B Simple Die Game: Win by Rolling 1-5 Before 6 - Odds of Winning?

[mathman]

TL;DR Summary

Throw die repeatedly

Game: throw die repeatedly until all numbers 1-5 show up ay least once (win), before a six shows up on the way (lose). Odds of winning?

Note: Someone posted this on another forum and I was able to come up with three different approaches (two very similar). Try to get them.

 

[Halc]

Trivially, some number has to come up last. Keep rolling until 5 of the 6 numbers have been seen.
Odds are identical that any number is that last one seen, so 1 in 6 odds that it's a 6 and you win.

Given the other approaches, do any give different odds?

This game has effectively been played for years in the Price is Right game show. The standard line is that with each roll, 'the odds are never against you', which of course they are overall.

 

[mathman]

Other approaches come up with same result.

 

[etotheipi]

Not a proof, but I tried

Spoiler: Thing

import random
count = 0
n = 100000
thrown = [0,0,0,0,0,0]
current = 0

for i in range(n):
    while(current!=5):
        current = random.randint(0, 5)
        thrown[current] = 1
        if(current==5):
            num = int("".join(str(x) for x in thrown), 2)
            if(num == 63):
                count+=1
    current = 0
    for i in range(0, len(thrown)):
        thrown[i] = 0

print(count/n)

and it gets pretty close to 0.166 with 100,000 throws.

 

[mathman]

Simple proof: ignore throws which repeat a number. The following considers only throws with a new number. First throw 5/6 not 6, second throw 4/5 not 6, etc. probability of not 6 after 5 throws is 5!/6!=1/6.