Easy math puzzles

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December 9th, 2020 at 8:21:08 AM permalink

NOTE: To play this puzzle, you must be familiar with the rules and components of the standard, Atlantic City version of Monopoly.

What is the most money you can possibly earn from the Go space in a single Monopoly game turn?

Explain how.

BONUS: What is the probability of it happening?
(This bonus probably isn't worth your time doing and, I confess, I don't know the answer since I keep thinking of new ways.)

Have you tried 22 tonight? I said 22.

Joeman

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December 9th, 2020 at 8:57:53 AM permalink

$600

1st roll is a 2 -- Community Chest:

2nd roll is another 2 -- Community Chest:

3rd roll is a 5 -- Chance:

Note that this does not account for the common, but horrible house rule of putting money from the bank into Free Parking.

ETA -- I like chevy's answer better than mine!

"Dealer has 'rock'... Pay 'paper!'"

chevy

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December 9th, 2020 at 9:03:34 AM permalink

roll 1+1 ----> Community Chest -----> Advance to Go, collect $200
roll 1+1 ----> Community Chest -----> Opening night at opera, Collect $50 from each of 7 (max) other players = $350
Roll 5 --------> Chance -----> Advance to Go, Collect $200. (or advance to Reading Railroad, collect $200 for p[assing GO)

Total = $750

Joeman

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December 9th, 2020 at 9:11:34 AM permalink

Quote: chevy

roll 1+1 ----> Community Chest -----> Advance to Go, collect $200
roll 1+1 ----> Community Chest -----> Opening night at opera, Collect $50 from each of 7 (max) other players = $350
Roll 5 --------> Chance -----> Advance to Go, Collect $200. (or advance to Reading Railroad, collect $200 for p[assing GO)

Total = $750

I didn't even think of the "Collect $50/opponent card." In my defense, I don't think I've ever played with more than 5 people, so it was never worth more than $200!

While we're at it, how about a:

More trivia than a puzzle, but... How many different property spaces (Street, RR, or Utility) are there to which a Chance card may direct your token? Can you name them?

One of them is tricky.

"Dealer has 'rock'... Pay 'paper!'"

Gialmere

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December 9th, 2020 at 9:16:46 AM permalink

The question asks what is the most money you can earn from passing/landing on Go in a single (not necessarily your first) turn.

Have you tried 22 tonight? I said 22.

chevy

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December 9th, 2020 at 9:23:16 AM permalink

Thanks Joeman. As to your bonus puzzle, I can think of

Directly Specified: Illinois, ST. Charles, Boardwalk, Reading
Indirectly specified : 2 nearest Utilities, 4 nearest railroads (only adds 3 to count)

If you include where you can end up, There is also Jail and GO....but you didn't list those as choices.

Would go back three to end on New York count? (I loved owning the oranges for exactly this reason by the way)

So maybe 9 + 2 + 1 = 12??

Joeman

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December 9th, 2020 at 9:32:12 AM permalink

Ah, I thought you meant how much starting from GO.

$800

Roll 1: start from Boardwalk and roll a hard 8 (or Park Place with a hard 10 or Short Line with a 12):
Pass Go and collect $200, then Chance:

+$200

Roll 2: 2 -- Community Chest:

+$200

Roll 3: 5 -- Chance:

+ $200 for a total of $800 all from passing Go.

"Dealer has 'rock'... Pay 'paper!'"

chevy

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December 9th, 2020 at 9:41:09 AM permalink

I took your original wording as "From the GO space" to mean that is where you were starting FROM.

I guess you meant it to mean earning as a result of the passing/landing on GO, without any other monies earned.

In that case I would get

Start on say Pennsylvania
Roll 4+4 to pass go (collect $200) and land on first community chest
----draw Adance to GO (Collect $200)
Roll 7 to land on Chance and Advance to GO (Collect $200)

Total = $600???

chevy

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December 9th, 2020 at 9:44:49 AM permalink

Joeman, Yep That's better then mine

Though I think your last roll is 7 since you have to get from GO to Chance....not from Community Chest to Chance??

ThatDonGuy

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December 9th, 2020 at 12:54:32 PM permalink

Quote: Gialmere
It's toughie Tuesday. You're hexed...

A hexagon with consecutive sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle.

What is the radius of the circle?

Sorry this is a little late...

Let 2a, 2b, and 2c be the measures of the angles opposite sides of 2, 7, and 11. Since there are two of each , 2 (2a + 2b + 2c) = 360 degrees, so c = 90 - (a + b).
Let R be the radius of the circle.
sin a = 2 / 2R; cos 2a = sqrt(4R^2 - 2) / 2R
sin b = 7 / 2R; cos 2b = sqrt(4R^2 - 49) / 2R
sin c = sin (90 - (a + b)) = cos (a + b) = cos a cos b - sin a sin b = 11 / 2R
(sqrt(4R^2 - 2) sqrt(4R^2 - 49) - 14) / 4R^2 = 11 / 2R
44 R^2 = 2R (sqrt(16 R^4 - 212 R^2 + 196) - 14)
22R + 14 = sqrt(16 R^4 - 212 R^2 + 196)
Square both sides, then reduce:
2R^3 - 87 R - 77 = 0 = (R - 7) (2 R^2 + 14 R + 11)
The only real root of this is R = 7, which is the solution

chevy

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December 9th, 2020 at 6:42:53 PM permalink

ThatDonGuy: Thanks for this, I could not get past my law of cosines approach to actually solve.

Comment in spoiler

I see what you did and follow it through.

Not sure if it matters as others may have moved on...but I think there may be a couple typos, not in the final equation or answer, just in typing up work.....(assuming I really am following your work)

You type cos(2a) and cos(2b) but I think you find cos(a) and cos(b) from sqrt(1-sin())...at least it seems that is what you plug in for.

And then I think cos(a) = sqrt(4R^2 - 2^2) = sqrt(4R^2-4)

That (4R^2-2) repeats two lines later

BUT it disappears because the equation after seems correct. (-212 = -4*49 - 4*4)....so I assume it was just transposing your work.

But it was clear what you intended and I thank you for ending my agony.

Gialmere

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December 9th, 2020 at 6:46:37 PM permalink

Quote: Joeman
Ah, I thought you meant how much starting from GO.

$800

Roll 1: start from Boardwalk and roll a hard 8 (or Park Place with a hard 10 or Short Line with a 12):
Pass Go and collect $200, then Chance:

+$200

Roll 2: 2 -- Community Chest:

+$200

Roll 3: 5 -- Chance:

+ $200 for a total of $800 all from passing Go.

Correct!

As mentioned, there are actually several ways to do this making finding an actual probability tedious. As Joeman points out, you must start on Boardwalk, Park Place or Short Line...

One Way
1) Roll doubles, pass Go ($200), land on Chance 7 and advance to boardwalk.
2) Roll a hard eight, pass Go ($200) land on Chance 7 and advance to Go ($200).
3) Roll 7, land on Chance 7 and take a ride on the Reading ($200).

Another Way
1) Roll doubles, pass Go ($200), land on Chance 7 and take a ride on the Reading ($200).
2) Roll midnight, land on Community Chest 17, advance to Go ($200).
3) Roll 7, land on Chance 7, advance to Go ($200).

I originally wanted to ask what the probability of going to jail on your first turn was (thinking it would be interesting mixing dice with cards) but quickly found it to be a can of worms.

----------------------------------------

If you're in a relationship and you can't quite work up the nerve to break things off, suggest a game of Monopoly.

Have you tried 22 tonight? I said 22.

chevy

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December 9th, 2020 at 8:07:22 PM permalink

Staying with Monopoly Theme

Question: What is the most money you can lose in a single turn of Monopoly?

(I have an answer, but past examples suggest I am not always right....)

Wizard
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Wizard

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December 10th, 2020 at 5:35:16 AM permalink

Quote: ThatDonGuy

22R + 14 = sqrt(16 R^4 - 212 R^2 + 196)
Square both sides, then reduce:
2R^3 - 87 R - 77 = 0 = (R - 7) (2 R^2 + 14 R + 11)
The only real root of this is R = 7, which is the solution

Since this problem is a bit dated, I'll dispense with the spoiler tags.

That said, how did you know you could make that reduction? I came to a similar equation that I didn't know how to solve without some kind of help.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Joeman

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December 10th, 2020 at 6:20:21 AM permalink

Quote: chevy
Thanks Joeman. As to your bonus puzzle, I can think of

Directly Specified: Illinois, ST. Charles, Boardwalk, Reading
Indirectly specified : 2 nearest Utilities, 4 nearest railroads (only adds 3 to count)

If you include where you can end up, There is also Jail and GO....but you didn't list those as choices.

Would go back three to end on New York count? (I loved owning the oranges for exactly this reason by the way)

So maybe 9 + 2 + 1 = 12??

You pretty much got it. And, yes, by "properties," I was asking specifically for those spaces that can be owned by a player.

1) Illinois Ave -- Via explicit instruction
2) St. Charles Place -- Via explicit instruction
3) Boardwalk -- Via explicit instruction
4) Reading RR -- Via explicit instruction or a "Nearest Railroad" from the chance next to Park Place
5) Pennsylvania RR -- Via a "Nearest Railroad" from the Chance next to Oriental Ave
6) B&O RR -- Via a "Nearest Railroad" from the Chance next to Illinois Ave
7) Electric Company -- Via a "Nearest Utility" from a Chance next to Oriental Ave or next to Park Place
8) Water Works -- Via a "Nearest Utility" from the Chance next to Illinois Ave

The 9th is the one that may be tricky to some folks. It's not Short Line, as SL is never the "Nearest Railroad" from any Chance space. As you stated, it is:

9) New York Ave -- Via a "Go Back 3 Spaces" from the Chance next to Illinois Ave

Quote: chevy
Staying with Monopoly Theme

Question: What is the most money you can lose in a single turn of Monopoly?

I would say $5,500, but you would get $200 for passing Go for a net loss of $5,300. Here's how:

First, you need to assume that both Boardwalk and Park Place have hotels and are owned by one of your opponents. Next, you need to be on B&O, Ventnor, Marvin Gardens, Pacific, Community Chest, or Short Line and roll doubles to land on Park Place (-$1500). Next roll would be a 2, landing on Boardwalk (-$2,000). Third roll would have to be an easy 8, landing on Chance and drawing an "Advance to Boardwalk" card (-$2,000).

This sequence would have you passing Go once, so you would collect $200 during your roll for a net loss of $5,300.

I think we need a separate "Easy Monopoly Puzzles" thread! ;)

"Dealer has 'rock'... Pay 'paper!'"

Wizard
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Wizard

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December 10th, 2020 at 6:56:25 AM permalink

Anyone know of a good site where we can play Monopoly against each other?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Gialmere

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December 10th, 2020 at 8:16:58 AM permalink

Quote: Wizard
That said, how did you know you could make that reduction? I came to a similar equation that I didn't know how to solve without some kind of help.

I'll post the official solve for the hexagon puzzle later today.

Have you tried 22 tonight? I said 22.

Gialmere

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December 10th, 2020 at 8:17:44 AM permalink

"Threes Away (Low)" is a game played with five standard dice. The object is to score the fewest points.

You begin your turn by rolling all five dice. You MUST set aside at least one of them for scoring but you may set aside as many as you wish. 3s score zero points. All other dice score their face values.

You then reroll any remaining dice and repeat the process until all your dice are scored. Low score wins.

Assuming optimal play, what is the expected value of your score?

If I'm not being clear with the rules, this video explains it in less than 90 seconds...

Have you tried 22 tonight? I said 22.

ThatDonGuy

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December 10th, 2020 at 8:32:40 AM permalink

Quote: Wizard
Since this problem is a bit dated, I'll dispense with the spoiler tags.

That said, how did you know you could make that reduction? I came to a similar equation that I didn't know how to solve without some kind of help.

When you square both sides:
484 R^2 + 616 R + 196 = 16 R^4 - 212 R^2 + 196
16 R^4 - 696 R^2 - 616 R = 0
2 R^3 - 87 R - 77 = 0

Okay, at this point I used Newton-Raphson with Excel to find the real positive root of 7.
Since 7 is a root, 2 R^3 - 87 R - 77 = (R - 7) multiplied by (a R^2 + b R + c).
"Obviously," a = 2
The R^2 coefficient = 0 = (-7 x 2) + (1 x b), which means b = 14
The R coefficient = -87 = (1 x c - 7 x b) = c - 98, which means c = 11
Check: (R - 7) (2 R^2 + 14 R + 11) = 2 R^3 + (14 - 14) R^2 + (11 - 98) R + ((-7) x 11) = 2 R^3 - 87 R - 77.

Also, there was a slight error in my statement which did not affect the solution:
2 R^2 + 14 R + 11 = 0
R = (-7 +/- 3 sqrt(3)) / 2
Both of the other roots are real, but neither is positive; 7 is the only positive solution.

chevy

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December 10th, 2020 at 9:13:30 AM permalink

Yes, I missed that Short Line is never the nearest railroad.

Quote: Joeman

I would say $5,500, but you would get $200 for passing Go for a net loss of $5,300. Here's how:

First, you need to assume that both Boardwalk and Park Place have hotels and are owned by one of your opponents. Next, you need to be on B&O, Ventnor, Marvin Gardens, Pacific, Community Chest, or Short Line and roll doubles to land on Park Place (-$1500). Next roll would be a 2, landing on Boardwalk (-$2,000). Third roll would have to be an easy 8, landing on Chance and drawing an "Advance to Boardwalk" card (-$2,000).

This sequence would have you passing Go once, so you would collect $200 during your roll for a net loss of $5,300.

I think we need a separate "Easy Monopoly Puzzles" thread! ;)

As to most money lost, My first solve was different than yours, but if I put our two together, I find an even larger loss.

Opponent owns ParkPlace/Boardwalk with Hotels
You own everything else....Have remaining 10 Hotels and all 32 houses built on them (game comes with only 32 houses and 12 hotels)

Roll doubles to land on Community Chest (in greens)
Draw the repair assessment card of $40 per house and $115 per hotel (32*40 + 115*10 = $2430)
Roll double (3+3) to land on Boardwalk = $2000
Roll easy 8 to pass GO and land on Chance (advance to Boardwalk) =$2000-200=$1800

Total = 6230

I originally had opponents own greens and blues....so land on Pacific (1400), Community chest (32 houses and 7 hotels....2085), boardwalk (2000). = $5485. I missed the advance to Boardwalk option.

Ace2

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December 10th, 2020 at 12:02:16 PM permalink

Quote: Gialmere

"Threes Away (Low)" is a game played with five standard dice. The object is to score the fewest points.

You begin your turn by rolling all five dice. You MUST set aside at least one of them for scoring but you may set aside as many as you wish. 3s score zero points. All other dice score their face values.

You then reroll any remaining dice and repeat the process until all your dice are scored. Low score wins.

Assuming optimal play, what is the expected value of your score?

If I'm not being clear with the rules, this video explains it in less than 90 seconds...

7.462890625

It�s all about making that GTA

Gialmere

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December 10th, 2020 at 2:32:42 PM permalink

Quote: Ace2
7.462890625

Sorry, no.

Have you tried 22 tonight? I said 22.

Gialmere

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December 10th, 2020 at 4:20:51 PM permalink

For those interested, here is the "official" solve for this week's Toughie Tuesday hexagon puzzle...

Question: A hexagon with consecutive sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle. Find the radius of the circle.

Each inscribed side of the hexagon subtends an angle at the center of the circle which is independent of its position in the circle. The sides are subject to the constraint that the sum of the angles subtended at the center equals 2pi. Hence we may permute the sides of the hexagon, from {2, 2, 7, 7, 11, 11} to {2, 7, 11, 2, 7, 11}. Since the two sets of sides, {2, 7, 11}, are congruent, each can be inscribed in a semicircle of the same radius as the original circle.

Have you tried 22 tonight? I said 22.

ThatDonGuy

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December 10th, 2020 at 6:10:58 PM permalink

Quote: Gialmere

"Threes Away (Low)" is a game played with five standard dice. The object is to score the fewest points.

You begin your turn by rolling all five dice. You MUST set aside at least one of them for scoring but you may set aside as many as you wish. 3s score zero points. All other dice score their face values.

You then reroll any remaining dice and repeat the process until all your dice are scored. Low score wins.

Assuming optimal play, what is the expected value of your score?

The expected value of a single die roll is 3, so, presumably, the strategy is to keep all dice 1,2,3 and reroll all dice 4,5,6, unless all of the dice in a roll are 4,5,6, in which case, keep the lowest one and roll the rest.
Let E(n) be the expected value of rolling n dice using these rules.

1 die: E(1) = 3

2 dice
Expected value of two high dice: 1/9 x 6 + 3/9 x 5 + 5/9 x 4 = 41/9
0 low: 1/4 x (41/9 + E(1)) = 17/9
1 low: 1/2 x (1 + E(1)) = 2
2 low: 1/4 x 2 = 1/2
E(2) = 79/18

3 dice
Expected value of three high dice: 1/27 x 6 + 7/27 x 5 + 19/27 x 4 = 117/27
0 low: 1/8 x (117/27 + E(2)) = 157/144
1 low: 3/8 x (1 + E(2)) = 97/48
2 low: 3/8 x (2 + E(1)) = 15/8
3 low: 1/8 x 3 = 3/8
E(3) = 193/36

4 dice
Expected value of four high dice: 1/81 x 6 + 15/81 x 5 + 65/81 x 4 = 341/81
0 low: 1/16 x (341/81 + E(3)) = 3101/5184
1 low: 1/4 x (1 + E(3)) = 229/144
2 low: 3/8 x (2 + E(2)) = 115/48
3 low: 1/4 x (3 + E(1)) = 3/2
4 low: 1/16 x 4 = 1/4
E(4) = 32,837/5184

5 dice
Expected value of five high dice: 1/243 x 6 + 31/243 x 5 + 211/243 x 4 = 335/81
0 low: 1/32 x (335/81 + E(4)) = 54,277/165,888
1 low: 5/32 x (1 + E(4)) = 190,105/165,888
2 low: 5/16 x (2 + E(3)) = 1325/576
3 low: 5/16 x (3 + E(2)) = 665/288
4 low: 5/32 x (4 + E(1)) = 35/32
5 low: 1/32 x 5 = 5/32
E(5) = 608,191 / 82,944 = about 7.33255

Gialmere

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December 10th, 2020 at 6:33:44 PM permalink

Quote: ThatDonGuy

The expected value of a single die roll is 3, so, presumably, the strategy is to keep all dice 1,2,3 and reroll all dice 4,5,6, unless all of the dice in a roll are 4,5,6, in which case, keep the lowest one and roll the rest.
Let E(n) be the expected value of rolling n dice using these rules.

1 die: E(1) = 3

2 dice
Expected value of two high dice: 1/9 x 6 + 3/9 x 5 + 5/9 x 4 = 41/9
0 low: 1/4 x (41/9 + E(1)) = 17/9
1 low: 1/2 x (1 + E(1)) = 2
2 low: 1/4 x 2 = 1/2
E(2) = 79/18

3 dice
Expected value of three high dice: 1/27 x 6 + 7/27 x 5 + 19/27 x 4 = 117/27
0 low: 1/8 x (117/27 + E(2)) = 157/144
1 low: 3/8 x (1 + E(2)) = 97/48
2 low: 3/8 x (2 + E(1)) = 15/8
3 low: 1/8 x 3 = 3/8
E(3) = 193/36

4 dice
Expected value of four high dice: 1/81 x 6 + 15/81 x 5 + 65/81 x 4 = 341/81
0 low: 1/16 x (341/81 + E(3)) = 3101/5184
1 low: 1/4 x (1 + E(3)) = 229/144
2 low: 3/8 x (2 + E(2)) = 115/48
3 low: 1/4 x (3 + E(1)) = 3/2
4 low: 1/16 x 4 = 1/4
E(4) = 32,837/5184

5 dice
Expected value of five high dice: 1/243 x 6 + 31/243 x 5 + 211/243 x 4 = 335/81
0 low: 1/32 x (335/81 + E(4)) = 54,277/165,888
1 low: 5/32 x (1 + E(4)) = 190,105/165,888
2 low: 5/16 x (2 + E(3)) = 1325/576
3 low: 5/16 x (3 + E(2)) = 665/288
4 low: 5/32 x (4 + E(1)) = 35/32
5 low: 1/32 x 5 = 5/32
E(5) = 608,191 / 82,944 = about 7.33255

Also incorrect.

[I'm kicking myself for not making this next week's Toughie Tuesday.]

Have you tried 22 tonight? I said 22.

unJon

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December 10th, 2020 at 6:56:44 PM permalink

Quote: ThatDonGuy
Quote: Gialmere

"Threes Away (Low)" is a game played with five standard dice. The object is to score the fewest points.

You begin your turn by rolling all five dice. You MUST set aside at least one of them for scoring but you may set aside as many as you wish. 3s score zero points. All other dice score their face values.

You then reroll any remaining dice and repeat the process until all your dice are scored. Low score wins.

Assuming optimal play, what is the expected value of your score?

The expected value of a single die roll is 3, so, presumably, the strategy is to keep all dice 1,2,3 and reroll all dice 4,5,6, unless all of the dice in a roll are 4,5,6, in which case, keep the lowest one and roll the rest.
Let E(n) be the expected value of rolling n dice using these rules.

1 die: E(1) = 3

2 dice
Expected value of two high dice: 1/9 x 6 + 3/9 x 5 + 5/9 x 4 = 41/9
0 low: 1/4 x (41/9 + E(1)) = 17/9
1 low: 1/2 x (1 + E(1)) = 2
2 low: 1/4 x 2 = 1/2
E(2) = 79/18

3 dice
Expected value of three high dice: 1/27 x 6 + 7/27 x 5 + 19/27 x 4 = 117/27
0 low: 1/8 x (117/27 + E(2)) = 157/144
1 low: 3/8 x (1 + E(2)) = 97/48
2 low: 3/8 x (2 + E(1)) = 15/8
3 low: 1/8 x 3 = 3/8
E(3) = 193/36

4 dice
Expected value of four high dice: 1/81 x 6 + 15/81 x 5 + 65/81 x 4 = 341/81
0 low: 1/16 x (341/81 + E(3)) = 3101/5184
1 low: 1/4 x (1 + E(3)) = 229/144
2 low: 3/8 x (2 + E(2)) = 115/48
3 low: 1/4 x (3 + E(1)) = 3/2
4 low: 1/16 x 4 = 1/4
E(4) = 32,837/5184

5 dice
Expected value of five high dice: 1/243 x 6 + 31/243 x 5 + 211/243 x 4 = 335/81
0 low: 1/32 x (335/81 + E(4)) = 54,277/165,888
1 low: 5/32 x (1 + E(4)) = 190,105/165,888
2 low: 5/16 x (2 + E(3)) = 1325/576
3 low: 5/16 x (3 + E(2)) = 665/288
4 low: 5/32 x (4 + E(1)) = 35/32
5 low: 1/32 x 5 = 5/32
E(5) = 608,191 / 82,944 = about 7.33255

A 3 is zero points. Nevermind. I misread your post.

Last edited by: unJon on Dec 10, 2020

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

DogHand

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December 10th, 2020 at 8:08:55 PM permalink

Quote: ThatDonGuy
<snip>2 R^3 - 87 R - 77 = 0

Okay, at this point I used Newton-Raphson with Excel to find the real positive root of 7.
Since 7 is a root...<snip>

Both of the other roots are real, but neither is positive; 7 is the only positive solution.

ThatDonGuy,

Here's an easier way to get to your "Both of the other roots are real, but neither is positive; 7 is the only positive solution." conclusion: use Descartes' rule of signs!

In the cubic polynomial, the cubed term has a positive coeffiecient, the squared term is missing, and both the linear and constant terms have negative coefficients. This means that the coefficient sign changes only once, so the cubic has exactly one positive real root. That means that once you found +7, you are guaranteed that it's the only positive real root.

Wikipedia's article: https://en.m.wikipedia.org/wiki/Descartes%27_rule_of_signs

Hope this helps!

Dog Hand

chevy

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December 10th, 2020 at 9:09:59 PM permalink

Quote: Gialmere
Quote: ThatDonGuy

The expected value of a single die roll is 3, so, presumably, the strategy is to keep all dice 1,2,3 and reroll all dice 4,5,6, unless all of the dice in a roll are 4,5,6, in which case, keep the lowest one and roll the rest.
Let E(n) be the expected value of rolling n dice using these rules.

1 die: E(1) = 3

2 dice
Expected value of two high dice: 1/9 x 6 + 3/9 x 5 + 5/9 x 4 = 41/9
0 low: 1/4 x (41/9 + E(1)) = 17/9
1 low: 1/2 x (1 + E(1)) = 2
2 low: 1/4 x 2 = 1/2
E(2) = 79/18

3 dice
Expected value of three high dice: 1/27 x 6 + 7/27 x 5 + 19/27 x 4 = 117/27
0 low: 1/8 x (117/27 + E(2)) = 157/144
1 low: 3/8 x (1 + E(2)) = 97/48
2 low: 3/8 x (2 + E(1)) = 15/8
3 low: 1/8 x 3 = 3/8
E(3) = 193/36

4 dice
Expected value of four high dice: 1/81 x 6 + 15/81 x 5 + 65/81 x 4 = 341/81
0 low: 1/16 x (341/81 + E(3)) = 3101/5184
1 low: 1/4 x (1 + E(3)) = 229/144
2 low: 3/8 x (2 + E(2)) = 115/48
3 low: 1/4 x (3 + E(1)) = 3/2
4 low: 1/16 x 4 = 1/4
E(4) = 32,837/5184

5 dice
Expected value of five high dice: 1/243 x 6 + 31/243 x 5 + 211/243 x 4 = 335/81
0 low: 1/32 x (335/81 + E(4)) = 54,277/165,888
1 low: 5/32 x (1 + E(4)) = 190,105/165,888
2 low: 5/16 x (2 + E(3)) = 1325/576
3 low: 5/16 x (3 + E(2)) = 665/288
4 low: 5/32 x (4 + E(1)) = 35/32
5 low: 1/32 x 5 = 5/32
E(5) = 608,191 / 82,944 = about 7.33255

Also incorrect.

[I'm kicking myself for not making this next week's Toughie Tuesday.]

I had approached it same method as ThatDonGuy though did not crank through the numbers as he did. If that is not correct, I can only suggest the following possible modification to ThatDonGuy's method (which I am not sure I implemented correctly)

(***Building using ThatDonGuy solution as starting point***)

Possible the strategy is different on the rounds. if I initially roll a 3,2,6,6,6 ....I certainly keep the 3.... but do I keep the 2? I have 4 rolls to improve it to a 1=1 or 3=0 (exp. val. = 1/2)

Am I right in thinking re-rolling that die until I get 1 or 3 or get to 4th re-roll has the expected value of

1/3 * (1/2) + 2/3 * { 1/3*(1/2) +2/3 * [ 1/3 * (1/2) + 2/3 * (3) ] } = 67/54 which is less than 2....so should re-roll

likewise getting a 2 with 3 re-rolls left , I can improve

1/3*(1/2) +2/3 * [ 1/3 * (1/2) + 2/3 * (3) ] = 87/54 (also less than 2)

But with 2 re-rolls expected value of re-rolling a 2 is

1/3 * (1/2) + 2/3 * (3) = 13/6. > 2

So I think proper strategy is to only accept a 2 if there will be less than 4 dice remaining....or it is the lowest value rolled and must be set aside????????????????????

------------
Comment: Same principle applied to accepting a 1 with 4 re-roll....expected value of re-rolling to improve to a 3=0 is

1/6 * (0) + 5/6 * { 1/6*(0) +5/6 * [ 1/6 * (0) + 5/6 * (3) ] } = 125/72 which is greater than 1, so no reroll.
-------------------

So modification of ThatDonGuy solution
*** E(1), E(2), E(3) stay same
*** E(4) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 low would be .5, 1, 1.5, 2 respectively

*** E(5) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 5 low would be .5, 1, 1.5, 2, 2.5 respectively

SO with credit to ThatDonGuy's framework, I get

4 dice
Expected value of four high dice: 1/256 x 6 + 15/256 x 5 + 65/256 x 4 + 175/256 x 2= 711/256
0 low: 1/16 x (711/256 + E(3)) = 18751/36864
1 low: 1/4 x (1/2 + E(3)) = 211/144
2 low: 3/8 x (1 + E(2)) = 97/48
3 low: 1/4 x (3/2 + E(1)) = 9/8
4 low: 1/16 x 2 = 1/8
E(4) = 5.24476539930556

5 dice
Expected value of five high dice: 1/1024 x 6 + 31/1024 x 5 + 211/1024 x 4 + 781/1024 x 2= 2567/1024
0 low: 1/32 x (2567/1024 + E(4)) = 0.242237514919705
1 low: 5/32 x (1/2 + E(4)) = 1.211187574598524
2 low: 5/16 x (1 + E(3)) = 1145/576
3 low: 5/16 x (3/2 + E(2)) = 530/288
4 low: 5/32 x (2 + E(1)) = 25/32
5 low: 1/32 x 5/2 = 5/64

Final Answer
E(5) = 6.140925089518229

???????

chevy

Threads: 3
Posts: 151

Joined: Apr 15, 2011

December 10th, 2020 at 9:24:45 PM permalink

***I don't think my previous suggestion is complete***

I don't think I implemented when to accept a 2 correctly.

if I roll 3,3,3,2,2.....I would keep the 3's. but then I also keep the 2's since I won't have enough re-rolls to improve them.......and I think that is yet to be accounted for.

???????

Gialmere

Threads: 45
Posts: 3050

Joined: Nov 26, 2018

December 10th, 2020 at 10:54:35 PM permalink

Quote: chevy
I had approached it same method as ThatDonGuy though did not crank through the numbers as he did. If that is not correct, I can only suggest the following possible modification to ThatDonGuy's method (which I am not sure I implemented correctly)

(***Building using ThatDonGuy solution as starting point***)

Possible the strategy is different on the rounds. if I initially roll a 3,2,6,6,6 ....I certainly keep the 3.... but do I keep the 2? I have 4 rolls to improve it to a 1=1 or 3=0 (exp. val. = 1/2)

Am I right in thinking re-rolling that die until I get 1 or 3 or get to 4th re-roll has the expected value of

1/3 * (1/2) + 2/3 * { 1/3*(1/2) +2/3 * [ 1/3 * (1/2) + 2/3 * (3) ] } = 67/54 which is less than 2....so should re-roll

likewise getting a 2 with 3 re-rolls left , I can improve

1/3*(1/2) +2/3 * [ 1/3 * (1/2) + 2/3 * (3) ] = 87/54 (also less than 2)

But with 2 re-rolls expected value of re-rolling a 2 is

1/3 * (1/2) + 2/3 * (3) = 13/6. > 2

So I think proper strategy is to only accept a 2 if there will be less than 4 dice remaining....or it is the lowest value rolled and must be set aside????????????????????

------------
Comment: Same principle applied to accepting a 1 with 4 re-roll....expected value of re-rolling to improve to a 3=0 is

1/6 * (0) + 5/6 * { 1/6*(0) +5/6 * [ 1/6 * (0) + 5/6 * (3) ] } = 125/72 which is greater than 1, so no reroll.
-------------------

So modification of ThatDonGuy solution
*** E(1), E(2), E(3) stay same
*** E(4) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 low would be .5, 1, 1.5, 2 respectively

*** E(5) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 5 low would be .5, 1, 1.5, 2, 2.5 respectively

SO with credit to ThatDonGuy's framework, I get

4 dice
Expected value of four high dice: 1/256 x 6 + 15/256 x 5 + 65/256 x 4 + 175/256 x 2= 711/256
0 low: 1/16 x (711/256 + E(3)) = 18751/36864
1 low: 1/4 x (1/2 + E(3)) = 211/144
2 low: 3/8 x (1 + E(2)) = 97/48
3 low: 1/4 x (3/2 + E(1)) = 9/8
4 low: 1/16 x 2 = 1/8
E(4) = 5.24476539930556

5 dice
Expected value of five high dice: 1/1024 x 6 + 31/1024 x 5 + 211/1024 x 4 + 781/1024 x 2= 2567/1024
0 low: 1/32 x (2567/1024 + E(4)) = 0.242237514919705
1 low: 5/32 x (1/2 + E(4)) = 1.211187574598524
2 low: 5/16 x (1 + E(3)) = 1145/576
3 low: 5/16 x (3/2 + E(2)) = 530/288
4 low: 5/32 x (2 + E(1)) = 25/32
5 low: 1/32 x 5/2 = 5/64

Final Answer
E(5) = 6.140925089518229

???????

Closer, but also incorrect.

Have you tried 22 tonight? I said 22.

chevy

Threads: 3
Posts: 151

Joined: Apr 15, 2011

December 10th, 2020 at 11:31:13 PM permalink

Quote: chevy
***I don't think my previous suggestion is complete***

I don't think I implemented when to accept a 2 correctly.

if I roll 3,3,3,2,2.....I would keep the 3's. but then I also keep the 2's since I won't have enough re-rolls to improve them.......and I think that is yet to be accounted for.

???????

So trying to modify ThatDonGuy starting point one last time and then I give up....(His solution was clean and systematic and I am spiraling it out of control for an "Easy Thursday" question...)

*****FINAL ANSWER = E(5) = 6.097234090169271******

-----------------------------------------------------
So RE-modification of ThatDonGuy solution

*** E(1), E(2), E(3) stay same
*** E(4) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 low would be .5, 1, 1.5, 2 respectively
And if #remaining dice would be 3,2,1 accept a roll of 2

*** E(5) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 5 low would be .5, 1, 1.5, 2, 2.5 respectively
And if #remaining dice would be 3,2,1 accept a roll of 2

SO with credit to ThatDonGuy's framework, I get

4 dice
Expected value of four high dice: 1/256 x 6 + 15/256 x 5 + 65/256 x 4 + 175/256 x 2= 711/256
0 low: 1/16 x (711/256 + E(3)) = 18751/36864
1 low: 1/4 x (1/2 + [1/64 * (2+2+2) + 9/64 * (2+2+E(1)) + 27/64 * (2+E(2)) + 27/64*E(3))] ) = 1.6337890625
2 low: 3/8 x (1 + [1/16 * (2+2) + 6/16*(2+E(1)) + 9/16*(E(2)))] = 2.09765625
3 low: 1/4 x (3/2 + (1/4 * 2 + 3/4 *E(1)) = 17/16
4 low: 1/16 x (2 = 1/8
E(4) = 5.427598741319444

5 dice
Expected value of five high dice: 1/1024 x 6 + 31/1024 x 5 + 211/1024 x 4 + 781/1024 x 2= 2567/1024
0 low: 1/32 x (2567/1024 + E(4)) = 0.247951083713108
1 low: 5/32 x (1/2 + E(4)) = 0.926187303331163
2 low: 5/16 x (1 + [1/64 * (2+2+2) + 9/64 * (2+2+E(1)) + 27/64 * (2+E(2)) + 27/64*E(3))] = 2.198486328125
3 low: 5/16 x (3/2 + [1/16 * (2+2) + 6/16*(2+E(1)) + 9/16*(E(2)))] ) = 1.904296875
4 low: 5/32 x (2 + (1/4 * 2 + 3/4 *E(1)) ) = 95/128
5 low: 1/32 x 5/2 = 5/64

Final Answer
E(5) = 6.097234090169271

???????

HAS TO BE A CLEANER WAY (if this is correct)

Gialmere

Threads: 45
Posts: 3050

Joined: Nov 26, 2018

December 11th, 2020 at 8:50:10 AM permalink

Quote: chevy
Quote: chevy
***I don't think my previous suggestion is complete***

I don't think I implemented when to accept a 2 correctly.

if I roll 3,3,3,2,2.....I would keep the 3's. but then I also keep the 2's since I won't have enough re-rolls to improve them.......and I think that is yet to be accounted for.

???????

So trying to modify ThatDonGuy starting point one last time and then I give up....(His solution was clean and systematic and I am spiraling it out of control for an "Easy Thursday" question...)

*****FINAL ANSWER = E(5) = 6.097234090169271******

-----------------------------------------------------
So RE-modification of ThatDonGuy solution

*** E(1), E(2), E(3) stay same
*** E(4) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 low would be .5, 1, 1.5, 2 respectively
And if #remaining dice would be 3,2,1 accept a roll of 2

*** E(5) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 5 low would be .5, 1, 1.5, 2, 2.5 respectively
And if #remaining dice would be 3,2,1 accept a roll of 2

SO with credit to ThatDonGuy's framework, I get

4 dice
Expected value of four high dice: 1/256 x 6 + 15/256 x 5 + 65/256 x 4 + 175/256 x 2= 711/256
0 low: 1/16 x (711/256 + E(3)) = 18751/36864
1 low: 1/4 x (1/2 + [1/64 * (2+2+2) + 9/64 * (2+2+E(1)) + 27/64 * (2+E(2)) + 27/64*E(3))] ) = 1.6337890625
2 low: 3/8 x (1 + [1/16 * (2+2) + 6/16*(2+E(1)) + 9/16*(E(2)))] = 2.09765625
3 low: 1/4 x (3/2 + (1/4 * 2 + 3/4 *E(1)) = 17/16
4 low: 1/16 x (2 = 1/8
E(4) = 5.427598741319444

5 dice
Expected value of five high dice: 1/1024 x 6 + 31/1024 x 5 + 211/1024 x 4 + 781/1024 x 2= 2567/1024
0 low: 1/32 x (2567/1024 + E(4)) = 0.247951083713108
1 low: 5/32 x (1/2 + E(4)) = 0.926187303331163
2 low: 5/16 x (1 + [1/64 * (2+2+2) + 9/64 * (2+2+E(1)) + 27/64 * (2+E(2)) + 27/64*E(3))] = 2.198486328125
3 low: 5/16 x (3/2 + [1/16 * (2+2) + 6/16*(2+E(1)) + 9/16*(E(2)))] ) = 1.904296875
4 low: 5/32 x (2 + (1/4 * 2 + 3/4 *E(1)) ) = 95/128
5 low: 1/32 x 5/2 = 5/64

Final Answer
E(5) = 6.097234090169271

???????

HAS TO BE A CLEANER WAY (if this is correct)

Sorry but, still incorrect.

Have you tried 22 tonight? I said 22.

unJon

unJon

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Posts: 4809

Joined: Jul 1, 2018

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December 11th, 2020 at 8:54:34 AM permalink

Quote: Gialmere
Quote: chevy
Quote: chevy
***I don't think my previous suggestion is complete***

I don't think I implemented when to accept a 2 correctly.

if I roll 3,3,3,2,2.....I would keep the 3's. but then I also keep the 2's since I won't have enough re-rolls to improve them.......and I think that is yet to be accounted for.

???????

So trying to modify ThatDonGuy starting point one last time and then I give up....(His solution was clean and systematic and I am spiraling it out of control for an "Easy Thursday" question...)

*****FINAL ANSWER = E(5) = 6.097234090169271******

-----------------------------------------------------
So RE-modification of ThatDonGuy solution

*** E(1), E(2), E(3) stay same
*** E(4) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 low would be .5, 1, 1.5, 2 respectively
And if #remaining dice would be 3,2,1 accept a roll of 2

*** E(5) has to be modified to be expected value of high to include 6,5,4,2
And all the EV terms for 1,2,3,4 5 low would be .5, 1, 1.5, 2, 2.5 respectively
And if #remaining dice would be 3,2,1 accept a roll of 2

SO with credit to ThatDonGuy's framework, I get

4 dice
Expected value of four high dice: 1/256 x 6 + 15/256 x 5 + 65/256 x 4 + 175/256 x 2= 711/256
0 low: 1/16 x (711/256 + E(3)) = 18751/36864
1 low: 1/4 x (1/2 + [1/64 * (2+2+2) + 9/64 * (2+2+E(1)) + 27/64 * (2+E(2)) + 27/64*E(3))] ) = 1.6337890625
2 low: 3/8 x (1 + [1/16 * (2+2) + 6/16*(2+E(1)) + 9/16*(E(2)))] = 2.09765625
3 low: 1/4 x (3/2 + (1/4 * 2 + 3/4 *E(1)) = 17/16
4 low: 1/16 x (2 = 1/8
E(4) = 5.427598741319444

5 dice
Expected value of five high dice: 1/1024 x 6 + 31/1024 x 5 + 211/1024 x 4 + 781/1024 x 2= 2567/1024
0 low: 1/32 x (2567/1024 + E(4)) = 0.247951083713108
1 low: 5/32 x (1/2 + E(4)) = 0.926187303331163
2 low: 5/16 x (1 + [1/64 * (2+2+2) + 9/64 * (2+2+E(1)) + 27/64 * (2+E(2)) + 27/64*E(3))] = 2.198486328125
3 low: 5/16 x (3/2 + [1/16 * (2+2) + 6/16*(2+E(1)) + 9/16*(E(2)))] ) = 1.904296875
4 low: 5/32 x (2 + (1/4 * 2 + 3/4 *E(1)) ) = 95/128
5 low: 1/32 x 5/2 = 5/64

Final Answer
E(5) = 6.097234090169271

???????

HAS TO BE A CLEANER WAY (if this is correct)

Sorry but, still incorrect.

I have an idea on modeling this but won�t have time until the weekend. Request that you do not post answer yet.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Ace2

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December 11th, 2020 at 9:07:05 AM permalink

Quote: Ace2
Quote: Gialmere

"Threes Away (Low)" is a game played with five standard dice. The object is to score the fewest points.

You begin your turn by rolling all five dice. You MUST set aside at least one of them for scoring but you may set aside as many as you wish. 3s score zero points. All other dice score their face values.

You then reroll any remaining dice and repeat the process until all your dice are scored. Low score wins.

Assuming optimal play, what is the expected value of your score?

If I'm not being clear with the rules, this video explains it in less than 90 seconds...

7.462890625

I had a small error in my first calculation. If this one is confirmed as correct, I�ll show my method

6.253979...

It�s all about making that GTA

charliepatrick

charliepatrick

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December 11th, 2020 at 1:16:26 PM permalink

First to get a lower bound assume you don't have to keep any dice and can re-roll them all.
On the last round the average of (0 1 2 4 5 6) is 3.
So after the second roll it's worth keeping 0 1 or 2 but not 4 5 or 6.
Thus the average per die prior to the second roll is (0 1 2 3 3 3) = 2.
So after the first roll it's worth keeping 0 or 1, but not 4 5 or 6. With a 2 it's best to re-roll as that gives you more dice, so less likely to roll all high ones and have to keep one you don't want to.
Thus the average per die prior to the first roll is (0 1 2 2 2 2) = 1.5.
So the average total is 5*1.5 = 7.5.

Now consider having to keep at least one dice.
On the second roll the effect of this depends on how many dice you have already held on round one.
Thus you need to work out the expected total for 2 to 4 dice (with 1 you have to keep it, so the average is 3).
Two Dice - expected total (for both) = 4.38888...
Three Dice - expected total (for all) = 6.166666...
Four Dice - expected total (for all) = 8.075617...

Work backwards to first round e.g..
5 dice held - average total = 2.5 (as they're either 0 or 1)
4 dice held = 2 + 3
3 dice held = 1.5 + 4.388888 = 5.888888
etc.

Giving an average of 7.761633.

gordonm888
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gordonm888

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December 11th, 2020 at 2:03:59 PM permalink

From Don's calculation:

3 dice
Expected value of three high dice: 1/27 x 6 + 7/27 x 5 + 19/27 x 4 = 117/27
0 low: 1/8 x (117/27 + E(2)) = 157/144
1 low: 3/8 x (1 + E(2)) = 97/48
2 low: 3/8 x (2 + E(1)) = 15/8
3 low: 1/8 x 3 = 3/8
E(3) = 193/36

Let's look at the "3 low" case

333 Tot = 0, Perms = 1, Tot x Perms = 0
322 Tot =4, Perms = 3, Tot x perms = 12
321 Tot =3, Perms =6, Tot x perms = 18
311 Tot = 2, Perms = 3, Tot x perms = 6
222 Tot =6, Perms =1, Tot x perms = 6
221 Tot = 5, Perms = 3, Tot x perms = 15
211 Tot = 4, Perms =3, Tot x perms = 12
111 Tot =3, Perms = 1, Tot x perms = 3

So Σ Perms = 21
Σ (tot x perms) = 72

So average total points = 72/21 =3.428571 not 3 as is shown in Don's calculation.

so 3 low = 1/8*72/21 = 9/21 = 3/7 NOT 3/8

I suspect this same methodology error may be present in E(4) and E(5).

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ThatDonGuy

Threads: 123
Posts: 6747

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December 11th, 2020 at 2:14:16 PM permalink

Quote: gordonm888
From Don's calculation:

3 dice
Expected value of three high dice: 1/27 x 6 + 7/27 x 5 + 19/27 x 4 = 117/27
0 low: 1/8 x (117/27 + E(2)) = 157/144
1 low: 3/8 x (1 + E(2)) = 97/48
2 low: 3/8 x (2 + E(1)) = 15/8
3 low: 1/8 x 3 = 3/8
E(3) = 193/36

Let's look at the "3 low" case

333 Tot = 0, Perms = 1, Tot x Perms = 0
322 Tot =4, Perms = 3, Tot x perms = 12
321 Tot =3, Perms =6, Tot x perms = 18
311 Tot = 2, Perms = 3, Tot x perms = 6
222 Tot =6, Perms =1, Tot x perms = 6
221 Tot = 5, Perms = 3, Tot x perms = 15
211 Tot = 4, Perms =3, Tot x perms = 12
111 Tot =3, Perms = 1, Tot x perms = 3

So Σ Perms = 21
Σ (tot x perms) = 72

So average total points = 72/21 =3.428571 not 3 as is shown in Don's calculation.

so 3 low = 1/8*72/21 = 9/21 = 3/7 NOT 3/8

I suspect this same methodology error may be present in E(4) and E(5).

You left out:
332 Tot = 2, Perms = 3, Tot x Perms = 6
331 Tot = 1, Perms = 3, Tot x Perms = 3
So Σ Perms = 27
Σ (tot x perms) = 81
Average total points = 81 / 27 = 3
Which also makes sense, as it should be the sum of the expected values of each die individually; each die with a "low" number can have a value of 1, 2, or 0, so the expected value for each die is 1, and the expected value for three dice is 3.

gordonm888
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gordonm888

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December 11th, 2020 at 5:48:47 PM permalink

Quote: ThatDonGuy
Quote: gordonm888
From Don's calculation:

3 dice
Expected value of three high dice: 1/27 x 6 + 7/27 x 5 + 19/27 x 4 = 117/27
0 low: 1/8 x (117/27 + E(2)) = 157/144
1 low: 3/8 x (1 + E(2)) = 97/48
2 low: 3/8 x (2 + E(1)) = 15/8
3 low: 1/8 x 3 = 3/8
E(3) = 193/36

Let's look at the "3 low" case

333 Tot = 0, Perms = 1, Tot x Perms = 0
322 Tot =4, Perms = 3, Tot x perms = 12
321 Tot =3, Perms =6, Tot x perms = 18
311 Tot = 2, Perms = 3, Tot x perms = 6
222 Tot =6, Perms =1, Tot x perms = 6
221 Tot = 5, Perms = 3, Tot x perms = 15
211 Tot = 4, Perms =3, Tot x perms = 12
111 Tot =3, Perms = 1, Tot x perms = 3

So Σ Perms = 21
Σ (tot x perms) = 72

So average total points = 72/21 =3.428571 not 3 as is shown in Don's calculation.

so 3 low = 1/8*72/21 = 9/21 = 3/7 NOT 3/8

I suspect this same methodology error may be present in E(4) and E(5).

You left out:
332 Tot = 2, Perms = 3, Tot x Perms = 6
331 Tot = 1, Perms = 3, Tot x Perms = 3
So Σ Perms = 27
Σ (tot x perms) = 81
Average total points = 81 / 27 = 3
Which also makes sense, as it should be the sum of the expected values of each die individually; each die with a "low" number can have a value of 1, 2, or 0, so the expected value for each die is 1, and the expected value for three dice is 3.

You are correct. I came late to this discussion and have been trying to understand what has gone wrong. I did not mean to muddle it up further. Is Charlie's solution the correct solution?

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

gordonm888
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gordonm888

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December 11th, 2020 at 5:48:50 PM permalink

Duplicate

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

charliepatrick

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December 12th, 2020 at 3:12:01 AM permalink

Quote: gordonm888
You are correct. I came late to this discussion and have been trying to understand what has gone wrong. I did not mean to muddle it up further. Is Charlie's solution the correct solution?

Obviously if you have a nice low one this isn't an issue, but the problem is when you have to keep a high one. One approach might have been to look at the average of each die, but this isn't correct as you need to factor in having to keep a high die. For instance consider what happens if your first roll is 33336. It's also worth working out what would happen if you didn't have to keep any dice, as this sets a lower bound for the answer.

chevy

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December 12th, 2020 at 10:23:34 AM permalink

Quote: charliepatrick
Obviously if you have a nice low one this isn't an issue, but the problem is when you have to keep a high one. One approach might have been to look at the average of each die, but this isn't correct as you need to factor in having to keep a high die. For instance consider what happens if your first roll is 33336. It's also worth working out what would happen if you didn't have to keep any dice, as this sets a lower bound for the answer.

charliepatrick : Question about your choices (in spoiler)

I agree the complication is in having to hold the dice. I also agreed with your original assessment on holding the 2 break point. But my difficulty was implementing that. In your solution you have
----------
Work backwards to first round e.g.
5 dice held - average total = 2.5 (as they're either 0 or 1)
4 dice held = 2 + 3
----------
5: I agree
4: It seems by using "3" you are assuming you re-roll the last die. But what if first roll is 1,1,1,1,2. In that case don't you keep the 2? as it is less than the expected value of re-rolling and getting EV=3?

So I was trying to separate the 2 (1/4 of the time for last die) from the 4,5,6. (3/4 of time for last die)

Hold 4 dice (3=0 or 1=1) :
(2 + (1/4 * (keep 2) + 3/4 *E(re-roll 1 die)) ) =
2 + (2/4 + 9/4)
19/4???

I admit I am still trying to sort through your other numbers....maybe this effect is accounted for elsewhere, (but it seems not in the 5 dice held case as you are assuming all 0 or 1 to get EV 2.5).

Anyway, that line stood out as different than my understanding?

charliepatrick

charliepatrick

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December 12th, 2020 at 1:42:22 PM permalink

Quote: chevy
charliepatrick : Question about your choices (in spoiler)...

I agree...

I haven't rerun the numbers but can see if all the other numbers are 2s after the first roll, then you're better off standing. The chances of this are fairly small, so it makes the answer slightly smaller than my first answer: about 7.752715.

Gialmere

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December 12th, 2020 at 5:36:10 PM permalink

Quote: unJon

I have an idea on modeling this but won�t have time until the weekend. Request that you do not post answer yet.

We're just waiting for unJon to make his post.

Have you tried 22 tonight? I said 22.

unJon

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December 13th, 2020 at 9:50:00 AM permalink

Quote: Gialmere
We're just waiting for unJon to make his post.

Sorry. Weekend got away from me. Don�t hold anything up for me.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Gialmere

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December 13th, 2020 at 10:07:02 AM permalink

Quote: Ace2

I had a small error in my first calculation. If this one is confirmed as correct, I�ll show my method

6.253979...

Correct!

Very good.

If desired, I'll post the official solve after Ace2 posts his method.

Have you tried 22 tonight? I said 22.

Ace2

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December 13th, 2020 at 3:11:11 PM permalink

Quote: Gialmere
Quote: Ace2

I had a small error in my first calculation. If this one is confirmed as correct, I�ll show my method

6.253979...

Correct!

Very good.

If desired, I'll post the official solve after Ace2 posts his method.

Though I attempted to find a slick formulaic solution, it's quite easy to get the answer via brute force in Excel. There are only 6^2 + 6^3 + 6^4 + 6^5 = (6^6 -1)/(6-1) - 6 -1 = 9,324 permutations to review

Working backwards and using correct strategy the expected values are:
Roll 1 die : 3.0000000 = a
Roll 2 dice: 4.38888� = b
Roll 3 dice: 5.23380� = c
Roll 4 dice: 5.83386... = d

Using the same method we used to find the aforementioned values, we find the expected value for rolling 5 dice as follows:
1) List all 6^5 = 7,776 permutations. Each permutation MUST be in ascending order. This is easy to do in Excel (let me know if you need the method)
2) Make 5 calculations for each permutation:
Sum of all dice 1-5
Sum of dice 1-4 plus a
Sum of dice 1-3 plus b
Sum of dice 1-2 plus c
Die 1 plus d
3) Take the minimum of the five calculations for each permutation
4) Take the average of the 7,776 minimums to get the final answer of 6.25398�

Incidentally, if efficiency was a key consideration, you could use the same method on just a few hundred combinations instead of 9,324 permutations. For instance, rolling 5 dice there are 7,776 permutations yet only 252 combinations. However, with a computer I think it's easier to generate all permutations and run the calculation on all of them instead of figuring/grouping the combinations. If you only had a pen, paper and slide rule, then it would be much easier to use combinations

Last edited by: Ace2 on Dec 13, 2020

It�s all about making that GTA

Gialmere

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December 14th, 2020 at 6:27:37 AM permalink

Yes. The interesting thing about this one is that you can't make a rule saying to reroll a die if it is x or higher.

Instead, if you roll a 3-3-3-2-2, you'd keep all the dice since the two 2s add up to less than 4.3.

If you roll a 3-3-2-2-2, however, you'd reroll the three 2s since they add up to more than 5.2.

Have you tried 22 tonight? I said 22.

Gialmere