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31 or die

 
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bonanova
Daedalian Member



PostPosted: Mon Mar 14, 2011 9:52 pm    Post subject: 1 Reply with quote

Here's a game for two players using a single fair die. First player rolls the die and notes the value of the top face. Play then alternates by rotating the die a quarter turn so that one of the side faces becomes the up face. Its value is then added to a running total. Player wins by making 31, or loses by going over 31.

This game is similar to the scam game of 31 that uses the 24 A-6 cards from a deck of cards. But it differs in two ways: [1] your plays here are limited [you can't play the top or bottom number next] but [2] you never run out of cards.

Regarding the first roll, one and only one number permits a forced win. What number is that?
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L'lanmal
Daedalian Member



PostPosted: Tue Mar 15, 2011 1:37 am    Post subject: 2 Reply with quote

I did not come up with any better strategy than to build the guaranteed losing states of the ordered pair (number below target, unusable pair on the dice). Which is actually not all that time consuming, but is error-prone by hand, since it can only really be verified by repeating the process. However, the pattern repeats . . .

Roll a 4, to bring you 27 from winning. Try to ensure that the opponent is a multiple of 9 away from winning whenever possible.

If opponent chooses a 3-6, choose a 6-3. Continue to match 9s as allowed.
If opponent chooses a 1 or 2, respond with a 4. The opponent cannot then immediately pick a 4 or 3 to land you a multiple of 9 away and this is an eventually losing situation for him. Left as an exercise to the reader.
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L'lanmal
Daedalian Member



PostPosted: Tue Mar 15, 2011 1:56 am    Post subject: 3 Reply with quote

Winning play chart, generated seive fashion. (Started at the top, filled-in plays that put your opponent in a losing situation, if there were none, then that was losing too.)
L denotes a losing situation.


SPOILER SPACE

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For a current total of n, the value of 31-n (mod 9) is shown on the left.
Being exactly 1 away from goal is losing for you if there is a 1 or 6 on the die. However, 1+9k for k>1 is not losing, as you can choose 5.
Code:
Die number  1/6  |  2/5  | 3/4
0            L       L      L
1            5       1      1/5       
2            2/3     3      2
3            3/4    3/4     L
4            4       4      L
5            5       L      5
6            3       6/3    6
7            4/3/2   4/3    2
8            4       4      L
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lostdummy
Daedalian Member



PostPosted: Wed Mar 16, 2011 9:48 am    Post subject: 4 Reply with quote

I'm not exactly sure what "forced win" means in this case, but based on results I got I would presume that it means "allows player who rolled first to win".

In that case, only it first player rolled 4 , he can win (if he play correctly later on of course).
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