[Guest]

flrd had an interesting dice question in a recent topic. And Ian's answer is clearly correct as well as phillip's simulation. I believe it was Ian's observation, that seems like a much more difficult question, that I don't necessarily have an answer for. But I think it deserves its own topic.

Two players are going to play a game of dice. Player 1, rolls a series of numbers on a standard die until he either stands pat, or rolls a 1. If he rolls a 1, he scores 0. If he stands pat, his score to beat, is the total of numbers he has rolled before he stood pat. Once player 1 is done, player 2 must roll until he either accumulates more points than player 1, or he rolls a 1, and scores a 0. The total points aren't worth anything, except to keep score. The sole goal for each player is to win the game.

So, player 2 has not particular strategy to follow, he either beats player 1's score, or he loses. But, player 1's strategy is very quite interesting, and not easy to solve. We know player 1 is a decided underdog to win the game. Going second is worth a lot in this game. But, what strategy can player 1 use to minimize player 2's advantage?

I'm fairly certain, player 1's goal, before he stands, should be much smaller than 20.

[Guest]

Haha, yes, this is a great question, much more difficult than the original, but I think it is still solvable without a simulation. I was thinking along these lines when discussing the first question, but never came up with something as eloquently defined in my mind. I will take a stab at this later.

[Guest]

My guess is player 1 has to balance the probability of achieving a certain score without rolling one and the probability of player 2 beating that score before rolling one. But I don't know how to do that without monstruous calculations, so I'll leave to the masters

[Guest]

i would like to try to solve this one mathematically before i run a simulation.

i think in this case, player 1 should stop after a particular number of rolls rather than a particular score.

in the previous sim, i found 4 or 5 rolls to be about the same. mathematically,

(5/6)^3.8 aprox = .5

so you may want a combo here. stop at 4 rolls if score is higher than 20, else stop at 5.

[Guest]

after running a simulation, i found that nearly any strategy for player 1 wins, short of only rolling twice or stopping after 10. oddly though i found the optimal strategy to be to go for a score of 25. this gives a 57.7% chance of player 1 winning. any other strategy produces a lower percentage.

[Guest]

Well, the average number of rolls that count towards the score would be 5, and as 2-6 have equal probability of landing, I'm pretty sure the average score itself would be 20?

In which case simply play to 20 or more then stop.

[Guest]

i found a small bug in my code. originally i had it such that if player 1 scored 0, player 2 would just take a draw. i now have it such that player 2 rolls at least once. now the results are the opposite. player 2 wins every time. in fact i could never find a case where player 1 wins. the best strategy for player 1 turns out to be going for a score of 10, with about a 33.4% chance of winning.

[Guest]

i found a small bug in my code. originally i had it such that if player 1 scored 0, player 2 would just take a draw. i now have it such that player 2 rolls at least once. now the results are the opposite. player 2 wins every time. in fact i could never find a case where player 1 wins. the best strategy for player 1 turns out to be going for a score of 10, with about a 33.4% chance of winning.

I'm not buying a number this high. It seems to me the the best you can do is about 25%. The odds of winning should be N*(1-N), where N=your chances of getting to a certain number. I can't see how your odds of winning could be much higher than 25%. If we picked a number where your odds are about 50%, you would be about 25% to win. I haven't figured out the exact number, but somewhere around 14 or 15 feels like the 50-50 number.

[Guest]

after eliminating the possibility of a draw all together you seem to be right, about 25-28% chance of player 1 winning. still I'm curious as to why that's the case. with the optimal strategy, any player should score an average of 8 points. being able to roll second shouldn't decrease this.

[Guest]

after eliminating the possibility of a draw all together you seem to be right, about 25-28% chance of player 1 winning. still I'm curious as to why that's the case. with the optimal strategy, any player should score an average of 8 points. being able to roll second shouldn't decrease this.

Going second is huge. Every time Player 1 busts (rolls a 1), player 2 just rolls once, and either wins or ties. It's' why the house still has an edge in Black Jack even though they never get to double down or get 1.5/1 on black jack. I can't think of a better number than 50%. Here are a few different strategies.

If you go for a number that you can get to 1/3 of the time your odds of winning are (1/3)*(2/3)= 2/9 which is less than 25%. Likewise, if you pick a number where you can get there 2/3 of the time, your odds are (2/3)*(1/3)=2/9 also. If your number is 40% then you odds are (.4)*(.6)=.24. Close, but not better than 25%. The bigger number you go for, the better your chances of winning if you get there, but the better your chances of busting. If you pick an easy number to get to, like 6, you don't bust much, but player 2 will pass you easily. Damned if you do, damned if you don't. So, I think the answer will be the number that is exactly 50% to get to.

[Guest]

I created a simulation for this game:

For each number 1 to 30 which player 1 stops above, ran 50,000 game trials and determined the winner. Here is the probability that player 1 wins for each number player 1 decides to stop above:

Num P1 Stops Above: Probability of winning

1: 0.24082

2: 0.26206

3: 0.27704

4: 0.29422

5: 0.31

6: 0.3251

7: 0.33302

8: 0.34364

9: 0.35076

10: 0.35378

11: 0.35728

12: 0.3627

13: 0.35978

14: 0.3598

15: 0.36042

16: 0.3602

17: 0.35714

18: 0.35534

19: 0.35356

20: 0.35358

21: 0.35

22: 0.35014

23: 0.34654

24: 0.33622

25: 0.33104

26: 0.32622

27: 0.32682

28: 0.31944

29: 0.31624

30: 0.31062

As you can see from above, my answer is:

Player 1 should stop once he reaches a value greater than 12

Source code attached (as .txt, but its really an .m file for Matlab)

Either way, this would make a great casino game. Reminds me of blackjack in a way

dicegame.txt

Edited July 28, 2010 by seg9585