expected minimum over four rounds with 2 dice

import scipy.stats
import numpy as np

# Python ranges are [x,y)
dice = scipy.stats.rv_discrete( values=(range(1,7), np.repeat(1/6.0, 6)))

twoDice = scipy.stats.rv_discrete( values=(range(2,13), [1/36.0,2/36.0, 3/36.0, 4/36.0, 5/36.0, 6/36.0, 5/36.0, 4/36.0, 3/36.0, 2/36.0, 1/36.0] ))

def getExpect(r,dice_dist):
    # See https://stats.stackexchange.com/questions/220/how-is-the-minimum-of-a-set-of-random-variables-distributed
    # for reasoning of the following formula:
    minOfDistCdf =  map(lambda x: 1-dice_dist.sf(x)**r, dice_dist.xk)
    # Convert the previous CDF to a PMF, by subtracing the next item in the list with the previous one.
    minOfDistPmf = [minOfDistCdf[0]]+[y - x for x,y in zip(minOfDistCdf,minOfDistCdf[1:])]
    new_dist = scipy.stats.rv_discrete( values=(dice_dist.xk, minOfDistPmf))
    return new_dist.expect()

def simulate(r,d,dice_dist, n):
    # r = rounds, d = amount of dice, dice_dist = ..., n = amount of simulation rounds.
    minima = list()
    for i in range(0,n):
        minSoFar = 1000000000 # This is really poor and lazy programming :D But it's for a lazy person anyway.
        for i in range(0,r):
            thisThrow = sum(np.random.choice(dice_dist.xk, d))
            minSoFar = min([minSoFar,thisThrow])
        minima.append(minSoFar)

return sum(minima) / float(n)

print simulate(4,2,dice,1000)

print getExpect(4,twoDice)

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