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edited May 28, 2020 at 19:50

30.5k
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62
218

Here is programming solution for this problem.

+------------+-------------------------+
| # of Rolls | Expected Turn to Finish |
+------------+-------------------------+
|          1 |                   1920.53 |
|          2 |                   1112.55 |
|          3 |                    910.14 |
|          4 |                    89.14 |
|          5 |                    78.79 |
|          6 |                    78.66 |
|          7 |                    78.92 |
|          8 |                    89.27 |
|          9 |                    89.52 |
+------------+-------------------------+

The table above shows if you roll

6 times without thinking the outcome of the rolls at the end, your chance to win is the highest with $7.66$$8.66$.

But if we consider stopping rolling after some sum values we got, let's see if something changes or not: https://pastebin.com/mSPgCi9m

I put # as the number of rolls to stop rolling, stop as stopping rolling after some values for your roll you got, and expected values for those.

As a result,

regardless of # of rolls, just roll until you get at least total roll value of $17$ then stop rolling, otherwise continue rolling seems the optimal way to play this game, which makes the game to win as expected around $7.32$$8.362$ turn.

with the little tweak as @Jaap Scherphuis suggested, I tried a couple of tricks for the end value where you get close to 50 with some threshold value, and found that having

at least total roll value of $16$ then stop rolling, with little tweak where you continue rolling if you are $47$, gets us $8.345$.

still little improvement though :)

Here is programming solution for this problem.

+------------+-------------------------+
| # of Rolls | Expected Turn to Finish |
+------------+-------------------------+
|          1 |                   19.53 |
|          2 |                   11.55 |
|          3 |                    9.14 |
|          4 |                    8.14 |
|          5 |                    7.79 |
|          6 |                    7.66 |
|          7 |                    7.92 |
|          8 |                    8.27 |
|          9 |                    8.52 |
+------------+-------------------------+

The table above shows if you roll

6 times without thinking the outcome of the rolls at the end, your chance to win is the highest with $7.66$.

But if we consider stopping rolling after some sum values we got, let's see if something changes or not: https://pastebin.com/mSPgCi9m

I put # as the number of rolls to stop rolling, stop as stopping rolling after some values for your roll you got, and expected values for those.

As a result,

regardless of # of rolls, just roll until you get at least total roll value of $17$ then stop rolling, otherwise continue rolling seems the optimal way to play this game, which makes the game to win as expected around $7.32$ turn.

Here is programming solution for this problem.

+------------+-------------------------+
| # of Rolls | Expected Turn to Finish |
+------------+-------------------------+
|          1 |                   20.53 |
|          2 |                   12.55 |
|          3 |                   10.14 |
|          4 |                    9.14 |
|          5 |                    8.79 |
|          6 |                    8.66 |
|          7 |                    8.92 |
|          8 |                    9.27 |
|          9 |                    9.52 |
+------------+-------------------------+

The table above shows if you roll

6 times without thinking the outcome of the rolls at the end, your chance to win is the highest with $8.66$.

But if we consider stopping rolling after some sum values we got, let's see if something changes or not: https://pastebin.com/mSPgCi9m

I put # as the number of rolls to stop rolling, stop as stopping rolling after some values for your roll you got, and expected values for those.

As a result,

regardless of # of rolls, just roll until you get at least total roll value of $17$ then stop rolling, otherwise continue rolling seems the optimal way to play this game, which makes the game to win as expected around $8.362$ turn.

with the little tweak as @Jaap Scherphuis suggested, I tried a couple of tricks for the end value where you get close to 50 with some threshold value, and found that having

at least total roll value of $16$ then stop rolling, with little tweak where you continue rolling if you are $47$, gets us $8.345$.

still little improvement though :)

Source Link

created May 28, 2020 at 9:37

Oray

30.5k
6
62
218

Here is programming solution for this problem.

+------------+-------------------------+
| # of Rolls | Expected Turn to Finish |
+------------+-------------------------+
|          1 |                   19.53 |
|          2 |                   11.55 |
|          3 |                    9.14 |
|          4 |                    8.14 |
|          5 |                    7.79 |
|          6 |                    7.66 |
|          7 |                    7.92 |
|          8 |                    8.27 |
|          9 |                    8.52 |
+------------+-------------------------+

The table above shows if you roll

6 times without thinking the outcome of the rolls at the end, your chance to win is the highest with $7.66$.

But if we consider stopping rolling after some sum values we got, let's see if something changes or not: https://pastebin.com/mSPgCi9m

I put # as the number of rolls to stop rolling, stop as stopping rolling after some values for your roll you got, and expected values for those.

As a result,

regardless of # of rolls, just roll until you get at least total roll value of $17$ then stop rolling, otherwise continue rolling seems the optimal way to play this game, which makes the game to win as expected around $7.32$ turn.