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Subject: Dice and probability question. rss

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Beau Bocephus Blasterfire
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What are the chances that a person rolling 2D6 rolls a higher sum than a person rolling 3D6? 4D6?

I'd like to minimize the chance that something happens, but at the same time have a chance of it happening. I want two people to be involved in the rolling process as one person will be the attacker and the other the defender. Thank you for any help with this matter.
 
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Russell InGA
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2d6 beating 4d6 is almost never going to happen.

As a rough guess I'd say that 2d6 beats 3d6 about 1 in 5 or 1 in 6.

One easy way to solve this is to write a computer simulation and then do 100,000 trials.
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Andrew Walters
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http://anydice.com/

output 2d6>4d6

0 96.41
1 3.59

...and equal 2.49%.

Those are smaller numbers than I would have expected, actually.

Anydice.com won't do *everything*, but it will do a lot and will even let you play with custom dice. If you find yourself asking questions like this with any frequency spend some time on that site and learn its lingo and you can then save some time and discover interesting things.
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Andy Mills
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Probability that you roll X on 2d6 * Probability that a 4d6 roll beats it:

2: 1/36 * 1
3: 2/36 * 1
4: 3/36 * 1295/1296
5: 4/36 * 1291/1296
6: 5/36 * 1281/1296
7: 6/36 * 1261/1296
8: 5/36 * 1226/1296
9: 4/36 * 1170/1296
10: 3/36 * 1096/1296
11: 2/36 * 986/1296
12: 1/36 * 861/1296

Add those all up. I'll let you do the arithmetic.

Edit: it occurs with probability 43839/46656, which is about 93.96%.
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Franz Kafka
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manydills wrote:
Probability that you roll X on 2d6 * Probability that a 4d6 roll beats it:

2: 1/36 * 1
3: 2/36 * 1
4: 3/36 * 1295/1296
5: 4/36 * 1291/1296
6: 5/36 * 1281/1296
7: 6/36 * 1261/1296
8: 5/36 * 1226/1296
9: 4/36 * 1170/1296
10: 3/36 * 1096/1296
11: 2/36 * 986/1296
12: 1/36 * 861/1296

Add those all up. I'll let you do the arithmetic.

Edit: it occurs with probability 43839/46656, which is about 93.96%.


For your calculation on 10, I came up with 1090/1296 instead of 1096/1296.
 
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Andy Mills
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JosefK wrote:
manydills wrote:
Probability that you roll X on 2d6 * Probability that a 4d6 roll beats it:

2: 1/36 * 1
3: 2/36 * 1
4: 3/36 * 1295/1296
5: 4/36 * 1291/1296
6: 5/36 * 1281/1296
7: 6/36 * 1261/1296
8: 5/36 * 1226/1296
9: 4/36 * 1170/1296
10: 3/36 * 1096/1296
11: 2/36 * 986/1296
12: 1/36 * 861/1296

Add those all up. I'll let you do the arithmetic.

Edit: it occurs with probability 43839/46656, which is about 93.96%.


For your calculation on 10, I came up with 1090/1296 instead of 1096/1296.


Yep - good call. I agree. That changes the probability by 18/46656, so it should be 43821/46656 = 93.92%.
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Don Weed
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A game that already uses this kind of convention to a degree is GMT's Iron and Oak. Civil War naval game.
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Beau Bocephus Blasterfire
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statman8 wrote:
A game that already uses this kind of convention to a degree is GMT's Iron and Oak. Civil War naval game.


Thanks for the information. What are you opinions on such a convention if any?
 
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