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Gialmere
Gialmere
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July 4th, 2022 at 11:50:29 AM permalink
Quote: DogHand

Answers:

#1: (b) is best, followed by (a)

#2: Roll the X; the Large Straight probability is 11/36

#3: Keep just the 2's

Dog Hand
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Quote: gordonm888



1.1 With 4,4,4,5,5 you should (b) Keep all five dice. Your score is 25 points. You can only be beaten by a small or large straight or Yahtzee or by a hand that totals 26 or higher.

1.2 Your second best option is (a) keep the 3 fours and roll for 4oaK or Yahtzee.

2. With a 1-2-4-5-X (X not a three) you should reroll the die with X. This gives you 2 rolls to hit a 3, you miss 25,36 of the time and make it 11/36 = 0.305556 of the time.

If you retain 2-4-5, your total probability of a large straight is 0.24537 in this way:

Roll a 3 + (1 or 6) and stop. P= 4/36 = 0.111111
Roll 3 + (2,4,5),reroll one die and make (1,6) P=6/36*2/6=0.055556
Roll two 3s, reroll one die and make (1,6) P= 1/36*2/6= 0.009259
Roll a (1,6) + (2,4,5), reroll one die and make a 3 P= 9/36*1/6=0.055556
Roll a (2,4,5) + (2,4,5), reroll both dice and make a 3 + (1,6) P= 9/36* 2/36= 0.013889
Total chance of Large Straight = 0.24537

3a) Keeping the pair of 2s is better than keeping 1-1-2-2.

With a pair of 2-2, you have three chances to roll a 2, twice.
With 1-1,2-2 you have one chance to roll a either a 1 or 2, twice.

3*1*2 > 1*2*2 so keep the 2-2 and reroll 3 dice.

3b) With a 1-1-2-2-3 at the beginning of the game you should roll for a full house both times, because its worth 25 points. If you miss, you should count your hand as 2 points in the ONEs category

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Correct!

For the most part

From Riddler...

Let’s walk carefully through two possible strategies.

First, suppose you keep all your dice. This one’s easy: You’ll score 25 points for sure for your full house. For the other strategy to be preferable, it needs to net more than 25 points, on average.

So suppose you risk it and reroll your two 5s. There are 36 (or 6×6) ways those two dice could fall. In only one of those cases will they both land as 4s, giving you your 50-point Yahtzee. In five other cases, where your two rerolled dice match, you’ll have another full house and earn 25 points. In the other 30 cases, you’ll wind up with a three- or four-of-a-kind, which is worth only the sum your five dice. The average of the sum of your dice in those 30 cases is 19. So, putting this all together, your expected score when rerolling your two 5s is (1/36)(50) + (5/36)(25) + (30/36)(19) ≈ 20.69 points.

In fact, you’re better off keeping the numbers you have in front of you than you are rerolling any combination of your dice. A handful of solvers wrote programs to calculate the expected score of possible rerolls. Every option, from rerolling a single die to rerolling all five, scores less than 25 points in expectation. But surprisingly to me, if you were forced to reroll, your best bet would be to keep one 4 and one 5 and just let the other three rip. You’d score an average of 22.76 points in that case, since you can catch a few straights that way.

You should just reroll your X in hopes of getting a 3.

Why? There are two future rolls — your second and third rolls — we need to consider. If we reroll just the X, there is a 1/6 chance we complete our large straight with the first roll, and a 5/6 chance that we don’t. If we still need another roll, there is again a 1/6 chance we complete our large straight and a 5/6 chance that we don’t. Overall, that’s a (1/6) + (5/6)(1/6) = 11/36 ≈ .306 chance you complete your large straight using this strategy.

If we reroll the 1 and the X — keeping 2, 4 and 5 — it’s a little more complicated. For your first reroll, solver Guy Moore explained, there are four outcomes, each of which has an optimal next step for the second reroll.

--On the first reroll, there’s a 1/4 chance that you get two “useless” numbers — 2, 4 or 5. If you do get one of those useless numbers, the best thing to do is reroll each die, with a 1/9 chance of completing the large straight.

--On the first reroll, there’s a 7/36 chance to get one 3 and another useless number. If that happens, you should keep the 3 and throw the remaining die, which has a 1/3 chance of getting your remaining 1 or 6.

--On the first reroll, there’s a 4/9 chance to get at least one 1 or 6 and no 3. In that event, you should keep either the 1 or the 6 and reroll the remaining die, which has a 1/6 chance of hitting your remaining 3.

--And, finally, there’s a 1/9 chance to get a 3 and a 1 or a 6 on the first reroll. That completes our large straight, so there’s nothing to do next!

Putting those probabilities together, you have a (1/4)(1/9) + (7/36)(1/3) + (4/9)(1/6) + (1/9)(1) ≈ 27.8 percent chance to hit that large straight. That’s less than the 30.6 chance you had rerolling just one die. In other words, the strategies succeed with 11/36 and 10/36 probabilities, respectively.

According to the Yahtzee calculator, for Part #1 you should keep the 2s and reroll the other three dice at both decision points.

For Part #2, you should keep the 2s and reroll the other three dice after your first roll. Surprisingly, the second best option is to keep only the 3. The third is to reroll all the dice.

If it's after your second roll you should keep both pairs and reroll the 3. If it's after your third roll, score the aces.
Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
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July 5th, 2022 at 7:58:09 AM permalink
It's toughie Tuesday. Let's do magic...



The faro shuffle is also called a perfect shuffle. With a standard deck, the cards are divided into two piles of 26 cards. The two piles are then shuffled together in such a way that they perfectly interlace with a card from one pile followed by a card from the other pile followed by a card from the first pile and so on with no clumps of two or three cards from the same pile.

Since the way the cards are combined is known, it is possible to track and/or manipulate the position of cards in the deck. Because of this, the faro shuffle is studied by a motley assortment of card cheats, magicians and mathematicians.

A faro shuffle where the top card of the deck remains on top is called an "out-shuffle". If the top card becomes the second from the top, it's called an "in-shuffle". (See diagram.)

How many faro out-shuffles in a row are required to cycle the deck back to its original sequence?

How many in-shuffles would it take?


Have you tried 22 tonight? I said 22.
charliepatrick
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Gialmere
July 5th, 2022 at 12:26:55 PM permalink
^I'm not sure how to do it without resorting to brute force, or in my case a spreadsheet!
The first case sorts itself out after 8 iterations, #2 moves to position 3, 5, 9, 17, 33, 14, 27, 2.
The second case takes all 52 permutations to get back to the starting position, after 26 shuffles it has reversed the pack.
GM
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Gialmere
July 5th, 2022 at 1:47:51 PM permalink
Quote: Gialmere


The faro shuffle is also called a perfect shuffle. With a standard deck, the cards are divided into two piles of 26 cards. The two piles are then shuffled together in such a way that they perfectly interlace with a card from one pile followed by a card from the other pile followed by a card from the first pile and so on with no clumps of two or three cards from the same pile.

Since the way the cards are combined is known, it is possible to track and/or manipulate the position of cards in the deck. Because of this, the faro shuffle is studied by a motley assortment of card cheats, magicians and mathematicians.

A faro shuffle where the top card of the deck remains on top is called an "out-shuffle". If the top card becomes the second from the top, it's called an "in-shuffle". (See diagram.)

How many faro out-shuffles in a row are required to cycle the deck back to its original sequence?

How many in-shuffles would it take?

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I agree wth charliepatrick's reply above.

It is interesting to consider this problem for a deck consisting of 2n cards. The number of out-shuffles required for 2n cards is the same as the number of in-shuffles required for 2n-2 cards, because an out-shuffle is the same as keeping the top and bottom cards fixed and doing an in-shuffle of the remaining 2n-2 cards. The number of shuffles required does not follow an obvious pattern, but https://mathweb.ucsd.edu/~ronspubs/83_05_shuffles.pdf contains the formula and a proof of the fact that the number of in-shuffles required is always at most 2n.
Last edited by: GM on Jul 5, 2022
ThatDonGuy
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July 5th, 2022 at 4:42:38 PM permalink
Speaking of Toughie Tuesdays, I've got a couple, including one for the real math boffins as I am not sure a solution even exists.

Anyway...
Using only a straightedge and compass, construct an angle:
(a) of 18 degrees;
(b) whose sine is 3/10.
Note that (a) and (b) are different angles.

Bonus points if you can tell me what upcoming event involves an angle whose sine is 3/10 - actually, it's an angle twice that large.
aceside
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Gialmere
July 5th, 2022 at 7:33:53 PM permalink
“The number of out-shuffles required for 2n cards is the same as the number of in-shuffles required for 2n-2 cards.”

Interesting! Very related to gambling. Let me use one deck of card to verify this.

The number of out-shuffles required for 2x26 cards is 8;
The number of in-shuffles required for 2x26 cards is 52;
The number of in-shuffles required for 2x26-2=50 cards is 8.
Last edited by: aceside on Jul 6, 2022
aceside
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July 5th, 2022 at 7:49:03 PM permalink
Quote: ThatDonGuy

Speaking of Toughie Tuesdays, I've got a couple, including one for the real math boffins as I am not sure a solution even exists.

Anyway...
Using only a straightedge and compass, construct an angle:
(a) of 18 degrees;
(b) whose sine is 3/10.
Note that (a) and (b) are different angles.

Bonus points if you can tell me what upcoming event involves an angle whose sine is 3/10 - actually, it's an angle twice that large.
link to original post


Are you British? Why do I see so many British words in your question?

For your question (a), I’ve looked into my compass on my iPhone and found all 360-degree marks on it, so you can just easily draw a 18 degree angle using these marks.
Last edited by: aceside on Jul 5, 2022
ChesterDog
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July 5th, 2022 at 9:44:38 PM permalink
Quote: ThatDonGuy

Speaking of Toughie Tuesdays, I've got a couple, including one for the real math boffins as I am not sure a solution even exists.

Anyway...
Using only a straightedge and compass, construct an angle:
(a) of 18 degrees;
(b) whose sine is 3/10.
Note that (a) and (b) are different angles.

Bonus points if you can tell me what upcoming event involves an angle whose sine is 3/10 - actually, it's an angle twice that large.
link to original post




Here's a way to find the sine of 18 degrees, which is one fourth of the quantity of the square root of five minus one.

And the second picture roughly shows the compass and straightedge construction of that angle.



To construct an angle whose sine is 3/10, first make a circle of diameter 10. Then place the compass's point on an end (A) of a diameter and draw an arc of radius 3 intersecting the circle at B. Then connect B to the other end (C) of the diameter. The sine of angle ACB is 3/10.

Which event has a connection of that angle? Is it MLB's All Star Game?
Gialmere
Gialmere
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July 5th, 2022 at 10:47:37 PM permalink
Quote: charliepatrick

^I'm not sure how to do it without resorting to brute force, or in my case a spreadsheet!

The first case sorts itself out after 8 iterations, #2 moves to position 3, 5, 9, 17, 33, 14, 27, 2.
The second case takes all 52 permutations to get back to the starting position, after 26 shuffles it has reversed the pack.

link to original post


Quote: GM

I agree with charliepatrick's reply above.

It is interesting to consider this problem for a deck consisting of 2n cards. The number of out-shuffles required for 2n cards is the same as the number of in-shuffles required for 2n-2 cards, because an out-shuffle is the same as keeping the top and bottom cards fixed and doing an in-shuffle of the remaining 2n-2 cards. The number of shuffles required does not follow an obvious pattern, but https://mathweb.ucsd.edu/~ronspubs/83_05_shuffles.pdf contains the formula and a proof of the fact that the number of in-shuffles required is always at most 2n.
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Correct!!

Good show.

Note that by combining in and out shuffles many card tricks can be created. Of course, learning how to accurately faro shuffle while appearing to be casually shuffling a deck all while keeping up a conversation (perhaps telling jokes to an audience) would be a time consuming task to say the least.
------------------------------------------------------

Ian
Have you tried 22 tonight? I said 22.
DogHand
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aceside
July 6th, 2022 at 5:17:04 AM permalink
Quote: aceside

Quote: ThatDonGuy

Speaking of Toughie Tuesdays, I've got a couple, including one for the real math boffins as I am not sure a solution even exists.

Anyway...
Using only a straightedge and compass, construct an angle:
(a) of 18 degrees;
(b) whose sine is 3/10.
Note that (a) and (b) are different angles.

Bonus points if you can tell me what upcoming event involves an angle whose sine is 3/10 - actually, it's an angle twice that large.
link to original post


Are you British? Why do I see so many British words in your question?

For your question (a), I’ve looked into my compass on my iPhone and found all 360-degree marks on it, so you can just easily draw a 18 degree angle using these marks.
link to original post



aceside,

In geometry, a "compass" is the device on the right in the picture below:



It is used to draw arcs or full circles.

Hope this helps!

Dog Hand

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