most simple of math Q's

yes

In 2-200 you're only excluding 1 number (1) out of 200,
in (1-100)+(1-100) you're excluding 1 number (1) out of 200 of being possible.
(100/199*49%) win chance in 2-200
and (100/199*49)% win chance in (1-100)+(1-100)
Same calculation, same probability.

+ Show Spoiler +

NO...this is a binomial distribution problem if I'm reading what ur asking correctly.

if random 1-100 r independent then no it's NOT the same distribution of random 2-200..think about it this way, odds of random 2-200 getting 100 or 200 exactly are both 1/199...odds of 2 random 1-100 trials adding to exactly 200 is 1/10000 since it's (1/100*1/100) whereas 2 random 1-100 adding to 100 could be done by 30+70, 60+40,55+45,etc each specific outcome has a probability of 2/10000 (1/10000+1/10000 since 30+70 is 1/10000 and 70+30 is 1/10000)..this follows a binomial distribution w/ 10k states which is a royal bitch to calculate right now for me.

But as a classic easier example would be 2 D6 vs 1 D12 w/ the 1 removed from the die so only 2-12 are possible
If you roll a 4 then the odds ur roll is greater then the sum of 2 D6 die you get .028+.056 or 1/36+2/36 (odds of getting 2+odds of getting 3) for a total probability of .084, 8.4%
For a die of 2-12 contains 12-2+1 sides so 11 sides, your odds of rolling below 4 on one roll of the die is .091+.091 or 1/11+1/11 or 2/11 which is 18%.

http://hyperphysics.phy-astr.gsu.edu/hbase/math/dice.html#c2

at a glance i have to disagree with the above.
if the dice are assumed independent of each other, there should be no difference.

im pretty sure this still forms a pretty uniform distribution that mimics the first, which is all were worried about

it does not form a uniform distribution it forms a standard distribution

ah you're right.
the 2-200 is uniform, 2d 1-100 is standard.

decided to figure it out for op,....here's an easyish way to do it.

Go again to rolling 2 six sided die. If you look at the distribution on the link I posted a patter occurs which is. Odds of getting a 2=1/36 (1+1) 3=2/36 (2+1 or 1+2), odds of getting a 4=3/36 (3+1,1+3,2+2) so odds of getting a 5=4/36 (4+1,1+4,2+3,3+2). See the pattern? Odds of getting x is equal to (x-1)/(sides^2)..this pattern follows up to the midway point. For values above the midway point the formula becomes probably of getting y is found by (2*median value-y)=x x willl be less then the median and will be the number u should use.. Since that was poorly worded the odds of getting an 8 (on 2 6 sided die) would be found by 2*7-8=6.the probabily of getting 6 is 5/36 (1+5,5+1,4+2,2+4,3+3) same as getting an 8 (4+4,5+3,3+5,6+2,2+6) and 10 would be 7*2-10=4, probability of a 4 is 3/36 which is the same as a 10.

Anyhow that's the pattern. So how to do less then. Odds of getting less then 6 is just odds of 5+odds of 4+odds of 3+odds of 2. which is 4/36+3/36+2/36+1/36. So odds of less then x=(sum from 1 to x-2)/(total combinations) odds of below a 7 would be (5+4+3+2+1)/36 this breaks and must be altered for above the median but thats irrelevant since 50 is below the median value.

For your problem the odds are below a 50 on 2 100 sided die. So follow the pattern (48+47+...1)/10000 or 1176/10000 or 11.76%. Easy way to add consecutive numbers 1 to x is x*(x+1)/2 so 48 to 1 added together is 48*49/2 or 1176.

fwiw the odds of getting below a 50 on the 2-200 is 49/199 or 24.6% not even close to the same probability.

What are random1 and random2?

Is random1 uniform on the discrete set {1,2,3,...,100}?
Is random2 uniform on the set {1,2,3,...,200}?

random 1 discrete set, natural numbers 1-100
random 2 discrete set, integers 2-200

least that's my assumption

Oh I see. So we have X is uniform on {1,...,100} and Y is uniform on {2,...,200}. Then:

P(X+X <= 50) = P(X <= 25) = 25/100 = 0.25

P(Y <= 50) = 49/199 ~= 0.246

Or are we assuming that we draw X again, independently, i.e. Z is uniform on {1,...,100} and we want P(X+Z <= 50)?

palak got it, thx

@nixxxbg that doesn't work. Both numbers on random1 don't have to be below 25. You could have x1=47, x2=1 so X=48 or x1=35 x2=12 ->X=47...both those satisfy that. Also you could have for example x1=10 x2=40 making X=50 and not meeting the requirement.

Sorry, I was confused.

Let X~Uniform{1,...,100} and Z~Uniform{1,...,100} and suppose they are independent. Then:

P(X+Z <= 50) = sum_j_2_50 P(X+Z = j) = sum_j_2_50 sum_k_1_100 P(X=k)*P(Z = j-k) = 1/100 * (1/100 + 2/100+ ... + 49/100) = 1225/10000 = 0.1225

I guess you had the right answer already. Sorry for the confusion.

^u only go from 1-48, not 49...49/100 doesn't work since that would mean the sum would total at least 50.

U can either add 1-48 or 2-49 depending if u want to add single die value or sum. The answer is obv the same but u can not add 1-49

thanks palak for the write up

lol

am i stupid for not understanding anything of anything here?

On October 23 2011 18:09 TalentedTom wrote: yes

''Hi im TalentedTom its a coin flip so ill answer and hope to be right so ill look smart''
xD