Enumeration of increasing sequences of 2 dice sums

[Lenus]

I tackle the following game analysis:

2 players, two 6-sided dice. Bigger sum of points win.
First roller has an advantage, as he wins even if 2nd player's dice sum equals to his.
As the game is played with doubling cube (potentially increasing the odds before any roll), I tried to enumerate the list of possible sequences.

For instance, if P1 rolls 12, the game ends (1 roll)
However, the. sequences like
2,3,6,12
4,5,8,9,12 etc are plausible.

I tried brutal force calculation and came up with 938 distinct sequences all of them ending with 12, obviously.

I am just trying to double check my calculation and learn at the same time, how to calculate the number of sequences without brutal force?

Any ideas?

 

[PeroK]

What are the rules of this game? Is it first to throw a double-six that wins?

 

[Lenus]

Rules are: if you roll lower or equal number than the opponent, you lose.
Dice are rolled until one side loses, therefore one game can consist of several rolls.

Obviously, player rolling 66 wins automatically, because ties lose, and even 66 won't save the bet.

Example,

p1: 7 points,
p2: 8 points,
p1: 10 points,
p2: 10 points (or less)

P1 wins.

p1: 5 points,
p2: 7 points
p1: 12 points

P1 wins.

So, as long as your current roll is higher than the previous roll, the bet is still on.
If someone rolls 65 (11pts) then the opponent only has 1/36 chances to win the bet.

Game is played with doubling cube, which makes the strategy quite complicated, that's why I was trying to enumerate the possible number of sequences first.

Obviously the longest one would be: 2,3,4,5,6,7,8,9,10,11,12 (11 rolls)

 

[PeroK]

What's your objective? It seems to me that you have only 12 scenarios, based on the last throw. What went on before is not relevant.

P(n) = probability next player wins the game when the previous roll was n.

You know that P(11) = 1/36.

And, P(10) = 3/36 - (2/36)(1/36)

Why is it important to enumerate the sequences, when P(n) is the key to the betting?

 

[Lenus]

You see, the introduction of doubling cube (which is of arithmetical progression, btw) complicates the strategy.
The simplest examples of all are:

1. Should a first player double the stakes before first roll?
2. If he did so, and rolled 5 pts, should p2 redouble thus giving away a chance to be redoubled again, or simply try to roll more than 5 points without increasing the odds further?

By enumeration I was hoping to enlist all possible scenarios for various decisions.

And what went before sometimes could be relevant, as different cube values would have different takepoints, e.g. i could take a cube set to 3 to roll higher than 9, but pass the cube if the value of it was only 2.

 

[PeroK]

You see, the introduction of doubling cube (which is of arithmetical progression, btw) complicates the strategy.
The simplest examples of all are:

1. Should a first player double the stakes before first roll?
2. If he did so, and rolled 5 pts, should p2 redouble thus giving away a chance to be redoubled again, or simply try to roll more than 5 points without increasing the odds further?

By enumeration I was hoping to enlist all possible scenarios for various decisions.

And what went before sometimes could be relevant, as different cube values would have different takepoints, e.g. i could take a cube set to 3 to roll higher than 9, but pass the cube if the value of it was only 2.

Please explain how this doubling cube works. Do the players have a choice when to double?

 

[PeroK]

Assuming the doubling is like Backgammon, I might have a solution if you are interested.

 

[Lenus]

Many thanks for your help.
Yes I would be interested in your solution for backgammon cube (2,4,8,16,32,64), especially when to issue an initial double / redouble.

In this game the cube looks like a die, so increments are just one point every time (1,2,3,4,5,6...)
Consequently, you have a 25% takepoint for the cube at 2, but only 16,7% for the cube at 3, and 10% for the cube at 4, as opposed to 25% for every cube decision with backgammon cube.

Now there are two variations of the game in question.
1. Player can double before his roll
2. Player can double before any roll

Of course, player must own the cube to use it, so if he had redoubled earlier, he loses this chance to use the cube until opponent uses it.

 

[PeroK]

Whether it's doubling or simply increasing thr stake may not make any difference. Here's the logic:

If you throw first, you should double for 7 or higher. Your opponent has two options: to accept a loss at the original stake or play on at double the stake.

If you threw 9-12 your opponent should accept the loss.

If you threw 8, your opponent should accept the double and play on (but will still, on average, lose close to the original stake - 0.89 by my calculation).

If you threw 7, your opponent should play on (and obviously redouble if they throw anything higher than 7). The expected win is about 0.35. Likewise, your strategy is then only to play on for an 8.

There's no point in doubling at anything lower than 7. Although it's quite close for throwing a 6.The way I solved it was to start at 11 and work backwards, calculating the expected gain at each value. For example. If you throw 7 and double:

Probability your opponent throws 9-12 = 0.28, they redouble and you lose. Total expected loss = -0.56

If they throw 8, they redouble, you play on at 4 points and lose 1.78 on average. The probability of an 8 is 0.14, so the expecetd loss is -0.25.

But, you win with probability of 0.58 for an expected gain of 1.16.

So, total average gain is 0.35.

An interesting problem.

 

[Lenus]

PeroK,

This is great. Thanks a lot for your time.

 

[PeroK]

I did have another thought about whether it's better for the player going first to double before they throw and it looks like it is. If you wait until you throw and then double for a 7 or higher, then that has an expected gain of about 0.22. But, if you double before you throw (and your opponent has to accept), your gain increases to 0.25.

So, that looks like a better strategy.

Also, if you throw a 2, it's better for your opponent to double again before they throw. But, if you throw at least a 3, then it isn't.

Although, I might like to redo all the calculations just to make sure!

 

[Lenus]

Wow, that's interesting. Thanks again.

Thats what was bothering me for weeks ))

Could I have a glance at your actual calculations when you re finished. Although I am not a mathematician, I love to see the method used in action.