A non-bankroll related risk of ruin question

[numeri]

First of all, this has nothing to do with the often-asked risk of ruin bankroll question.

A friend and I are talking about a game where two players take turn rolling a die. (The actual game is "Pass the Pig", but that's not important.) Each player can roll repeatedly, adding points each time. One event, however, causes the player to lose all of his points. Let's say that rolling a 6 will cause a player to lose his points. At any point, the player can stop and "stand pat" with his current total.

The goal of the game is to be the first to 100 points or have the highest total after 10 rounds.

So the question is, what is the best point to stop on a given round? A lot depends on the player and his lead/deficit in relation to his opponent. The other component is how much is he willing to risk.

With the game I have described, (roll a die repeatedly, stopping at any point, and a 6 leaves the player with no points for that round) the EV of any roll is 2.5, with a standard deviation of sqrt(2)~1.414.

I really don't know where to go from here, though I feel I'm making it more difficult than it needs to be. Any thoughts?

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With the game I have described, (roll a die repeatedly, stopping at any point, and a 6 leaves the player with no points for that round) the EV of any roll is 2.5, with a standard deviation of sqrt(2)~1.414.


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This is only true on the first roll, each sucessive roll the 6 becomes more devastating. 5/6 your total will go up 3, (mean of 1,2,3,4,5). 1/6 your total loses all of its current value. The score you should stop at is when 1/6*X is higher than 5/6 *3. So 15 points should be where you should stop rolling. However, if your opponent has more points than you, and rolls before you, you may have to break this guideline on later rounds.

[numeri]

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With the game I have described, (roll a die repeatedly, stopping at any point, and a 6 leaves the player with no points for that round) the EV of any roll is 2.5, with a standard deviation of sqrt(2)~1.414.


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This is only true on the first roll, each sucessive roll the 6 becomes more devastating. 5/6 your total will go up 3, (mean of 1,2,3,4,5). 1/6 your total loses all of its current value. The score you should stop at is when 1/6*X is higher than 5/6 *3. So 15 points should be where you should stop rolling. However, if your opponent has more points than you, and rolls before you, you may have to break this guideline on later rounds.

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Nice. That was easier than I was making it. Now the challenge is how to alter it if we're behind.

Round Ten Strategy should be the most simple.
Strategy when behind Round 10: You must always roll. If you are first to act, and stay pat, the person ahead of you will stay pat as well and still be ahead of you. If you are behind and act second, staying pat will cause you to lose.
Strategy when ahead Round 10: First to act, the number of points you are ahead by should be the determining factor of your choice, assuming your opponent uses perfect strategy when behind.
1. Ahead by 6 or more, stay, you are guranteed a win.

2. Ahead by 5, examine effects: Roll: Win 5/6 Lose 1/6 Stay: Win 5/6 Tie 1/6
So clearly staying is the best option.

3. Ahead by 4, examine effects: Roll: Win 29/36 Tie 1/36 Lose 6/36
Stay: Win 24/36 Tie 6/36 Lose 6/36
In this case it appears you should roll.

4. Ahead by 3, examine effects: Roll: Tie 2/36 Lose 7/36 Win 27/36
Stay: Win 18/36 Tie 6/36 Lose 12/36

Following this pattern, it appears we should roll when first to act and ahead by 4 or less.

I think this is the right logic to use, but I may be an idiot so correct me if you Edit: see anything wrong.
I need to finish my sentences.

Round 1: Roll Dice (both)
Round 2: Roll Dice (both)
Round 3: Roll Dice (both)
Rounds 4-9: Coming soon, if someone else wants to help me out feel free.
Round Ten: See my earlier post.

[numeri]

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1. Ahead by 6 or more, stay, you are guranteed a win.

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I think I may not have been clear enough describing the game. You can continue to roll and add on additional rolls to your total for that round and stop at any point. For example, here are some possible outcomes for a particular round:

1) 1,4,5,stop => total = 10
2) 5,2,6 => total = 0
3) 2,4,2,5,stop => total = 13

So being ahead by 6 on the final round doesn't give you a guaranteed win by any means. In fact, there is no total that gives you a guaranteed win if you are first to act in round 10.

We can certainly roll at least once in round 10, because we only lose our points accumulated in that round by rolling a 6, so we can't lose anything if we roll a 6. How much we roll depends on our lead. If we're up by 50, we have no pressure. If we're up by 2, we need some points.

Sorry I'm not offering much help here. I haven't had time to think about this much - just throwing ideas out.

[numeri]

EDIT: Nevermind...

EDIT2: Hmmm... maybe the geometric distribution could apply?

Consider the special case when we are ahead first to act in round 10. We would like to have enough of a buffer to diffuse the chance that our opponent will catch up. Since our opponent will not stop at any point without catching up, we can consider his turn as following a geometric distribution. (This is the number of trials before the first failure - exactly what would happen if our opponent continued to roll until rolling a 6.)

In general, the geometric distribution is:

P(n) = p(1-p)^n, where p is the probability of a failure. (1/6 in our example.) The mean of a geometric distribution is (1-p)/p , which would be 5 in our example.

Since we expect villain to roll 5 times on average, and gain 3 points for each roll, we would expect villain to get... 15 that final roll. Hmmm... So we should attempt to get 15 points ahead, right?

[numeri]

So, this is really a complicated question. I did some research, and came up with a couple of interesting sites. The difficulty comes from the dependence on your opponent's score.

A zip (http://cs.gettysburg.edu/~tneller/papers/pig.zip) containing a pdf of some heavy math. This is a slightly different form, but similar.

Some simulations (http://www.derepas.com/petco/) comparing different "threshold" numbers.

I think I may be in over my head! /images/graemlins/laugh.gif

[TennesseeKid]

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I think I may not have been clear enough describing the game.

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There are more than six ways to score with different variables involved.

A double razor back is not worth 2 leaning jowlers--

Plus, there is also one worth zero points that does not lose your turn.