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Quote:chevyI 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.

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)

Quote:chevyQuote: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.

Quote:GialmereQuote:chevyQuote: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.

Quote:Ace2Quote: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...

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.

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).

Quote:gordonm888From 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.

Quote:ThatDonGuyQuote:gordonm888From 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?