python 2.6 - how to print same randomint after if/elif? -

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i want randint printed same in if/elif statements each time executes without new integer returning. there way instead of having write code after while statement or fine?

from random import * def randomchance():     return randint(1, 100) def randomknife():     return randint(1, 8) skins = ['blue', 'purple', 'pink', 'red'] knives = ['karambit', 'flip', 'bowie', 'butterfly', 'm9 bayonet', 'bayonet', 'daggers', 'falchion'] print 'welcome cs go case lottery' skin in skins:     print "available", skin, knife in knives:     print "available", knife, print '(blue = common, purple = uncommon, pink = rare, red = epic)' keys = 10 while keys >0:     resp=raw_input("enter 'yes' open case: ")     if (resp == str('yes') or resp == str('yes')):         print 'opening case...'         if (randomchance() >= 35):             print 'you\'ve won a', skins[0]         elif (randomchance() >= 20):             print 'you\'ve won a', skins[1]         elif (randomchance() >= 10):             print 'you\'ve won a', skins[2]         elif (randomchance() >= 5):             print 'you\'ve won a', skins[3]         elif (randomchance() >= 1):             if randomknife == 1:                 print 'you\'ve won a', knifes[0]             elif randomknife() == 2:                 print 'you\'ve won a', knifes[1]             elif randomknife() == 3:                 print 'you\'ve won a', knifes[2]             elif randomknife() == 4:                 print 'you\'ve won a', knifes[3]             elif randomknife() == 5:                 print 'you\'ve won a', knifes[4]             elif randomknife() == 6:                 print 'you\'ve won a', knifes[5]             elif randomknife() == 7:                 print 'you\'ve won a', knifes[6]             elif randomknife() == 8:                 print 'you\'ve won a', knifes[7]         keys -= 1              elif(resp == str('no') or resp==str('no')):         resp1=raw_input('would exit? enter no exit: ')         if resp1 == 'no' or "no":             exit()     else:         print "yes or no. answers only" else:     print 'you\'ve run out of keys!'

@konstantin has right answer how choose random number once before testing conditions.

as second question (how use less code test conditions), take @ rewrite/cleanup below. in particular, used random.choice pick knife, , used list of thresholds (35, 20, etc.) pick skins.

i fixed couple bugs, if resp1 == 'no' or "no": should if resp1 == 'no' or resp1 == "no": (but used resp1.lower() instead).

and couldn't stand 'would exit? enter no exit: ', made "yes" exit instead. :-)

import random  skins = ['blue', 'purple', 'pink', 'red'] knives = ['karambit', 'flip', 'bowie', 'butterfly', 'm9 bayonet', 'bayonet', 'daggers', 'falchion']  print 'welcome cs go case lottery'  print "available skins: %s" % ', '.join(skins) print "available knives: %s" % ', '.join(knives) print '(blue = common, purple = uncommon, pink = rare, red = epic)'  _ in range(10):     resp = raw_input("enter 'yes' open case: ")     while resp.lower() != 'yes':         if resp.lower() == 'no':             if raw_input('would exit? ').lower() == 'yes':                 exit()         else:             print "yes or no answers only."         resp = raw_input("enter 'yes' open case: ")      print 'opening case...'      chance = random.randint(1, 100)      i, n in enumerate([35, 20, 10, 5]):         if chance >= n:             print "you've won %s skin." % skins[i]             break      if chance < 5:         print "you've won %s knife." % random.choice(knives)  print "you've run out of keys!"

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