Dice math question
May 24th, 2012 at 12:32:29 AM
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Here's the scenario. You have 6 coins. Each coin allows you to roll two 6-sided dice. If the total is 7 or higher, you may roll again and subtract 2 from the result, this number is added to your score. You may also spend one coin to add one additional die to any roll before the roll is made. You may add a maximum of one additional die per roll, at the cost of one coin each.
Your goal is to score 16 or higher. You can play the game only once. How do you proceed? What are the odds of success?
This is a simplified example from a tabletop wargame. Calculating my odds of success/failure is beyond my math ability. I've found some threads of hope in the binomial distribution by attempting to calculate the odds of at least N initial rolls resulting in 7 or higher, but I don't know how to apply that result to the secondary roll.
Thanks for any links or advice!
Your goal is to score 16 or higher. You can play the game only once. How do you proceed? What are the odds of success?
This is a simplified example from a tabletop wargame. Calculating my odds of success/failure is beyond my math ability. I've found some threads of hope in the binomial distribution by attempting to calculate the odds of at least N initial rolls resulting in 7 or higher, but I don't know how to apply that result to the secondary roll.
Thanks for any links or advice!
I chose to spend all 6 coins separately, not adding dice to any of my rolls. This gives me 6 chances to roll 7 or higher, each with an average score of 5. Taking into account the chance of not rolling 7 or higher, the average score per roll is 0.58*5=2.9. Giving me an average score of 17.4 over 6 throws. Is this correct? Is this the best way to play if I can only play the game once?
May 25th, 2012 at 3:14:14 PM
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I'm a bit surprised that nobody chimed in, since we have a handful of members who jump on dice questions.
Anyhow, if I understand correctly, you have two choices:
1. Roll 2 dice 6 times.
2. Roll 3 dice 3 times.
I assume that the rules are the same for 2 or 3 dice, in that you must roll a 7 or higher to continue and if you do, you re-roll all of the dice and subtract 2 from the total.
If you choose option 1, you have a 58.7582928% chance of success with an average total of 17.5.
If you choose option 2, you have a 87.4471978% chance of success with an average total of 23.139.
The reason that option 2 is so much better is that you have a much higher likelihood of rolling a 7 or higher: 196/216 vs 21/36 chance with 2 dice.
Anyhow, if I understand correctly, you have two choices:
1. Roll 2 dice 6 times.
2. Roll 3 dice 3 times.
I assume that the rules are the same for 2 or 3 dice, in that you must roll a 7 or higher to continue and if you do, you re-roll all of the dice and subtract 2 from the total.
If you choose option 1, you have a 58.7582928% chance of success with an average total of 17.5.
If you choose option 2, you have a 87.4471978% chance of success with an average total of 23.139.
The reason that option 2 is so much better is that you have a much higher likelihood of rolling a 7 or higher: 196/216 vs 21/36 chance with 2 dice.
I heart Crystal Math.
May 25th, 2012 at 3:23:03 PM
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Quote: CrystalMathThe reason that option 2 is so much better is that you have a much higher likelihood of rolling a 7 or higher: 196/216 vs 21/36 chance with 2 dice.
That's true, and so obviously so that I wonder if the OP didn't get the rules wrong. When the additional die is added to the score, is it done before or after the 7 is rolled? In other words, is it done up front, or only at the end, after the one or more 2-die rolls are carried out?
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
