The Floor is Lava!

The dragon Smaug says he will set you free if you roll a 7 or higher.

If you intend to survive, which would you rather roll: Two 10-sided dice (your score is the sum) or one 20-sided die?

Facts:
1) The 10-sided dice are fair with sides numbered 0 to 9 .
2) The 20-sided die is fair with sides numbered 1 to 20 .

It doesn't matter Two 10-sided dice One 20-sided die

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7 solutions

Zandra Vinegar Staff
Sep 4, 2015

With a D 20 D20 , you have a 14 20 \frac{14}{20} chance of rolling a 7 or higher. So, the probability of success is .70 \fbox{.70} on 2 D 10 2D10 .

Rolling the 2 D 10 D10 , there are 10 possible rolls with the first die, each of which is equally likely, and then 10 possible rolls for the second die. So there are 1 0 2 = 100 10^2 = 100 possible events total, each of which is equally likely, and we must figure out how many of them are successes.

Given each possible roll on die one, how many ways could die 2 come up that would mean success (a total of 7 or greater on both dice)?

roll a 0 on die 1 ==> you must roll a 7, 8, or 9 on die 2 ----- 3 ways
roll a 1 on die 1 ==> you must roll a 6, 7, 8, or 9 on die 2 ----- 4 ways
roll a 2 on die 1 ==> you must roll a 5, 6, 7, 8, or 9 on die 2 ----- 5 ways
roll a 3 on die 1 ==> you must roll a 4, 5, 6, 7, 8, or 9 on die 2 ----- 6 ways
roll a 4 on die 1 ==> you must roll anything but a 0, 1, or 2 on die 2 ----- 7 ways
roll a 5 on die 1 ==> you must roll anything but a 0 or 1 on die 2 ----- 8 ways
roll a 6 on die 1 ==> you must roll anything but a 0 on die 2 ----- 9 ways
roll a 7 on die 1 ==> you can roll anything on die 2 on die 2 ----- 10 ways
roll a 8 on die 1 ==> you can roll anything on die 2 on die 2 ----- 10 ways
roll a 9 on die 1 ==> you can roll anything on die 2 on die 2 ----- 10 ways

So 72 ways total. So there are 72 events that are successful out of 100 possible events. Therefore the probability of succeeding is .72 \fbox{.72} on 2 D 10 2D10 .

Fun question. I found it to be a bit quicker to count the complement first for the 2 D 10 D10 cases. It turns out the the number of ways d ( n ) d(n) of getting a total of n n for 0 n 9 0 \le n \le 9 is d ( n ) = n + 1 , d(n) = n + 1, so n = 0 6 d ( n ) = 7 ( 7 + 1 ) 2 = 28 , \sum_{n=0}^{6} d(n) = \frac{7*(7 + 1)}{2} = 28, and so the number of ways of getting 7 7 or more is 100 28 = 72. 100 - 28 = 72.

P.S.. In "Facts: 2)" I think it should read "... with 20 sides numbered 1-20 . " \textbf{1-20}."

Brian Charlesworth - 5 years, 9 months ago

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It is slightly surprising that the 2D10 has a higher probability. This ultimately is because the D20 has a much fatter tail, despite having a higher expected value.

Calvin Lin Staff - 5 years, 9 months ago

So if you calculate in the same way you did for 7,8,9 the answer doesn't match...ie summation 0 to 7 (n+1) + summation 0 to 8 (n+1) + summation 0 to 9 (n +1)...i think i missed something there...can you please point it out ?

Karan Mirani - 5 years, 9 months ago

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I think that's because my "formula" d ( n ) = n + 1 d(n) = n + 1 only works up to n = 9. n = 9. For n > 9 n \gt 9 the number of ways starts to decrease, with d ( 10 ) = 9 , d ( 11 ) = 8 , d(10) = 9, d(11) = 8, and in general d ( n ) = 19 n d(n) = 19 - n for 10 n 18. 10 \le n \le 18. So n = 10 18 d ( n ) = 9 + 8 + 7 + . . . . + 2 + 1 = 45 , \sum_{n=10}^{18} d(n) = 9 + 8 + 7 + .... + 2 + 1 = 45, which along with d ( 7 ) = 8 , d ( 8 ) = 9 d(7) = 8, d(8) = 9 and d ( 9 ) = 10 d(9) = 10 add to 72 72 as expected.

Note that the distribution is symmetric about n = 9 , n = 9, as d ( n ) = d ( 18 n ) . d(n) = d(18 - n).

Brian Charlesworth - 5 years, 9 months ago

Isn't the 0 a ten? It is when I roll for damage.

Stephen Petersen - 4 years, 8 months ago

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If you read the question, it states that

The 10-sided dice are fair with sides numbered 0 to 9.

This quesiton is not about dealing damage to Smaug.

Calvin Lin Staff - 4 years, 8 months ago
P. Stuart Cohen
Oct 5, 2016

Faster and easier looking only at the probability of losing! With the 20 sided, that would be 6 out of 20. 30 %. With the two 10 sided, there are 100 possible combinations, of which 28 lose. (one way to roll zero, two ways to roll 1, three ways to roll 2, four ways to roll 3, five ways to roll 4, six to roll 5, and seven to roll 6. That adds up to 28 losing possibilities. ) 28 % chance to lose. Better odds than 30 %.

On the two 10-sided, I think there are two ways to roll zero, (0,0 and vice versa) , two ways to roll 1 (0,1 & 1,0), four ways to roll 2 (0,2 & 1,1 - both vice versa), four ways to roll 3 (0,3 & 1,2 both vice versa)... And similarly eight ways to roll 6 (0,6, 1,5, 2,4 , 3,3 - and each vice versa). This makes 32/100 combinations less than 7 as a sum. So prob. of getting 7 or higher should be 68/100. This is less than probability of finding a 7 or higher on a 20-sided die. Hence 20 sides should be the answer. What do you think?

Ali Shehzad - 4 years, 8 months ago
Achille 'Gilles'
Sep 8, 2015

Roll the pair of 10 sided dices.

  • With the pair of 10 sided dice, 72 combinations out of 100 add up 7 or higher (odds are 72%)

  • With the 20 sided dice, 14 out of 20 faces are 7 or higher (odds are 70%)

probably :) a lot of people got this wrong because of vague wording. It should say "roll a sum of seven or greater" not "roll a seven or greater".

P Cooper - 4 years, 7 months ago
Fred Rackow
Oct 11, 2016

The blue dice pictured are actually 8-sided dice :)

There are 55 different combinations of the numbers 0 to 9 on the 2 dice. 39 of them have sums of 7 or greater. 39/55 = 70.9090909% The 20-sided die has 14 numbers greater than or equal to 7. 14/20 = 70%

Rutvik Bapat
Oct 4, 2016

For a 20-sided die with sides numbered 1 to 20 , the probability to come up with a number greater than or equal to 7 = 14 20 \frac{14}{20} = 0.7 .

However. for a 10-sided die with sides numbered 0 to 9 , the probability to come up with a number greater than or equal to 7 will be sum of the all kinds of combinations like so,

Assuming the first die gets a 9, then any one of the 10 numbers on the second die would fetch a sum greater than or equal to 7 . Probability for the aforementioned case = 1 10 \frac{1}{10} * 10 10 \frac{10}{10} = 10 100 \frac{10}{100} .

Similarly if we were to sum the probabilities for all the numbers from 9 to 0, we would end up with an expression like so, 10 100 \frac{10}{100} + 10 100 \frac{10}{100} + 10 100 \frac{10}{100} + 9 100 \frac{9}{100} + 8 100 \frac{8}{100} + 7 100 \frac{7}{100} + 6 100 \frac{6}{100} + 5 100 \frac{5}{100} + 4 100 \frac{4}{100} + 3 100 \frac{3}{100} = 72 100 \frac{72}{100} = 0.72 .

Ergo, looking at the results, I would rather take my chances with rolling the two 10-sided dice if asked by the dragon Smaug.
Robert Price
Oct 1, 2016

|I know that with two d6 dice, 7 is the most likely outcome, because there are more combinations that add up to 7 than any other number. On that basis, I went for the two d10

Mick McCarthy
Sep 16, 2015

The two 10 sided dice give you a better chance of rolling a 7 or better There are 28 ways out of 100 - or 28% - chance not to roll a 7 with two 10 sided dice.

First die 0 second die 0-6 seven ways

First die 1 second die 0-5 six ways

First die 2 second die 0-4 five ways

First die 3 second die 0-3 four ways

First die 4 second die 0-2 three ways

First die 5 second die 0-1 two ways

First die 6 second die 0 one way

                                         = 28/100

So if you wouldn't roll a 7 28% of the time that means you will roll a 7 or better 72% of the time

With the one 20 sided die you have a 14/20 - or 70% - chance of rolling a 7 or better

The correct answer is the D20. It says "roll a 7 or higher" NOT "roll a total of 7 or higher". Totals and digits are not the same thing, and the question clearly implies digits.

Bruce Walton - 4 years, 8 months ago

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