Prob. question.

[capsulecorpjx]

Ok, person A randomly gets a number from 1 to 100.
Person B randomly gets a number from 1 to 70.

What are the chances that Person A's number is > Person B's number?

30% of the time, A will win regardless of what B gets.
The rest is 50/50? So would it be 30 % + 45%? or 75% of the time A will win?

[Edit]: It can't be exactly 30%, it must be at least > 50%, I'm guessing its 75%

[Rassis]

I can prove by simulation that P(Xa) > P(Xb) is approximately equal to 0,65 but need some more time to do it analytically or perhaps I leave it to some other colleague in this forum.

[Rassis]

Let me add two more results that might bring some light on the way to the analytical solution:

If 1 <= Xa <= 100 and 1 <= Xb <= 100, then P(Xa) > P(Xb) = 0,50
If 1 <= Xa <= 100 and 1 <= Xb <= 50, then P(Xa) > P(Xb) = 0,75

[zaza]

I don't know where you get the 45% from.
30% of the time, A wins outright. Of the remaining 70%, it is split 50-50 which is to say 35% each.

Hence you get 65% in favour of A, 35% to B. Actually, it is slightly less than this because there is also the probability that they get the same number. But close enough.

zaza

[capsulecorpjx]

I don't know where you get the 45% from.
30% of the time, A wins outright. Of the remaining 70%, it is split 50-50 which is to say 35% each.

Hence you get 65% in favour of A, 35% to B. Actually, it is slightly less than this because there is also the probability that they get the same number. But close enough.

zaza

Thx! Stupid mistakes. I always make them.