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
.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Fun question. I found it to be a bit quicker to count the complement first for the 2 D 1 0 cases. It turns out the the number of ways d ( n ) of getting a total of n for 0 ≤ n ≤ 9 is d ( n ) = n + 1 , so ∑ n = 0 6 d ( n ) = 2 7 ∗ ( 7 + 1 ) = 2 8 , and so the number of ways of getting 7 or more is 1 0 0 − 2 8 = 7 2 .
P.S.. In "Facts: 2)" I think it should read "... with 20 sides numbered 1-20 . "
Log in to reply
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.
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 ?
Log in to reply
I think that's because my "formula" d ( n ) = n + 1 only works up to n = 9 . For n > 9 the number of ways starts to decrease, with d ( 1 0 ) = 9 , d ( 1 1 ) = 8 , and in general d ( n ) = 1 9 − n for 1 0 ≤ n ≤ 1 8 . So ∑ n = 1 0 1 8 d ( n ) = 9 + 8 + 7 + . . . . + 2 + 1 = 4 5 , which along with d ( 7 ) = 8 , d ( 8 ) = 9 and d ( 9 ) = 1 0 add to 7 2 as expected.
Note that the distribution is symmetric about n = 9 , as d ( n ) = d ( 1 8 − n ) .
Isn't the 0 a ten? It is when I roll for damage.
Log in to reply
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.
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?
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".
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%
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 = 2 0 1 4 = 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 0 1 * 1 0 1 0 = 1 0 0 1 0 .
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, 1 0 0 1 0 + 1 0 0 1 0 + 1 0 0 1 0 + 1 0 0 9 + 1 0 0 8 + 1 0 0 7 + 1 0 0 6 + 1 0 0 5 + 1 0 0 4 + 1 0 0 3 = 1 0 0 7 2 = 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. |
|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
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.
Problem Loading...
Note Loading...
Set Loading...
With a D 2 0 , you have a 2 0 1 4 chance of rolling a 7 or higher. So, the probability of success is . 7 0 on 2 D 1 0 .
Rolling the 2 D 1 0 , 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 = 1 0 0 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 . 7 2 on 2 D 1 0 .