GameBanshee Forums

Tricky wrote:How do you get to 36% Dottie? It's an odd percentage..

Tricky wrote:Leaving out the percentages, outcomes for double drawings are false/false, false/true, true/false, true/true. A 1/4th chance of not getting any winnings at all. On the 40% game, wouldn't that mean 1/4th of the 60%? 15%?

Tricky wrote:Hmm, I should've made this into a poll..

Tricky wrote:That depends on the computer. Pseudo generated random numbers are a major security leak. My old Pentium 3 had a small temperature measurement tool that measured the thermal vibrations of a small metal element and based its random calculations on that.

Loki[D.d.G] wrote:Good point. It'll be pretty interesting to see just how random a random generated computer number is. After all, it is a machine coupled with a software that conforms to their own specific rules.

Tricky wrote:That depends on the computer. Pseudo generated random numbers are a major security leak. My old Pentium 3 had a small temperature measurement tool that measured the thermal vibrations of a small metal element and based its random calculations on that.

I would go with one @ 60% chance of winning. I understand that in normal circumstances every throw of a die is 50/50, but in the given circumstance it's either one at 60% or two at 40% each. The one is above the average, the two are below the average, as Fable said, so you have a slightly better chance with the one 60-percenter, imo.

import random
winOne=0
winTwo=0
for i in range(0,1000000): #A MILLION DICE ROLLS
	if random.random() < 0.6:
		winOne+=1
	if random.random() < 0.4:
		winTwo+=1
	if random.random() < 0.4:
		winTwo+=1
print 'A 60% roll: ', winOne
print 'Two 40% rolls: ', winTwo

vicsuns-macbook-pro-109 :D esktop Vicsun$ python diceroll.py 
A 60% roll:  600592
Two 40% rolls:  799387
vicsuns-macbook-pro-109 :D esktop Vicsun$ python diceroll.py 
A 60% roll:  600707
Two 40% rolls:  800127
vicsuns-macbook-pro-109 :D esktop Vicsun$ python diceroll.py 
A 60% roll:  600286
Two 40% rolls:  801798

Vicsun wrote:However, the chance of making either one of two 40% throws is above average, and also above 60%. As a matter of fact it's closer to 80%

fable wrote: You roll once, and your chance is 4 out of 10 of getting what you want. You roll the second time, and your chance is still 4 out of 10 of getting what you want. So how does that add up to an 8 out of 10 chance? Isn't it rather 2 attempts at a 4 out of 10 roll?

And what's the chance of making that 60% roll? 6 out of 10, correct?

Vicsun wrote:Exactly, and if you are successful at either one of the two attempts, it counts as a win If you are successful at both it counts as two wins.

Think of it the following way: you are rolling a 1d10. In the first case, you 'win' if you roll 6 or higher. In the second case, you 'win' if you roll 4 or higher, but you can try to roll 4 or higher a hundred times. It's incredibly likely that at least one of those hundred rolls will be higher than a 4. As a matter of fact, you'd get, on average, 40 wins. In the first case, you'd only get 0.6 wins on average.