Two Dice Puzzle: Part Deux
I'm really not inclined to film it but anyone else can do so royalty free.
It may be filmed by Alan, and Alan may be the player.
Only one condition on the copyright: ALAN MAY NOT BE THE DICE PEEKER and ALAN MAY NOT TOUCH THE DICE OR CUP. EVER!!!
[Cue video shot of Dice Thrower. He looks a bit shady.]
[Thrower holds a cup open side up. It seems that there may be dice in the cup, but they are unseen.]
[Nearby there is a paytable. It's not in focus. On the green baize table there are some betting areas, but the marking is out of focus. All of this is important, as will become clear.]
[Alan approaches table stage left.]
[Camera angle switches to over Alan's right shoulder, Shady Dice Thrower now face on.]
Quote: AlanWhat's the game? Please explain.
Quote: ThrowerWell sir, it's a game of dice. I have two ordinary dice in this cup and I'm going to shake it and slam it on the table. I'm then going to peek under the cup. If neither of the dice shows a deuce, I'll lift the cup and all games are a push. However, If at least one of the dice shows a Deuce, I'll make an announcement and then all bets are on. Bets pay as per paytable. I'll take bets on all possible outcomes including what we refer to as the sucker bet lucky pair of deuces.
Quote: AlanI've heard of a similar game. Let's play... But wait...
[Alan gazes at the paytable and asks...]
Quote: AlanWhy are you not paying 35:1 on pair of deuces?
Quote: ThrowerWell sir, we know that one of the dice is a deuce, so we eliminate that die leaving one other die. That other die has only 6 faces, so you can see that my 8:1 payout is quite generous.
[Alan shrugs and lays $1 in a betting circle, which is seen to be labelled 'Pair of Deuces'.]
[Thrower shakes the cup. It rattles. He slams it down on the table over what seems to be a pair of dice.]
[Thrower peeks under the cup, not revealing anything to Alan.]
[Pause for theatrical effect]
[Thrower then simultaneously lifts the cup and says . . .]
Quote: ThrowerIt's a push. I'll shake again.
Quote: AlanOne Moment. Let me look at those dice.
[Alan looks at the dice but does not touch. There is a blue die showing a 6 and a red die showing a 3.]
Quote: AlanYou never said anything about the dice being different colors.
Quote: ThrowerDoes it make any difference to you sir? Please refer to the pay-table before betting.
[Thrower again shakes the cup. It rattles. He slams it down on the table over what seems to be a pair of dice.]
[Thrower peeks under the cup, not revealing anything to Alan.]
Quote: ThrowerWell sir, At least one of the dice is a deuce, bets are on, Would you like to select the outcome? I'm still taking bets.
Quote: AlanSo, at least one of the dice is a deuce? Can we eliminate that die?
Quote: ThrowerYes sir, absolutely, I'm happy to confirm that.
Quote: AlanI'm a bit wary, is it the blue die that's the deuce?
Quote: ThrowerI'm sorry sir, I'm not going to reveal that until all bets are placed.
Quote: AlanBut we can eliminate one of the dice, leaving just one six sided die under the cup?
Quote: ThrowerYes, sir, I'm happy to confirm that.
Quote: AlanI'm still wary about this having just one die to decide the outcome.
Quote: ThrowerWell, sir, I shouldn't really do this but....
[Thrower reaches under the cup and apparently removes one die, which he puts in his pocket. Neither of the dice are seen.]
Quote: AlanI'm still wary about this. Did you just remove the blue die?
Quote: ThrowerI'm sorry sir, I'm not going to reveal that until all bets are placed. But I did remove one die and that die was a deuce and I did leave one die untouched.
Quote: AlanI'm still wary about this. Did you just remove the red die?
Quote: ThrowerI'm REALLY sorry sir, but I'm really not going to reveal which die I removed until all bets are placed.
Quote: AlanBut we can eliminate one of the dice can't we, leaving just one six sided die under the cup to decide the result?
Quote: ThrowerAbsolutely sir, I'm happy to confirm that.
Quote: AlanHmmm. So, let me look again at that paytable...
[Alan Looks closely at the paytable. He sees that pair of Deuces pays 8:1. Actually, apart from a rubbish side bet all the payouts are 8:1]
[The camera pans to the paytable which reads as follows . . .]
Quote: Paytable
Red 2 : Blue 1 Pays 8:1
Red 2 : Blue 2 ( Pair of deuces ) Pays 8:1
Red 2 : Blue 3 Pays 8:1
Red 2 : Blue 4 Pays 8:1
Red 2 : Blue 5 Pays 8:1
Red 2 : Blue 6 Pays 8:1
Red 1 : Blue 2 Pays 8:1
Red 2 : Blue 2 ( Pair of deuces ) Pays 8:1 : Someone has struck that out in biro
Red 3 : Blue 2 Pays 8:1
Red 4 : Blue 2 Pays 8:1
Red 5 : Blue 2 Pays 8:1
Red 6 : Blue 2 Pays 8:1
Side bet
If you wish to cover both combinations of...
Red 2 : Blue 1 and Red 1 : Blue 2 with one just bet, please use the side-bet betting area where payout will be at 4:1
Every other outcome is a push.
[Camera angle goes back over Alan's shoulder.]
Quote: AlanSo, Why is one line struck out in biro?
Quote: ThrowerSorry sir, a bit of a cock-up by our printer. That pay result is already on the list.
Quote: AlanSo, Getting back to the game. That die under the cup might be Blue?
Quote: ThrowerYes sir, But I'm not allowed to reveal that until all bets are placed.
Quote: AlanSo, That die under the cup might be Red? And it definitely has six faces numbered 1 to 6?
Quote: ThrowerWell sir, I'm not allowed to reveal the color, but I can confirm that it has six faces and it might have landed on any one of those six faces.
Quote: AlanSo, That die under the cup might be any of 1, 2, 3, 4, 5, 6 without any sort of bias?
Quote: ThrowerWell sir, I'm still not allowed to reveal the color, but I can confirm that it has six faces and had an equal chance of landing on any one of those six faces.
Quote: AlanSo, That die under the cup might be any of Red1, Red2, Red3, Red4, Red5, Red6, Blue1, Blue2, Blue3, Blue4, Blue5, or Blue6 without any sort of bias?
Quote: ThrowerWell sir, I'm still not allowed to reveal the color, but I can confirm that it could have been any one of those outcomes under the cup, each with pretty much equal probability.
Quote: AlanWhat do you mean 'pretty much equal probability'?
Quote: ThrowerWell sir, If the die under the cup is indeed a deuce, I had a choice of which to put into may pocket. If it's indeed a deuce, I could have chosen which to leave behind. I'm a random kind of guy, so half the time I'd leave the blue one and half the time I'd leave the red one. So Blue2 and Red2 are each half as likely as any other outcome.
[Alan shrugs as a mathematical fog descends on him]
Quote: AlanIf you left a red die under the cup, it's not suddenly going to flip to being blue
[Thrower looks puzzled]
Quote:Nothing's going to flip sir, what's under the cup is under the cup. Might be red. Might be blue. Might be 1 or 2 or3 or 4 or 5 or 6
Quote: ThrowerRemember sir, I'm still happy to pay out as per the paytable.
Quote: ThrowerHowever sir, remember that I put one die in my pocket. So you are really only betting on the outcome of the other die. The result will be shown when I lift the cup. At that time, I will reach the first die from my pocket and I will set it face up as a deuce, just like it was when it was under the cup.
[Meanwhile a crowd has formed. There are ten other members of Alan's forum and they want a piece of the action.]
[The crowd and Alan all mutter in unison . . .]
Quote: Alan and his friendsThe result is to be decided by just the one die that's still under the cup
And that die has only six faces.
And the table is paying 8:1 on everything !!!!!
[A betting frenzy ensues. ten more bets are placed. All different. Every possibility is covered by a crisp dollar bill. Nobody bothers with the crappy side bet.]
[Pause for theatrical effect.]
Quote: ThrowerNo more bets.
[The thrower lifts the cup.]
[He pulls a die from his pocket and sets it as a deuce, next to the other die.]
[One player wins and has his original dollar and 8 more dollars shoved towards him (It's never revealed who won.)]
[The Thrower scoops up ten other losing bets]
[One of the assembled crowd thinks out loud . . .]
Quote: think bubbleSurely all of the eleven results on that paytable were equally likely?
[Camera zooms in on Thrower's big cheesy grin.]
Directors Cut.
Shortly after this film was made, there was uproar about the effect that having different colored dice had made to the payout rates.
The game is still in circulation, but now, it's only accepting bets on 'Pair of deuces' All mention of dice color has been struck off the pay-table. It still pay's 8:1 All eleven game outcomes where 'At least one of the dice is a deuce' are still in play and all occasions where neither die is a deuce are still pushes.
The Thrower has now sold the game design as a franchise for tens of millions of dollars.
Quote: indignant99Only in the minds of unclear thinkers.
"Although our intellect always longs for clarity and certainty, our nature often finds uncertainty fascinating." Carl von Clausewitz
"It took the madmen of yesterday for us to be able to act with extreme clarity today. I want to be one of those madmen. We must dare to invent the future." Thomas Sankara
Clarity seems an evenly matched tug-of-war between nature and nurture rather than a drawn-out, monotonous argument.
Quote: RSAwww, poor Alan had everyone lined up against him on his forum so he locked the thread. ;(
Didn't you know. It's interfering with all the Caesar's/Total Rewards/video poker talk.
Quote: OnceDear[A betting frenzy ensues. ten more bets are placed. All different. Every possibility is covered by a crisp dollar bill. Nobody bothers with the crappy side bet.]
We can film at my casino. I only have one table game.
It's called "Two For The Show CRAPS," and here's the table layout:
OnceDear, you may share half-ownership of this casino, if you like.
I gotta remove that "4-way Corner" bet. It'll bankrupt me.
Quote: indignant99Quote: OnceDear[A betting frenzy ensues. ten more bets are placed. All different. Every possibility is covered by a crisp dollar bill. Nobody bothers with the crappy side bet.]
We can film at my casino. I only have one table game.
It's called "Two For The Show CRAPS," and here's the table layout:
OnceDear, you may share half-ownership of this casino, if you like.
I gotta remove that "4-way Corner" bet. It'll bankrupt me.
Nice work Indi. Your casino is more generous than mine paying 9:1 compared to my 8:1. But still pretty darned lucrative.
I owe you another beer. See you on my Yacht later.
Quote: OnceDearNice work Indi. Your casino is more generous than mine paying 9:1 compared to my 8:1. But still pretty darned lucrative.
I owe you another beer. See you on my Yacht later.
Couldn't I have a jumbo Glenlivet, instead?
In all honesty, your script/screenplay tipped my scale to design this game. (I was kinda cogitating a table layout for a long time, now. You released my avalanche.)
If our casino gets customers, I have another game to introduce.
It's called "Lucky 'Leven CRAPS."
There is no f***ing way Alan will see that "his bet" - his dice-in-the-cup scenario - is right there on the felt.Quote: indignant99
It's not for lack of colour now, thanks to your layout.Quote: indignant99There is no f***ing way Alan will see that "his bet" - his dice-in-the-cup scenario - is right there on the felt.
You're right of course. On his forum he has already dismissed the script and your layout totally.
"Once upon a time... Am I going to fast for you Alan... Let's start again. Sorry Children, I cannot proceed until Alan catches up"
Well, though the video script was tongue in cheek, it deliberately tries to give a real world feel to the scenario, so no-one can accuse me of baffling with numbers or maths or workbooks. I did have to introduce colours, which is a pity, but hopefully someone who has half a brain and is close to Alan's ear will get him to stop looking at one f**king die and realise that nothing has to flip from one die to the other in animatronics animation, for him to simply not have enough information to say there are only six sides of one identified die to take into account.
Note that in the script, the thrower could have told Alan the colour of the dice in his pocket, only if Alan was betting on the pair. It would not have given him any more data, but of course would have influenced any other bettors.
At least we got that going for us.
Quote: WizardThis may have been asked before, but let me see if this will help Alan and the 1/6 camp.
A woman is chosen at random from all women in the world with two children. She is asked, "Do you have at least one boy?" She answers, "yes." What is the probability she has two boys?
I found an added wrinkle to this puzzle which makes it even more interesting -
A woman is chosen at random from all women in the world with two children. She has two children, one of whom is a son born on a Tuesday. What is the probability she has two boys?
This version of the puzzle was posed by Gary Foshee, a puzzle designer from Issaquah, Washington.
The assumptions are that boys and girls are equally likely, and that births on any particular day of the week are equally likely.
Quote: indignant99There is no f***ing way Alan will see that "his bet" - his dice-in-the-cup scenario - is right there on the felt.Quote: indignant99
Since this game is so favorable in player-edge to Alan,
he should always keep table max ($5000) bets on:
- Pair of Deuces. Only 5-to-1 against, but pays a sparkling 9-to-1.
- Red Street. Only pays even money (1-to-1), but damn, 5-of-6 probability to win!
- Blue Street. Only pays even money (1-to-1), but damn, 5-of-6 probability to win!
Quote: RSAwww, poor Alan had everyone lined up against him on his forum so he locked the thread. ;(
The problem with the name forums. The one who "gets up the skirt" feels the fire from both sides. In Alan's case, it's another thing to take the argument in completely different directions. Time to retire with dignity?
“In the land of Gibberish, the man who makes sense, the man who speaks clearly, clearly speaks nonsense.” ― Jarod Kintz, This Book Has No Title.
Quote: RSLooks like Alan is holding true to his "i'm not posting in this thread" post [not verbatim, don't ban me].
At least we got that going for us.
Lol. I'm still here.
Quote: KerkebetMy issue here is that something clearly remains unresolved. And, the onus lays on the side of the mathematicians to clear it up. Certainly, not on television reporter Alan alone.
Why the apparently protracted angst against one or two hold-outs? Off hand, I see nothing to gain from thus trying to convert Alan to the forum's way of thinking here; except, some sort of sensationalism for the site. But, the apparent angst continues.
Lastly, how do you know that you have the only right answer? There are other ways to look at the gist of the problem as written. There almost always are.
Angst is in the eye of the beholder. Plus, those who deny an objectively and demonstrably correct mathematical formula are often met with derision and flabbergaster. (That's my new word for the day: the noun form of "flabbergasted").
This isn't a subjective question like tax policy, landscape design, or how fast should you take "Anthropology". Many questions have multiple interpretations but this isn't one of them. As someone said early on, Alan's problem is more directly a reading comprehension one -- he doesn't think the words say what they actually say. Alan's repeated (and improper) reference to "the other die" is the clincher because there's no sensible way to refer to "the other die" when both dice are already 2.
Directly to your question, I am sure there is only one right answer because the question is quantifiable in terms of probability. It's not a distribution, it's a single value:
Given a fair toss of a pair of standard 6 sided dice, where Event A is "at least one die shows 2" and Event B is "both dice show 2" then P(B|A) = 1/11. I posted a derivation earlier in the thread. It may not be perfectly clear to someone who hasn't (yet) studied conditional probability, but that doesn't mean there are multiple correct answers.
Someone who hasn't studied conditional probability, or probability in general, may dispute the answer and even be willing to bet on it, just as Alan is. Which is not uncommon; there's an entire industry that caters to people who are willing to bet based on their misperception of probabilities.
Quote: Dalex64A woman is chosen at random from all women in the world with two children. She has two children, one of whom is a son born on a Tuesday. What is the probability she has two boys?
This version of the puzzle was posed by Gary Foshee, a puzzle designer from Issaquah, Washington.
The assumptions are that boys and girls are equally likely, and that births on any particular day of the week are equally likely.
Quote: Wizard481/999
That's what I got too, but I ended up with
Quote: MathExtremistDid you intentionally multiply the denominator in your answer or did you derive it that way? If the latter, how did you derive it?
I did the math in Excel, which gave me 0.481481481...
I could see three digits repeating so I multiplied by 999 to get 481. Too lazy to reduce it further I guess.
Quote: WizardQuote: Dalex64A woman is chosen at random from all women in the world with two children. She has two children, one of whom is a son born on a Tuesday. What is the probability she has two boys?
This version of the puzzle was posed by Gary Foshee, a puzzle designer from Issaquah, Washington.
The assumptions are that boys and girls are equally likely, and that births on any particular day of the week are equally likely.481/999
| Weekdays | Prob | Siblings Incl | Siblings Excl |
|---|---|---|---|
| (TT) Tues-Tues | 1/49 | BB BG GB | GG |
| (TO) Tues-Other | 6/49 | BB BG | GB GG |
| (OT) Other-Tues | 6/49 | BB GB | BG GG |
| (OO) Other-Other | 36/49 | none | all |
(12/13 * 1/2) = 1/39 + 18/39 = 19/39
Okay, now clear the air, of that Brain Fart I just emitted above.
| Weekdays | Prob | Siblings Incl | Siblings Excl |
|---|---|---|---|
| (TT) Tues-Tues | 1/13 | BB(1/3) _ BG(1/3) _ GB(1/3) | GG |
| (TO) Tues-Other | 6/13 | BB(1/3) _ BG(1/3) | GB GG |
| (OT) Other-Tues | 6/13 | BB(1/3) _ GB(1/3) | BG GG |
(1/13*1/3)+(6/13*1/3)+(6/13*1/3) + (1/13*2/3)+(12/13*1/3) = 13/27.
Quote: indignant9919/39. Wiz's answer can be reduced 481/999 = 13/27.
Weekdays Prob Siblings Incl Siblings Excl (TT) Tues-Tues 1/49 BB BG GB GG (TO) Tues-Other 6/49 BB BG GB GG (OT) Other-Tues 6/49 BB GB BG GG (OO) Other-Other 36/49 none all
( 1/13 * 1/3) +
(12/13 * 1/2) = 1/39 + 18/39 = 19/39
Except by reversing the order of factors, you end up with a different answer:
| Genders | Prob | Days Incl | Days Excl |
|---|---|---|---|
| (BB) | 1/4 | TT TO OT | OO |
| (BG) | 1/4 | TT TO | OT OO |
| (GB) | 1/4 | TT OT | TO OO |
| (GG) | 1/4 | none | all |
(1/3 * 1/3) +
(2/3 * 1/2) = 1/9 + 3/9 = 4/9 (!= 19/39)
So what went wrong?
Quote: Dalex64I found an added wrinkle to this puzzle which makes it even more interesting -
A woman is chosen at random from all women in the world with two children. She has two children, one of whom is a son born on a Tuesday. What is the probability she has two boys?
This version of the puzzle was posed by Gary Foshee, a puzzle designer from Issaquah, Washington.
The assumptions are that boys and girls are equally likely, and that births on any particular day of the week are equally likely.
By the way, this is the same exact variant I proposed back on page 17 of this thread (but using dice/color instead of sex/birthday).
Thanks.
Quote: AlanMendelsonWizard: If I am wrong, the same way you can visualize the coin problem by flipping heads and tails, and the children problem with stick figures of boys and girls, please use two dice to help me visualize the dice problem.
Thanks.
Posts in this thread have proposed many different ways for you to use two dice to help you visualize the problem. Have you actually performed them? If not, why should the Wizard type up another?
Quote: AlanMendelsonWizard: If I am wrong, the same way you can visualize the coin problem by flipping heads and tails, and the children problem with stick figures of boys and girls, please use two dice to help me visualize the dice problem.
Thanks.
You are incapable of visualizing 1/11. The entire time you'll just be saying, "How stupid are the WOVers....I'm looking at these dice, the 2 can't jump from one die to the next. They must be playing a trick on me!!!"
Quote: Dalex64I found an added wrinkle to this puzzle which makes it even more interesting -
A woman is chosen at random from all women in the world with two children. She has two children, one of whom is a son born on a Tuesday. What is the probability she has two boys?
This version of the puzzle was posed by Gary Foshee, a puzzle designer from Issaquah, Washington.
The assumptions are that boys and girls are equally likely, and that births on any particular day of the week are equally likely.
Some follow-up, with applications to the original problem:
"If you pick a random child out of all the households with two children and condition on the event that they're a boy born on a Tuesday, then the probability that the other child is also a boy is one half simply by virtue of the fact that we're assuming that the sexes and birthdates of the two children are independent. On the other hand, if we pick a random household and condition on the event that one of the children is a boy born on a Tuesday, the answer is not one half because that double counts the households where both children are boys born on Tuesdays." (comment by Justin Rising, emphases mine)
http://www.quora.com/What-is-the-reasoning-behind-Gary-Foshees-boy-born-on-a-Tuesday-problem-where-providing-additional-apparently-irrelevant-information-changes-the-resulting-probabilities
In the original dice problem, we are not picking a single die showing a 2 and asking, under the condition that at least one of the dice shows 2, what is the probability that the other die also shows 2. What we are doing is actually picking a pair of dice, under the condition that at least one of the dice shows 2, and asking what is the probability that both dice show 2. This is equivalent to selecting a random household vs a random child in the children problem above, and for exactly the same reason (double-counting), is why the probability of both dice being 2 in the original problem is not 1/6 (or 2/12). When you subtract out the double-counting, you get (2-1)/(12-1) or 1/11.
Quote: RSYou are incapable of visualizing 1/11. The entire time you'll just be saying, "How stupid are the WOVers....I'm looking at these dice, the 2 can't jump from one die to the next. They must be playing a trick on me!!!"
Yes, RS. I cannot visualize 1/11 using two dice, with at least one die showing a 2. Please show it to me. It should be simple to do.
There have been graphs shown, and there have been H and T shown for coin flips.
Well, how about using two dice to show me how it all boils down to 1/11 ??
I've been asking this for several days now. Instead I've been told, in so many words, that I am ignorant or stupid or words with underscores instead of letters. Now I am saying, please show me so I will finally understand. Please... take two dice and show me why the answer is 1/11.
Quote: MathExtremistPosts in this thread have proposed many different ways for you to use two dice to help you visualize the problem. Have you actually performed them? If not, why should the Wizard type up another?
I am sorry but I missed that post or multiple posts. Would you mind telling me the post number or copying the info? Thanks.
Quote: AlanMendelsonI am sorry but I missed that post or multiple posts. Would you mind telling me the post number or copying the info? Thanks.
For starters, go back through this thread and look at all the questions I've asked you about various dice probabilities. Try to answer them.
You can also find the posts I've made to this thread here:
https://wizardofvegas.com/member/mathextremist/posts/
Just give me a picture of two dice (it doesn't have to be a video) with at least one die showing a 2 and then explain to me how the answer is 1/11 for both dice showing a 2.
Yeah, I'm stupid. So humor me.
Wizard, you have a camera.
Quote: AlanMendelsonI'm sorry, ME, but I was able to use two dice to show why I think the answer is 1/6. Why can't you use two dice to show the answer is 1/11 ??
Just give me a picture of two dice (it doesn't have to be a video) with at least one die showing a 2 and then explain to me how the answer is 1/11 for both dice showing a 2.
Yeah, I'm stupid. So humor me.
Wizard, you have a camera.
Like I said before, you are incapable of recognizing 1/11. We have already said this before: look at all the faces (there are 12) on the two dice. Remove one of the deuces. You are left with 11 faces, one of which is a deuce. Now please stop trolling.
I already did. You ignored it.Quote: AlanMendelsonI'm sorry, ME, but I was able to use two dice to show why I think the answer is 1/6. Why can't you use two dice to show the answer is 1/11 ??
Just give me a picture of two dice (it doesn't have to be a video) with at least one die showing a 2 and then explain to me how the answer is 1/11 for both dice showing a 2.
Yeah, I'm stupid. So humor me.
https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/42/#post455861
If you'd like to not ignore it this time, please look at the following picture that shows all combinations of two dice.
Please answer:
a) How many combinations of two dice have "at least one die showing a 2"?
b) How many combinations of two dice have "both dice showing a 2"?
Quote: MathExtremist
Please answer:
a) How many combinations of two dice have "at least one die showing a 2"?
b) How many combinations of two dice have "both dice showing a 2"?
First I want to say that I am sure you are a very nice and intelligent and worldly man and I appreciate your help in making me understand.
The problem I AM HAVING is that I can't consider 11 different combinations with at least one die showing a 2 in a two-dice problem. And let me explain why.
With each die, when a 2 is showing, there can no longer be 11 different combinations. Let me show you:
2....1
2....2
2....3
2....4
2....5
2....6
So, when you tell me that at least ONE DIE is showing a 2, I look at my combinations above and see that there are only 6 -- not 11.
If we look at the mirror image of the combinations above I still get only 6 combinations -- not 11.
1....2
2....2
3....2
4....2
5....2
6....2
Now, it is also possible that when we say "at least one die" it can also mean that BOTH dice are showing a 2. Well, when I look at two dice, both showing a 2, I treat either of the dice as being the "variable die." And by that I mean, this:
2....2 (but it could have been a 1)
2....2
2....2 (but it could have been a 3)
2....2 (but it could have been a 4)
2....2 (but it could have been a 5)
2....2 (but it could have been a 6)
And what this indicates to me, is that even if both dice were showing a 2, you still have only a 1/6 of getting that 2-2 combination.
So this is why I ask: using two dice, show me how the answer is 1/11 ??
Can you visualise this.
You have two dice. Examine them closely.
They have, between them, 12 faces. Not 36.
On two of those faces we see a two.
We know one of those twos has been all used up in the face up die that the peeker saw.
We aren't sure which of those twos has been used up, so we cannot eliminate either die.
That means all of the eleven 'other' faces are 'in play'
Off those 11 other faces, one of them would be a deuce. That deuce would surface how often......
Oh to hell with this. Go look at indi's layout again. totally forget visualizing dice.
Look at the table layout. Study it. Study the payout rates.
Work out a strategy for making a profit.
In fact, why not spend some time reading and looking at what has been shown to you, rather than asking for something else to ignore.
Maybe even think on this........ How many skilled and profitable players have jumped in to take bets on offer here. Wouldn't you have expected a stampede? I've personally offered to stake my pension fund. If I'm an idiot for offering that risk, how many bigger idiots turned down the chance? Why do you think that is?
Note To ME.... he's indicated on his own forum that he didn't in any way, shape or form 'get' the two child problem. He didn't even acknowledge the enigma. So save your breathe there.
Note to face. Suspend me if you want.
Quote: AlanMendelson almost but didn't say
2....2 (but it could have been a 2...1 AND it could have been a 1...2)
2....2
2....2 (but it could have been a 2...3 AND it could have been a 3...2)
2....2 (but it could have been a 2...4 AND it could have been a 4...2)
2....2 (but it could have been a 2...5 AND it could have been a 5...2)
2....2 (but it could have been a 2...6 AND it could have been a 6...2)
And what this indicates to me, is that even if both dice were showing a 2, you still have only a 1/11 of getting that 2-2 combination.
Then you have to remember that although there are 6 possibilities, they do not occur with equal probability.
When you roll two dice and at least one of them is a two, the probability of each of the combinations occurring is 2/11, except for 2-2 which only occurrs 1/11
That is just the nature of the 11 ways that two dice can land showing at least one two.
You can see that there are two ways for the dice to land 2-1: Either 2-1 or 1-2. So it is fine if you only want to consider 2-1 as the only way to count the 2 and a 1 result, but there are two ways for you to get there. Same with the rest, except for 2-2.
So, when one die is a 2, the other one is:
A 2 one time in 11
A 1,3,4,5,6 each two times in 11
Let me label them for you:
Quote: AlanMendelson
(1) 2....1
(2) 2....2
(3) 2....3
(4) 2....4
(5) 2....5
(6) 2....6
(7) 1....2
(8) 2....2
(9) 3....2
(10) 4....2
(11) 5....2
(12) 6....2
But combinations 2 and 8 are double-counted, so there are only 11 different combinations. And because combination 2 and 8 are double-counted, there is only one way of getting that 2-2 pair. So there is one (1) way out of 11 combinations to get 2-2, using your own listing of the combination of 2 dice.
Quote: OnceDearsomeone tampered with your yellow dice.
Yeah, I'm seeing 10 combinations there...! :)
Quote: MathExtremistQuote: indignant9919/39. Wiz's answer can be reduced 481/999 = 13/27.
Weekdays Prob Siblings Incl Siblings Excl (TT) Tues-Tues 1/49 BB BG GB GG (TO) Tues-Other 6/49 BB BG GB GG (OT) Other-Tues 6/49 BB GB BG GG (OO) Other-Other 36/49 none all
( 1/13 * 1/3) +
(12/13 * 1/2) = 1/39 + 18/39 = 19/39
Except by reversing the order of factors, you end up with a different answer:
Genders Prob Days Incl Days Excl (BB) 1/4 TT TO OT OO (BG) 1/4 TT TO OT OO (GB) 1/4 TT OT TO OO (GG) 1/4 none all
(1/3 * 1/3) +
(2/3 * 1/2) = 1/9 + 3/9 = 4/9 (!= 19/39)
So what went wrong?
What went wrong, in your calculation, are the probabilities of the day-combinations.
In essence, you've elevated Tues-Tues to parity with T-O and O-T.
| Genders | Prob | Days Incl | Days Excl |
|---|---|---|---|
| (BB) | 1/3 | TT(1/13) _ TO(6/13) _ OT(6/13) | OO |
| (BG) | 1/3 | TT(1/13) _ TO(6/13) | OT OO |
| (GB) | 1/3 | TT(1/13) _ OT(6/13) | TO OO |
1/3 * (1/13 + 6/13 + 6/13) /
[1/3 * (1/13 + 6/13 + 6/13) + (1/3 * 7/13) + (1/3 * 7/13)] = 13/27.
Oops.Quote: OnceDearsomeone tampered with your yellow dice.
That's pretty good, miplet! Are those casino dice?Quote: mipletOver slept today so no video. I'll do one tomorrow. Here's a picture to hold you over.
Quote: mipletOver slept today so no video. I'll do one tomorrow. Here's a picture to hold you over.
I might suggest that mirror-image pip-counts, be shown in (different) mirror-image colors. Why? Because it makes it very obvious that a 3-5 is a distinct outcome versus a 5-3 -- you couldn't argue that it was really just one unique event, re-arranged.
Quote: MathExtremist
That stubborn gif file is not displaying for most viewers. I can help convert it to an image that WILL display properly.
Not quite. You're not using TWO DICE. This is a two dice problem.
Now, show me 11 combinations with at least 1 die showing a 2, with a total of only TWO DICE.
Quote: AlanMendelson
Not quite. You're not using TWO DICE. This is a two dice problem.
Now, show me 11 combinations with at least 1 die showing a 2, with a total of only TWO DICE.
Now that's talking stupid !!!!
This is where the video should show up. Miplet, where are you?!Quote: OnceDearQuote: AlanMendelson
Not quite. You're not using TWO DICE. This is a two dice problem.
Now, show me 11 combinations with at least 1 die showing a 2, with a total of only TWO DICE.
Now that's talking stupid !!!!
Edit: I just made a video.
Quote: indignant99Quote: MathExtremist
That stubborn gif file is not displaying for most viewers. I can help convert it to an image that WILL display properly.
http://www.imageurlhost.com/images/c8mfag0jewsaxvno1w9.jpg
Quote: AlanMendelson
Not quite. You're not using TWO DICE. This is a two dice problem.
Now, show me 11 combinations with at least 1 die showing a 2, with a total of only TWO DICE.
Alan,
I've been giving you a lot of latitude based on the past, your many good conversations, and your apparent confusion, but is it possible you're NOT trolling the forum at this point?
