Some Game-based Probability - take a guess, or work it out

[QuikSand]

Yeah, back in the day we used to do this sort of thing all the time around here. Somewhere around the lions-and-lambs puzzle, I think I sort of lost my heart. For some reason, trying again.

The Setup

I played a board game Doubles Wild (http://boardgamegeek.com/boardgame/5120/doubles-wild) for the first time recently. Specifics aren't essential - it's a decent, fairly light, abstract strategy game. I'd recommend it if you like something that might offset more complex or rigorous games. Wooden version available on Amazon (http://www.amazon.com/Enginuity-Games-0872-Doubles-Wild/dp/B00005J8FA) is attractive, and the gameplay is solid.

Anyway - there's an element of the game that is thoroughly probabilistic in nature, but the actual calculation of it requires a little more effort than I can muster myself. Seems like it sets up as a worthwhile "exercise" for the assembled minds here, or at least some of them who enjoy this sort of thing.

The "Challenge Roll"

The component of the game that's interesting is when one person tries to challenge another person in an attempt to take over a position on the game board. In a challenge, there are four six-sided dice used, and it works like this:

Attacker rolls 4 dice to try to get a high total, and may discard the roll and re-roll up to two more times. Defender then rolls 4 dice, seeking to either equal or exceed the attacker's total, and receives one more try than the attacker used to successfully defend.

So, there's a certain element of strategy involved for the attacker, making this a little bit more complicated than just plotting out all the possible outcomes from rolling four dice. The attacker needs to decide whether to simply accept an initial roll of, say, 13. 13 is a bit less than the expected roll for 4 dice, so that's a disappointing first roll, but by re-rolling he guarantees an extra chance by the defender to beat the score.

So, with that background, I can see a few different levels of this.

The FOFC Challenge

1) For anyone who would want to think about it... care to take a guess what is the percent chance of an attack being successful in this setup?

2) For the incrementally more advanced, a helpful bit would be to come up with a quick "strategy" for this game. If I'm attacking, what are the minimum rolls that I ought to keep on roll #1, and on roll #2?

3) Overall, what is the actual calculated percent chance that the attacker will win, assuming he follows perfect strategy?


Have at it. I have not made any meaningful attempt at working this out, but I think I know how I'd set it up with brute force and a spreadsheet. There's likely a formula shortcut that would work fine, I just don't think I know how to express it, myself.

[PilotMan]

Alright, I'll take the first stab at this, and it's more of an observation and less hard and fast math but here goes:

Using the table the I found here:

http://alumnus.caltech.edu/~leif/FRP/probability.html

And looking at the probabilities the following things occurred to me:

1. the extra roll the defender gets is a huge advantage and allows a much higher success rate than I figured just guessing. For example. The probability that I nail a 14 on one roll is just about 45%. A defender with an extra roll is a near guarantee to defend.

2. Success in attacking appears to be much greater scoring a 15 and up. As you will get that in about 1 in every 3 tries.

3. With each and every re-roll the number to stop on needs to be higher and higher, simply because the defender will always have a mathematical advantage to achieve the same score. For example, stopping on a 15 with the second roll would be a lower chance of winning than if you had stopped on a 14 with the first roll.

4. On a roll of 16 or higher with the first attempt the defender has a probability to defend less than 50% of the time, a roll of 17 or higher on a the second roll results in the defender defending less than 50% of the time, and a roll of 18 or higher on the third roll has the defender winning less than 50% of the time.

Fast rules then dictate.

Roll 1 stop on a 16 or better
Roll 2 17
Roll 3 doesn't matter, but the chances of winning are in your favor better than 18.

[TRO]

2/3

[QuikSand]

Roll 1 stop on a 16 or better

Off the top of my head, this sounds too constricting. If I roll a 15 on my first role, that sounds like a keeper to me. Maybe even an easy keeper.

My chances of improving on the 15 in roll #2 are obviously less than half, and I am sure to give the defender at least one extra roll to match/beat me.

I would guess that the break point for roll #1 might be between 13 and 14, or even one notch lower.

[digamma]

Off the top of my head, this sounds too constricting. If I roll a 15 on my first role, that sounds like a keeper to me. Maybe even an easy keeper.

My chances of improving on the 15 in roll #2 are obviously less than half, and I am sure to give the defender at least one extra roll to match/beat me.

I would guess that the break point for roll #1 might be between 13 and 14, or even one notch lower.

Re-reading the rules, I think I agree with you. When I initially read the rules, I was thinking more like Yahtzee where you could keep individual dice in order to improve your roll. In that scenario, I think 16 is a minimum. Where you discard the entire roll, I think it sinks back a few. Not to get off track, but that distinction seems key to me.

[gstelmack]

Let's break the first roll down.

Using the table linked above, if you stop on 13, your opponent has a 55% chance to beat you on their FIRST roll, and if they miss that a 55% chance on the second roll. Your odds of LOSING if you roll a 13 and stick on it are 55% + (55% of 45%) = 80% chance of losing.

For 14 you lose on 45% + (45% of 55%) = 70% chance of losing.

For 15 you lose on 34% + (34% of 66%) = 56% chance of losing.

Is that math right? Basically take the odds you get beat on the defenders first roll, and then what are the odds he'll take a second roll and beat you on it, right?

If that's correct, let's see what happens on a reroll.

Roll a 14 on a SECOND roll, now we're looking at 45% + (45% of 55%) + (45% of 30% based on calc above) = 45% + 25% + 13% = 83% chance of losing.

Roll a 15 on a SECOND roll, and it's 34% + (34% of 66%) + (34% of 44%) = 34% + 22% + 15% = 71% chance of losing.

Sounds like we ought to be able to get to a formula on this to fill in the spreadsheet Quik mentioned above, which would get us to the result we want.

What's interesting to me is in the 14 case, hitting it on a second roll means a 13% increase in chance of losing, while the 15 case is a 15% increase. That may be my rounding, but it sure shows that second and third rolls are a huge risk for the attacker, but we still need to include the odds that you'll actually improve your position on the second roll.

So on a 14, my odds of improving my chances on my second roll (hitting 15 or better) are only 34% (at 13 they are 55%). Even at 13 I only have a 1 in 2 chance of actually improving anything.

Wait, let me make sure of one thing. As attacker, do I get the BETTER of my two (or three) rolls, or am I stuck with the last roll I made? So if I roll a 14, then re-roll and get a 12, am I on 12 or am I on 14? That changes the analysis of the second and third rolls. If I'm stuck with the last roll I made, which is how I interpret "DISCARD" in the original instructions.

Regardless I hate the odds as an attacker, as overall the odds are going to be pretty poor to win an attack.

[PilotMan]

A 14 is the most likely outcome (44.4%) in a roll of 4 six-sided dice. Given that, it seems improbable that stopping on that roll after one throw would be deemed worthy, especially when the defender get's to throw twice.

You are essentially betting that your opponent will land heads twice in a row or tails twice in a row. With only two outcomes the probability of getting the desired number is 1 out of 2, with two throws things are not in your favor at all.

You should be able to expect at least a 14 on any of the 3 throws so banking on a higher throw should be a given, IMO.

A roll of 15 is 33.57% likely to happen on any one throw. Stopping after one throw could be done. You would have roughly 33% chance of success with your opponent getting 2 defense attempts.

The question that you would have to ask would be "What is the expected success rate in any given attack?" If you could set that up, and come up with an acceptable answer, considering loss rate the numbers would come down.

If, like baseball say, 30% is a good success rate then it would make sense that stopping at 15 or maybe gambling on a 14 would be correct. If you want a better than 50% success rate then you need to shoot higher.

[QuikSand]

Just in case I wasn't 100% clear, I want to make sure we all understand the rules at work here.

Roll #1, you throw 2-3-4-5 for a 14.

Your choices are:

A) Settle for the 14, and your foe gets two chances to equal or exceed 14; or

B) Forget you ever rolled the 14 (e.g. you don't get to go back and take the 14 if your next roll is a 10) and roll again.


If you're thinking that the 14 becomes a given, and another roll can improve but not diminish your score... please be advised. Thought I covered this with "discard" but in case I didn't, take note.

[Daimyo]

This is basically a decision tree that goes uncertainty > decision > uncertainty > decision > uncertainty so to solve it start from the end and work back to the beginning.

Assuming you decide to use your second re-roll then, right before rolling, what is your chance of winning? I get 16.8% (=for each number from 4 to 24 multiply the probability of getting that number times the probability of your opponent rolling less than that number 4 times).

That means that after your first re-roll you will only roll again if you think you have less than 16.84% chance of winning standing pat. I calculate that occurs if you roll less than a 15 (probability of opponent rolling less than a 15 three times = 17.22% which is greater than 16.84%). So after making your first re-roll you have a 16.84% chance of winning on less than a 15 and on greater than a 15 you stand pat and win when the opponent rolls less than you three straight times.

Assuming you decide to use your first re-roll then, right before rolling, what is your chance of winning? I get 29.27% (=for each number from 4 to 24 multiply the probability of getting that number times the the greater of [probability of rolling less than that number 3 times or 16.84%])

That means that after your first roll you will only roll again if you think you have less than 29.27% chance of winning standing pat. I calculate that occurs if you roll less than a 15 (probability of opponent rolling less than a 15 two times = 30.95% which is greater than 29.27%). So after making your first roll you have a 29.27% chance of winning on less than a 15 and on greater than a 15 you stand pat and win when the opponent rolls less than you two straight times.

Assuming you decide to attack then, right before rolling, what is your chance of winning? I get 41.43% (=for each number from 4 to 24 multiply the probability of getting that number times the the greater of [probability of rolling less than that number 2 times or 29.27%])

So based on that I think the strategy is deceptively simple. Stand pat only on a 15 or higher on any roll.

[QuikSand]

Ok, I get your approach generally. Reading through to see if I disagree on an element.

Assuming you decide to use your first re-roll then, right before rolling, what is your chance of winning? I get 29.27% (=for each number from 4 to 24 multiply the probability of getting that number times the the greater of [probability of rolling less than that number 3 times or 16.84%])

I think you need another layer here, though. I buy your logic on the last roll, so don't you need to factor that option into the chance to win assuming you have decided to move on to roll #2? Specifically, if you roll less than a 15 on roll #2, you override your calculated odds from above, and replace them with the 16.84% flat number, since you won't be keeping roll #2.

Right?

[Daimyo]

Yeah, that's what I did. That's what I meant by "the greater of [probability of rolling less than that number 3 times or 16.84%]" If roll less than a 15 your chances of winning are 16.84% because you use your second re-roll... 15 or greater you get the calculated value of your opponent rolling and not beating you. Let me see if I can paste my excel table:

<table><tr><td>Roll</td><td>Prob</td><td>Cumulative</td><td>1</td><td>2</td><td>3</td></tr><tr><td>4</td><td>0.08%</td><td>0.00%</td><td>29.27%</td><td>16.84%</td><td>0.00%</td></tr><tr><td>5</td><td>0.31%</td><td>0.08%</td><td>29.27%</td><td>16.84%</td><td>0.00%</td></tr><tr><td>6</td><td>0.77%</td><td>0.39%</td><td>29.27%</td><td>16.84%</td><td>0.00%</td></tr><tr><td>7</td><td>1.54%</td><td>1.16%</td><td>29.27%</td><td>16.84%</td><td>0.00%</td></tr><tr><td>8</td><td>2.70%</td><td>2.70%</td><td>29.27%</td><td>16.84%</td><td>0.00%</td></tr><tr><td>9</td><td>4.32%</td><td>5.40%</td><td>29.27%</td><td>16.84%</td><td>0.00%</td></tr><tr><td>10</td><td>6.17%</td><td>9.72%</td><td>29.27%</td><td>16.84%</td><td>0.01%</td></tr><tr><td>11</td><td>8.02%</td><td>15.90%</td><td>29.27%</td><td>16.84%</td><td>0.06%</td></tr><tr><td>12</td><td>9.65%</td><td>23.92%</td><td>29.27%</td><td>16.84%</td><td>0.33%</td></tr><tr><td>13</td><td>10.80%</td><td>33.56%</td><td>29.27%</td><td>16.84%</td><td>1.27%</td></tr><tr><td>14</td><td>11.27%</td><td>44.37%</td><td>29.27%</td><td>16.84%</td><td>3.87%</td></tr><tr><td>15</td><td>10.80%</td><td>55.63%</td><td>30.95%</td><td>17.22%</td><td>9.58%</td></tr><tr><td>16</td><td>9.65%</td><td>66.44%</td><td>44.14%</td><td>29.32%</td><td>19.48%</td></tr><tr><td>17</td><td>8.02%</td><td>76.08%</td><td>57.88%</td><td>44.04%</td><td>33.50%</td></tr><tr><td>18</td><td>6.17%</td><td>84.10%</td><td>70.74%</td><td>59.49%</td><td>50.04%</td></tr><tr><td>19</td><td>4.32%</td><td>90.28%</td><td>81.50%</td><td>73.58%</td><td>66.42%</td></tr><tr><td>20</td><td>2.70%</td><td>94.60%</td><td>89.49%</td><td>84.66%</td><td>80.08%</td></tr><tr><td>21</td><td>1.54%</td><td>97.30%</td><td>94.67%</td><td>92.11%</td><td>89.63%</td></tr><tr><td>22</td><td>0.77%</td><td>98.84%</td><td>97.70%</td><td>96.57%</td><td>95.45%</td></tr><tr><td>23</td><td>0.31%</td><td>99.61%</td><td>99.23%</td><td>98.85%</td><td>98.47%</td></tr><tr><td>24</td><td>0.08%</td><td>99.92%</td><td>99.85%</td><td>99.77%</td><td>99.69%</td></tr><tr><td></td><td></td><td></td><td>41.43%</td><td>29.27%</td><td>16.84%</td></tr></table>

[Daimyo]

Roll columns is the sum of the four dice
Prob is the probability of getting that roll
Cumulative is the probability of rolling less than that roll on a single try
1 is your chance of winning given that roll on your first roll
2 is your chance of winning given that roll on your second roll
3 is your chance of winning given that roll on your third roll

The totals at the bottom are your chance of winning before the role indicated by the column.

[QuikSand]

Ok gotcha. Right approach.

[MIJB#19]

So basically the strategy is to keep rerolling until you hit a score of more than average (here 14) or have to settle with your 3rd throw? AT least that's what I'm reading from Daimyo's table. Doing a small scale simulation with 1D3, that appeared to be the best way to maximize your slim chances to win. Interesting to see it works out pretty much the same way when you increase the number of dice.

From what I'm reading, the winning odds appear to be very low. In the 1D3 test case it was only 24%, I wouldn't be surprised if it is something like that in the 4D6 environment.


Also, now I am wondering: shouldn't the potential profit or to be lost value be included in the math? Afterall, if you're batting .240, you'll want to make sure the win/loss is more than 4 to 1 to make it worthwhile (or this has to be some sort of move that gives you a victorious game ending move).

[QuikSand]

Yes, I think having a decent idea of the overall chance of winning an attack is important of you actually play the game, since in the game it's at your discretion whether to do so. I had been assuming the chance of winning was something south of 25%, but I have only played the game for one sitting.

Didn't want to bring all that level of detail into the pure math offering here, though.

In-game, I think the bottom line is that you only engage in an attack if you are up against it, and the piece in question is of critical importance to your effort to rally. It's a desperation move, since there is a material loss involved in losing an attack.

[MIJB#19]

Ah yes, understood. Your approach is the other way around: once you know what the odds are, you'll know whether it's worth your while to even bother trying. That, uhm, makes a lot more sense too.