 . 25 . 35 . 45 . 65
Six	66 . 61 . 62 . 63 . 64 . 65 . 16 . 26 . 36 . 46 . 56

Very well. What is the maximum I can bet against you? Do you live in Las Vegas // when would you be coming to LV next? How would you want it to go down -- cup + 2 dice? Live craps table? Other?

Quote: indignant99

Yes, the announcer does announce a number every roll.
Yes, I will be the casino, paying 9-to-1 on a double of the announced number.

Announcement	Rolls that satisfied the announcement
Ace	11 . 12 . 13 . 14 . 15 . 16 . 21 . 31 . 41 . 51 . 61
Deuce	22 . 21 . 23 . 24 . 25 . 26 . 12 . 32 . 42 . 52 . 62
Three	33 . 31 . 32 . 34 . 35 . 36 . 13 . 23 . 43 . 53 . 63
Four	44 . 41 . 42 . 43 . 45 . 46 . 14 . 24 . 34 . 54 . 64
Five	55 . 51 . 52 . 53 . 54 . 56 . 15 . 25 . 35 . 45 . 65
Six	66 . 61 . 62 . 63 . 64 . 65 . 16 . 26 . 36 . 46 . 56

Haven't you forgotten something Indie?

There are only 36 ways the dice can land because we are not interested in their spatial position. Naming the dice left and right was just a way of working through the outcomes, while keeping some clarity.

E.g 21 is the same outcome as 12

Also, you are leaving the non-called die under the cup.

Of course, if you want to consider spoatial arrangement 1 1 is possible in two ways too. so each of your rows now has 12 combinations.

Quote: OnceDear

Quote: indignant99

Yes, the announcer does announce a number every roll.
Yes, I will be the casino, paying 9-to-1 on a double of the announced number.

Announcement	Rolls that satisfied the announcement
Ace	11 . 12 . 13 . 14 . 15 . 16 . 21 . 31 . 41 . 51 . 61
Deuce	22 . 21 . 23 . 24 . 25 . 26 . 12 . 32 . 42 . 52 . 62
Three	33 . 31 . 32 . 34 . 35 . 36 . 13 . 23 . 43 . 53 . 63
Four	44 . 41 . 42 . 43 . 45 . 46 . 14 . 24 . 34 . 54 . 64
Five	55 . 51 . 52 . 53 . 54 . 56 . 15 . 25 . 35 . 45 . 65
Six	66 . 61 . 62 . 63 . 64 . 65 . 16 . 26 . 36 . 46 . 56

Haven't you forgotten something Indie?

There are only 36 ways the dice can land because we are not interested in their spatial position. Naming the dice left and right was just a way of working through the outcomes, while keeping some clarity.

E.g 21 is the same outcome as 12

Also, you are leaving the non-called die under the cup.

Shhhhhh!

Quote: OnceDear

Haven't you forgotten something Indie?
... E.g 21 is the same outcome as 12 ...

You'd better tell me what I'm forgetting.
No, 21 and 12 are not the same outcome. They're two distinct rolls.

Quote: indignant99

You'd better tell me what I'm forgetting.
No, 21 and 12 are not the same outcome. They're two distinct rolls.

Didn't you once say there were 36 different results? Look at your table. Tell me again what would be announced if it was 2-5. Keep looking, because you are hardly going to announce
"At least one of the dice is a two and at least one of the dice is a 5: Do you want to bet on it being a pair?"

Let's face it, with whatever slant you want to take on it, some of the time you are going to look up the result and announce on this basis,

Quote:

Announcement	Rolls that satisfied the announcement
Ace	11 . 12 . 13 . 14 . 15 . 16
Deuce	22 . 21 . 23 . 24 . 25 . 26
Three	33 . 31 . 32 . 34 . 35 . 36
Four	44 . 41 . 42 . 43 . 45 . 46
Five	55 . 51 . 52 . 53 . 54 . 56
Six	66 . 61 . 62 . 63 . 64 . 65

And for some part of the time you are going to announce on this basis

Quote:

Announcement	Rolls that satisfied the announcement
Ace	11 . 21 . 31 . 41 . 51 . 61
Deuce	22 . 12 . 32 . 42 . 52 . 62
Three	33 . 13 . 23 . 43 . 53 . 63
Four	44 . 14 . 24 . 34 . 54 . 64
Five	55 . 15 . 25 . 35 . 45 . 65
Six	66 . 16 . 26 . 36 . 46 . 56

You can switch between look up tables at your leisure and no-one will notice or care.

Quote: RS

Very well. What is the maximum I can bet against you? Do you live in Las Vegas // when would you be coming to LV next? How would you want it to go down -- cup + 2 dice? Live craps table? Other?

I've never been to LV, and have no plans to visit.
How would it go down?
Where does Wizard shoot his educational videos? I'd like to do it there.
The Wizard would bring dice & cup, and shake, slam, peek, and announce.
If the full result is double-the-announcement, you win 9 from me.
If the full result isn't double-the-announcement, I win 1 from you.
108 total rolls (3 x 36). Alan and son video-record the session.
(Alan bets too - a 5-cent nickel - but only when "Deuce" is the announcement.)

Quote: indignant99

I've never been to LV, and have no plans to visit.
How would it go down?
Where does Wizard shoot his educational videos? I'd like to do it there.
The Wizard would bring dice & cup, and shake, slam, peek, and announce.
If the full result is double-the-announcement, you win 9 from me.
If the full result isn't double-the-announcement, I win 1 from you.
108 total rolls (3 x 36). Alan and son video-record the session.
(Alan bets too - a 5-cent nickel - but only when "Deuce" is the announcement.)

Indie, Please my friend. see the light. See my most recent post which was pretty much simultaneous with yours. See the error of your ways.

Or catch a flight to the UK and take my life savings off me at table min $0.50, Table max $16. You can even Martingale me. It should easily cover the travel costs.

Isn't editing tables a pain?

Quote: OnceDear

Didn't you once say there were 36 different results? Look at your table. Tell me again what would be announced if it was 2-5...

Of course there are 36 different results...

• 11 of them contain an ace.
• 11 of them contain a deuce.
• 11 of them contain a three.
• 11 of them contain a four.
• 11 of them contain a five.
• 11 of them contain a six.

There is overlap. Every set of eleven, shares 2 of its results, with every other set of eleven. On a 2-5 (or 5-2) the announcement is a random pick of either "deuce" or "five." The Wizard would be making that random choice. Either way, he'd be isolating a set of 11 outcomes. Out of those two sets of eleven outcomes (the "master-deuce" versus the "master-five") two outcomes are shared in common.

Quote: indignant99

Of course there are 36 different results...

• 11 of them contain an ace.
• 11 of them contain a deuce.
• 11 of them contain a three.
• 11 of them contain a four.
• 11 of them contain a five.
• 11 of them contain a six.

There is overlap. Every set of eleven, shares 2 of its results, with every other set of eleven. On a 2-5 (or 5-2) the announcement is a random pick of either "deuce" or "five." The Wizard would be making that random choice. Either way, he'd be isolating a set of 11 outcomes. Out of those two sets of eleven outcomes (the "master-deuce" versus the "master-five") two outcomes are shared in common.

So if it's 2-5 then at random, either deuce would be announced, or 5 would be announced?

How is that different to the situation where the announcer looks at the dice, randomly selects one of my two lookup tables, uses that lookup table, and makes his call? You HAVE TO translate from outcome to call in some way and that way is based on the outcome and the announcers own style of randomness.

tables here

Quote: indignant99

I've never been to LV, and have no plans to visit.
How would it go down?
Where does Wizard shoot his educational videos? I'd like to do it there.
The Wizard would bring dice & cup, and shake, slam, peek, and announce.
If the full result is double-the-announcement, you win 9 from me.
If the full result isn't double-the-announcement, I win 1 from you.
108 total rolls (3 x 36). Alan and son video-record the session.
(Alan bets too - a 5-cent nickel - but only when "Deuce" is the announcement.)

Sounds good to me. Although I'd prefer 5x36 rolls (180) to smooth out the variance a bit. My only main concern is payment -- idk how that stuff works? Like PayPal? Or can someone (wizard) be an escrow holder? (I'd prefer that over all.). And up to how much can I bet? I would flat bet the entire way through.

Quote: RS

Sounds good to me. Although I'd prefer 5x36 rolls (180) to smooth out the variance a bit. My only main concern is payment -- idk how that stuff works? Like PayPal? Or can someone (wizard) be an escrow holder? (I'd prefer that over all.). And up to how much can I bet? I would flat bet the entire way through.

Similar here, I would not insist on flat betting, though would flat bet if Indi so wished. I would rather insist on 100 or more resolved bets and would keep going till my substantial life savings were gone.

Quote: OnceDear

So if it's 2-5 then at random, either deuce would be announced, or 5 would be announced?

Yes, one or the other would be announced.

Quote: OnceDear

How is that different to the situation where the announcer looks at the dice, randomly selects one of my two lookup tables, uses that lookup table, and makes his call?

Whaddaya mean, use lookup tables? That'd be nuts. Wizard looks at the two dice, and simply announces one of his choice.

Quote: indignant99

Yes, one or the other would be announced.

Whaddaya mean, use lookup tables? That'd be nuts. Wizard looks at the two dice, and simply announces one of his choice.

Indi, stop it, you are cracking me up matey.... The announcer doesn't physically need to look up the results from some printed cards. Of course he can do that bit of working out in his head. But the announcement would still come from the sequence...

Roll dice
Look at dice
Choose one die at random
Announce that die's value.
The value/announcement pair would come from one of those two lookup tables.
D'ya wanna try it as a skype exercise for pennies. You can be banker, roller, announcer all in one. You can use your own flavour of random? We each start with say 200 pennies. I'll be the punter playing you heads up.
The first one to lose 200 pennies sends payment to the charity of his choice. We record the entire exercise for all to see (dunno how yet, but it cannot be too hard) Happy to use real dice or random.org which is time stamped.