Is it possible to calculate the probability the sum of one set of random variables equals the sum of a different set of random variables? I'm doing data analysis at the business I work for. 23 people were asked for input for two separate lists. The first list the answers could be any integer 6 to 10. The second list the answers could be any integer 0 to 15. After receiving all inputs, the average number chosen for each list was identical. For the average to be identical so was the sum. Is it possible to calculate the probability the two sums were equal even though they drew from different ranges?
The original question was how much would a prospective baby weigh. 23 people answered with a guess of pounds and additional ounces. The pounds were limited 6 to 10 and the ounces were limited 0 to 15. There was no discernible connection between the guess of pounds to the guess of ounces. 
Also, I need help with the calculation in addition to a yes or no answer. I need a calculation over results obtained through simulation but thank you for those results as well. 
I would comment on posts, however, I am not allowed to yet.
Also, can you please try to answer in Layman's terms please.
 A: Yes it is possible, and the other answers/comments have shown how you can calculate/approximate the probability.
But this is probably (I imagine) not what you want. If you want to show the probability that people respond similar or different on two different scales then it is a bit simplistic to compare them to people that just randomly pick uniform numbers. We are not speaking of random dice rolls. 
The current setup gives a low 1.5% probability that the two sets end up with the same sum (as answered and commented by others). So you might be tempted to assume that you have an extreme case of similarity (ie reject the hypothesis that the sums are not gonna end up the same). If you had a larger range for the pounds then you probably would have had the same equal sum (since few newborns are less than 6 or more than 10 pounds), but the theoretical naively calculated probability would be extremely low.



*

*estimate of ounces. This is probably gonna end up in the middle around 7.5 since not so many people know the precision and are just estimating like picking a uniformly distributed random number. (possibly a bit lower than 7.5 since people might tend to place the estimate behind the comma more low or leave it out)

*estimate of the pounds. Since the average newborn is close to between 7 an 8 you are very likely gonna end up with a mean somewhere close to 7.


This makes it much more likely that the sum of ounces and sum of inches are gonna match. (in comparison to the 1.5% based on uniform picks)

You could use your data to create estimates of the distributions how people will prospect birth weight (for the pounds I would model with a multinomial with the events 6,7,8,9,10, for the ounces I would model with a beta distribution). 
Based on those estimates of the distributions you could estimate the probability for any other new group of 23 people (from the same population) to again produce the same sum of the two different sets of random variables.
A: As stated above:  it's all in the assumptions.  If the numbers in each range are equally likely and the 23 folks choose independently, the brute force method to get an exact answer is to use probability generating functions.
Here is a Mathematica implementation:
pgf1 = Sum[t^i (1/5), {i, 6, 10}]^23;
pgf2 = Sum[t^i (1/16), {i, 0, 15}]^23;
pr1 = Table[Coefficient[pgf1, t^i], {i, 6*23, 10*23}];
pr2 = Table[Coefficient[pgf2, t^i], {i, 6*23, 10*23}];
Total[pr1 pr2]

with the answer being 
$$\frac{448626236408308149794568086913862442605587}{29514790517935282585600000000000000000000000}$$
or approximately 0.01520004813.
Update
Here is the equivalent in R:
library(polynom)
n = 23
p1 = polynomial(c(0,0,0,0,0,0,rep(1/5,5)))^n
p2 = polynomial(rep(1/16,16))^n

c1 = coef(p1)[(6*n+1):(10*n+1)]
c2 = coef(p2)[(6*n+1):(10*n+1)]

sum(c1*c2)
# 0.01520005

