# Die rolling and probability question

**URL:** <https://boards.straightdope.com/t/die-rolling-and-probability-question/228121>\
**Category:** Factual Questions\
**Created:** [February 8, 2004, 2:45pm UTC](https://boards.straightdope.com/t/die-rolling-and-probability-question/228121 "2004-02-08T14:45:13Z")\
**Posts on this page:** 7\
**Page:** 1

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**Author:** ![Coil](https://avatars.discourse-cdn.com/v4/letter/c/bb73d2/32.png) [@Coil](https://boards.straightdope.com/u/Coil)\
**Post date:** [February 8, 2004, 2:45pm UTC](https://boards.straightdope.com/t/die-rolling-and-probability-question/228121/1 "2004-02-08T14:45:13Z")

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I’d like to ask the math/statistics experts on the board for help on some die-rolling probabilities. Please help me understand how to do these types of calculations.

Rolling X dice and picking the X-1 highest/lowest. How do I calculate the expected result? How do I check the probability of the roll being less/more than a certain number?

For example rolling 3 dice and picking the two highest what is my chance of getting more than 8? How do I calculate the expected sum of that roll?

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**Author:** ![ccwaterback](https://avatars.discourse-cdn.com/v4/letter/c/df705f/32.png) [@ccwaterback](https://boards.straightdope.com/u/ccwaterback)\
**Post date:** [February 8, 2004, 8:25pm UTC](https://boards.straightdope.com/t/die-rolling-and-probability-question/228121/2 "2004-02-08T20:25:38Z")

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Actually calculating the probability of the sum of the dice gets quite involved with higher numbers of dice. This would be a somewhat easy computer program to write though.

X = number of dice; /\* X \> 0 _/  
int total[6_X]; /\* array is bigger than we need \*/

/\* lets do X = 3 for example \*/  
for (a=1, a\<7, a++) {  
for (b=1, b\<7, b++) {  
for (c=1, c\<7, c++) {  
sum = sum(a,b,c) - min(a,b,c);  
total[sum]++;  
}  
}  
}

print ("Number of dice = " X);

for (ax=X-1, ax\<=6\*(X-1), ax++) {  
print ("probability of roll = " ax " is " total[ax] " in " ((6\*(X-1)) - (X-1)) ));  
}

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**Author:** ![ccwaterback](https://avatars.discourse-cdn.com/v4/letter/c/df705f/32.png) [@ccwaterback](https://boards.straightdope.com/u/ccwaterback)\
**Post date:** [February 8, 2004, 8:51pm UTC](https://boards.straightdope.com/t/die-rolling-and-probability-question/228121/3 "2004-02-08T20:51:11Z")

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Update Patch #1: 🙂

The value in the last line: (6\*(X-1)) - (X-1)) should be (6\*(X-1)) - (X-2)), this is the total number of outcomes with X dice.

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**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\
**Post date:** [February 8, 2004, 8:59pm UTC](https://boards.straightdope.com/t/die-rolling-and-probability-question/228121/4 "2004-02-08T20:59:01Z")

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> [@Coil](#):
>
> Rolling X dice and picking the X-1 highest/lowest. How do I calculate the expected result?

This part is actually not that hard to compute, but the formula takes a little understanding. What you want to do is compute the expected value of the lowest (or highest) die. As an example I’ll do it for X = 3, and choosing the 2 lowest.

All together, there are 216 (= 6[sup]X[/sup]) combinations.  
There is 1 possible combo where 1 is the highest - 111.  
There are 7 where 2 is the highest - 112, 121, 122, 211, 212, 221, 222.  
There are 19 where 3 is the highest.

In general, there will be k[sup]X[/sup] - (k-1)[sup]X[/sup] where k is the highest. I can explain this more if you don’t see why.

So the expected value of the highest die is just SUM(k(k[sup]X[/sup] - (k-1)[sup]X[/sup]), k = 1…6) / 216. For X = 3, this is 119/24 = 4.958. Now you know the average total is 3.5X = 10.5. So the average after subtracting the highest die is 5.542.

It’s possible to simplify that sum somewhat, depending on what you want it to look like. I have a form with binomial coefficients, which may be more or less simple depending on how you like to do it. Anyway, if you instead want to use the X-1 highest dice, the expected value for the lowest die is:

SUM(k((7-k)[sup]X[/sup] - (6-k)[sup]X[/sup]), k = 1…6) / 216

For X = 3, the two highest dice have an expected value of 8.458. Not surprisingly, this is also equal to 7(X-1) - 5.542, where 5.542 is the answer from before.

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**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\
**Post date:** [February 9, 2004, 2:21am UTC](https://boards.straightdope.com/t/die-rolling-and-probability-question/228121/5 "2004-02-09T02:21:24Z")

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Unable to come up with a simple formula, I went ahead and wrote a program to do it. Put this in an HTML file, and it should run in your browser window. X=3 runs in a trivial amount of time, but it took about 12 seconds to run X=6, and I wasn’t patient enough for X=8.

```auto

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN">
<html lang="en-US" xml:lang="en-US"><head>
<meta http-equiv="Content-Type" content="text/html; charset=us-ascii" />
<meta http-equiv="Content-Language" content="en-us" />
<title>Rolling the Dice</title>
<script type="text/javascript">
function Compute() {
  calc = document.forms[0]
  X = calc.x.value
  n = calc.n.value
  k = calc.k.value
  die = new Array(X); for (j = 0; j < X; ++j) die[j] = 1
  L = total = 0; part = [0, 0, 0]
  Count = Math.pow(n, X)  
  do {
    min = max = die[0]; sum = 0
    for (j in die) {
      if (min > die[j]) min = die[j]
      if (max < die[j]) max = die[j]
      sum += die[j]; }
    total += (sum -= calc.remove[0].checked ? min : max)
    ++part[sum < k ? 0 : (sum > k ? 2 : 1)]
    ++die[0]
    for (j in die) if (die[j] > n) { die[L = j] = 1; ++die[parseInt(j)+1]; }
  } while (L < X - 1)
  calc.average.value = total / Count
  calc.less.value = part[0] / Count
  calc.equal.value = part[1] / Count
  calc.greater.value = part[2] / Count
}
</script></head><body>
<form action="" id="calculon"><p>
Number of dice (X): <input name="x" value="3" size="2" /><br />
Sides per die (n): <input name="n" value="6" size="2" /><br />
Goal value (k): <input name="k" value="8" size="2" /><br />
<input type="radio" name="remove" value="min" checked="checked" /> Remove lowest roll<br />
<input type="radio" name="remove" value="max" /> Remove highest roll<br />
<button type="button" onclick="Compute()">Compute!</button><br /><br />
Average: <input name="average" value="" size="10" /><br />
Fraction less than k: <input name="less" value="" size="10" /><br />
Fraction equal to k: <input name="equal" value="" size="10" /><br />
Fraction greater than k: <input name="greater" value="" size="10" /><br />
</p></form>
</body>
</html>

```

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<div class="post-metadata">

**Author:** ![Coil](https://avatars.discourse-cdn.com/v4/letter/c/bb73d2/32.png) [@Coil](https://boards.straightdope.com/u/Coil)\
**Post date:** [February 9, 2004, 8:00am UTC](https://boards.straightdope.com/t/die-rolling-and-probability-question/228121/6 "2004-02-09T08:00:38Z")

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Thanks **Achernar** that was great.

> [@](#):
>
> In general, there will be kX - (k-1)X where k is the highest. I can explain this more if you don’t see why.

May I ask you to expand on this, so that I hopefully may be able to understand and replicate the calculation myself.

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**Author:** ![MrWhatsIt](https://avatars.discourse-cdn.com/v4/letter/m/df705f/32.png) [@MrWhatsIt](https://boards.straightdope.com/u/MrWhatsIt)\
**Post date:** [February 9, 2004, 4:17pm UTC](https://boards.straightdope.com/t/die-rolling-and-probability-question/228121/7 "2004-02-09T16:17:39Z")

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FWIW, Coil, in your last post where you quote Achernar, the X’s do not appear as superscripts, so the formula in your quote is not correct.

If I may attempt to answer for Achernar, the easiest way to find the probability of the highest of X dice equalling a number K is by recognizing that this is equal to the probability of the highest die being K or less, minus the probability of the highest die being less than K.

The odds of one die being K or less is (K/N). For the highest of X dice to be K or less, all dice must be K or less, and the odds of that are (K/N) to the X. For example, the odds of the highest of 3 six-sided dice (can I assume we’re all gamers enough to just say 3d6?) being 5 or less is (5/6) cubed or 125/216. That is, there are 125 ways to roll 3d6 and not get any 6’s. But not all of those rolls include a 5. There are (4/6) cubed or 64 ways to roll 3d6 and not get any 5’s or 6’s. So the number of rolls that contain at least one 5 and no 6’s is 125 - 64 or 61, and the odds of 5 being the highest number rolled is 61/216.

If your web browser can run Java, [http://www.drizzle.com/~keitha/DieGraph.html](http://www.drizzle.com/~keitha/DieGraph.html) may be helpful, especially if you ever want to drop highest/lowest 2 or more dice from a larger roll.