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chevy
chevy
Joined: Apr 15, 2011
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Thanks for this post from:
Joeman
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
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
Itís all about making that GTA
Gialmere
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
Gialmere
Joined: Nov 26, 2018
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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
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
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
unJon 
Joined: Jul 1, 2018
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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
DogHand
Joined: Sep 24, 2011
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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
chevy
Joined: Apr 15, 2011
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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
chevy
Joined: Apr 15, 2011
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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.

???????

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