If one die has 2, 3, 4, 5, 6, and 7
And the other has 0, 1, 2, 3, 4,and 5
that gives the same values and odds, right?
That’s true, Biotop, but I also requested that each side of each die be labelled with a positive whole number (thus, no zeros allowed!).
The two dice should be1-2-2-3-3-4 and 1-3-4-5-6-8
That’s right, Little Nemo! What was your discovery process?
Nice, Little Nemo ! Note that
your solution can be derived with the recomposition recipe I mentioned in a spoiler above:
{1,2,3,4,5,6} = {0,2,4} + {1,2}
{1,2,3,4,5,6} = {0,1,2} + {1,4}
{1,2,3,4,5,6} + {1,2,3,4,5,6}
= ( {0,2,4} + {1,2} ) + ( {0,1,2} + {1,4} )
= ( {0,2,4} + {1,4} ) + ( {0,1,2} + {1,2} )
= {1,3,4,5,6,8} + {1,2,2,3,3,4}
That’s right, septimus! We can prove the uniqueness of the solution by going further with the same ideas:
Note that what septimus notates as “{a, b, c, …}” can be thought of as a polynomial X[sup]a[/sup] + X[sup]b[/sup] + X[sup]c[/sup] + …; then what septimus notates as “+” between these objects corresponds to familiar multiplication of polynomials.
An allowed die, then, amounts to a polynomial with natural number coefficients with 6 terms (in the sense that its output is 6 when its input is 1), all of positive degree (in the sense that its output is 0 when the input is 0).
If we look at the general space of polynomials with integer (possibly negative) coefficients, it has the wonderful property of unique (up to signs) factorization into prime factors, just like the integers themselves (because of “Gauss’s lemma”).
A standard die is X[sup]1[/sup] + X[sup]2[/sup] + X[sup]3[/sup] + X[sup]4[/sup] + X[sup]5[/sup] + X[sup]6[/sup]. This has prime factorization X[sup]1[/sup] * (X[sup]0[/sup] + X[sup]1[/sup]) * (X[sup]0[/sup] + X[sup]1[/sup] + X[sup]2[/sup]) * (X[sup]0[/sup] - X[sup]1[/sup] + X[sup]2[/sup]). [Note the negative coefficient in the last factor!]
By the unique factorization property, we must take two copies of each of these factors and partition them among our nonstandard dice.
Each of our dice will need a factor of X[sup]1[/sup] to ensure all its sides have positive values.
Furthermore, at X = 1, we have that (X[sup]0[/sup] + X[sup]1[/sup]) = 2 and (X[sup]0[/sup] + X[sup]1[/sup] + X[sup]2[/sup]) = 3, while all the other factors of note are 1. Thus, each of our dice will need one of each of these last two factors to achieve output 6 at input 1.
This leaves only one decision where our hands are not tied: do we split the two factors of (X[sup]0[/sup] - X[sup]1[/sup] + X[sup]2[/sup]) among the two dice, or do we give both to one die and none to the other?
The former choice gives us standard dice; the latter choice (which luckily has no negative coefficients in the result) gives us our Sicherman dice:
X[sup]1[/sup] * (X[sup]0[/sup] + X[sup]1[/sup]) * (X[sup]0[/sup] + X[sup]1[/sup] + X[sup]2[/sup]) = X[sup]1[/sup] + X[sup]2[/sup] + X[sup]2[/sup] + X[sup]3[/sup] + X[sup]3[/sup] + X[sup]4[/sup]
X[sup]1[/sup] * (X[sup]0[/sup] + X[sup]1[/sup]) * (X[sup]0[/sup] + X[sup]1[/sup] + X[sup]2[/sup]) * (X[sup]0[/sup] - X[sup]1[/sup] + X[sup]2[/sup])[sup]2[/sup] = X[sup]1[/sup] + X[sup]3[/sup] + X[sup]4[/sup] + X[sup]5[/sup] + X[sup]6[/sup] + X[sup]8[/sup]
Nice one! I got the same result as Little Nemo, as we’d expect from your later proof (I promise I didn’t spoil myself).[spoiler]My process was experimental, using the notion that the net result is a convolution between the two dice. The normal case convolves ****** with ****** to form this:
*
**
**
*
So I tried to figure out what other shapes we could convolve together to form this shape. I knew the endpoints had to be singles, since the convolved shape has sharp ends. And although I wasn’t certain, symmetry was a strong possibility. So after a few other tries, I came up with this:
*
**
**
*
And I just took a greedy approach to finding the other die; the first number had to be 1 of course, and the next had to be 3, and so on. It all worked out rather nicely.[/spoiler]
I starting from needing a single 2, so I knew I needed each die to have one 1 side. Then I needed two 3’s, three 4’s, etc. so I just worked upwards filling in the numbers I needed to make the results I wanted.
After receiving too many complaints, Monty Hall revises his approach: now contestants choose a curtain by pure chance. If you select a red card you get the Ferrari, around-the-world cruise with the super-model and all the lobster you can eat. If you select a black card, you get a year’s supply of dog poop.
But you have the option of seeing part of the deck before you take your choice of card. Starting with his finger atop an ordinary deck of 52, Monty asks “Pass or Play?” If you Play the top card tells you your prize. If you Pass, Monty exposes the card you would have gotten; if it’s black your chance is now 26-25 instead of 50-50; whatever the color of the exposed card he asks “Pass or Play?” for the 2nd card. Once you say “Play” your curtain is opened, no switching allowed. Whenever you Pass, you get to see the color of the card you passed up. (If you Pass 51 times in a row, you’re required to Play the final card.)
What is your best strategy to get the Ferrari?
Do the dates matter, as in 2014 vs 2015? Could you have asked this question ten years ago?
This riddle could not have been properly done even six months ago. Most people (including me) would not have said that the word I am thinking of shared the common factor.
On septimus’s puzzle: I don’t think it makes a difference. At any time when you’ve seen more black than red, the odds favor you, and at any time when you’ve seen more red than black, they’re against you, but it’s random which one you get. You can try to wait until the first time that you’ve seen more black than red, which would give you a slight edge, but it also might not happen at all, and if you’re waiting for that and it doesn’t happen, then you’re sure to be stuck with a black card.
That’s right, Chronos. Indeed,
At any moment, the cards you haven’t seen are all the same to you; any one of them is just as likely to be red as any other one. So it makes no difference to a strategy’s winning chances if, instead of being given the next card when you finally say “Play”, you are given the last card instead. Which means your chances of winning are just the same as if you were forced from the beginning to take the last card: 50-50.
Yes. It seems a neat puzzle to me because it is tempting to imagine that waiting to see a black card is a good idea: after all, 50% of the time your odds immediately improve to 26/51. Yet, as Indistinguishable elegantly proves, all strategies have the same success rate.
If your strategy is Pick as soon as remaining Reds outnumber Blacks, you’ll usually get an advantageous pick before the end but 4.7% of the time you’ll end up with the last card and it will be Red only 21% of the time in those scenarios.
You can reverse the strategy and Pick when Blacks outnumber Reds! You’ll usually make a disadvantageous pick, but will be forced to the last card 4.7% which will be Red 79% of those scenarios … for the same net 50.000000% chance.
Of course, many words share the common factor, but I prefer to focus on and select from the two dozen or so that are closest to the word I am thinking of. That’s why I chose Harvard as opposed to Princeton or Yale.
Is this tied to a certain event?
No. The word I am thinking of and the similar words I listed will not change, common factor-wise, for the foreseeable future.
Well, there must have been some event of some sort within the last six months that caused most people to switch from not considering your word to have the common factor to now considering it to have the common factor, no?
Ah, good job, Indistinguishable. I was pretty sure there’d be an elegant proof, but I just couldn’t put my finger on it.
It reminds me of an old one from Click and Clack: There’s a plane with 100 seats, and it’s sold out: There are 100 people waiting in line to board. Everyone has a seat assigned on their ticket, but the first person in line is a senile old lady who chooses her seat at random. Everyone else on the flight is too polite to comment about this: They’ll sit in their assigned seat if available, or if it’s already taken, they’ll just pick some other seat at random (and this process can of course repeat). You’re the last person in line. What is the probability that you’ll end up in your assigned seat?
There may be a more elegant way to put this, but re: Chronos’s Click and Clack conundrum:
Consider the string “The senile old lady randomly took the seat assigned to X, who randomly took the seat assigned to Y, who randomly took the seat assigned to Z…”. At some point, this must wrap around into “…who randomly took the seat assigned to the senile old lady”. Everyone outside this cycle takes their own seat, no problem. Everyone within it takes someone else’s seat (except in the degenerate case where the senile old lady takes her own seat).
You’ll get to sit in your assigned seat, therefore, if within the string of random seat-selections, the old lady’s assigned seat is chosen before yours is. Of course, by symmetry, either one is equally likely to get chosen first, and so your chances are 50-50.