# Child's game Probability Question

**URL:** <https://boards.straightdope.com/t/childs-game-probability-question/752236>\
**Category:** Factual Questions\
**Created:** [April 15, 2016, 1:39am UTC](https://boards.straightdope.com/t/childs-game-probability-question/752236 "2016-04-15T01:39:40Z")\
**Posts on this page:** 3\
**Page:** 2

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**Author:** ![Trinopus](https://avatars.discourse-cdn.com/v4/letter/t/2bfe46/32.png) [@Trinopus](https://boards.straightdope.com/u/Trinopus)\
**Post date:** [April 15, 2016, 8:35pm UTC](https://boards.straightdope.com/t/childs-game-probability-question/752236/21 "2016-04-15T20:35:11Z")

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> [@septimus](#):
>
> I ran 20,000,000 simulations for each value of Crow’s path-length from 1 to 6, and noted that the derived precision was better than the three sig-figs I reported here. The batch of 20 million simulations for length-6 took about 11 seconds on my slow laptop.

Whoo! Your machine is a damn sight faster than mine! Good work!

> [@lisiate](#):
>
> Yes, if you roll a colour you pick a fruit of that colour, if there is one left. If there isn’t that’s your turn over.

This is why infinite sequences are (mathematically) possible, and why an explicit analysis would be a pain. This is (in the abstract) an infinite game, not a finite one.

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**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\
**Post date:** [April 15, 2016, 8:52pm UTC](https://boards.straightdope.com/t/childs-game-probability-question/752236/22 "2016-04-15T20:52:51Z")

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I wrote some code to the analysis exactly by boiling the game down to 345 (or 414) distinct states. Each state has a small number of parent states and the transition probabilities between states are pretty easy to compute. You can also normalize state exit probabilities to handle the infinite loop problem. Then just sum the probabilities of the 5 (or 6) winning states.

The unsurprising answer is:  
0.6313573066006350 for 5 crow moves  
0.7684969326194143 for 6 crow moves

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<div class="post-metadata">

**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\
**Post date:** [April 15, 2016, 9:09pm UTC](https://boards.straightdope.com/t/childs-game-probability-question/752236/23 "2016-04-15T21:09:26Z")

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Here’s the code (Python). It’s pretty cludgy, but it runs pretty much instantly. I’m sure it could be tightened up quite a bit.

```auto

def find_parents(a,b,c,d,x):
    if a==d:
        if a<4:
            parents = [[a+1,b,c,d,x]]
        else:
            parents = []
    elif a==c:
        if a<4:
            parents = [[a+1,b,c,d,x],[a,b,c,d+1,x]]
        else:
            parents = [[a,b,c,d+1,x]]
    elif a==b and c==d:
        if a<4:
            parents = [[a+1,b,c,d,x],[a,b,c+1,d,x]]
        else:
            parents = [[a,b,c+1,d,x]]
    elif a==b and c!=d:
        if a<4:
            parents = [[a+1,b,c,d,x],[a,b,c+1,d,x],[a,b,c,d+1,x]]
        else:
            parents = [[a,b,c+1,d,x],[a,b,c,d+1,x]]
    elif b==d:
        if a<4:
            parents = [[a+1,b,c,d,x],[a,b+1,c,d,x]]
        else:
            parents = [[a,b+1,c,d,x]]
    elif b==c:
        if a<4:
            parents = [[a+1,b,c,d,x],[a,b+1,c,d,x],[a,b,c,d+1,x]]
        else:
            parents = [[a,b+1,c,d,x],[a,b,c,d+1,x]]
    elif c==d:
        if a<4:
            parents = [[a+1,b,c,d,x],[a,b+1,c,d,x],[a,b,c+1,d,x]]
        else:
            parents = [[a,b+1,c,d,x],[a,b,c+1,d,x]]
    else:
        if a<4:
            parents = [[a+1,b,c,d,x],[a,b+1,c,d,x],[a,b,c+1,d,x],[a,b,c,d+1,x]]
        else:
            parents = [[a,b+1,c,d,x],[a,b,c+1,d,x],[a,b,c,d+1,x]]
    if x > 0 and a>0:
        parents += [[a,b,c,d,x-1]]
    return parents
    
def find_prob_from_parent(child, parent):
    a,b,c,d,x = child
    q,r,s,t,y = parent
    denom = 6 - int(q==0) - int(r==0) - int(s==0) - int(t==0)
    if q > a:
        numer = parent[:-1].count(q) + 1
    elif r > b:
        numer = parent[:-1].count(r) + int(r==q)
    elif s > c:
        numer = parent[:-1].count(s) + int(s==q)
    elif t > d:
        numer = parent[:-1].count(t) + int(t==q)
    else:
        numer = 1
    return numer/denom
    
probstatemat = [[[[[-1 for x in range(8)] for d in range(5)] for c in range(5)] for b in range(5)] for a in range(5)]
probstatemat[4][4][4][4][0] = 1.0
def probstate(a,b,c,d,x):
    if probstatemat[a]**[c][d][x] == -1:
        ps = find_parents(a,b,c,d,x)
        probstatemat[a]**[c][d][x] = sum([find_prob_from_parent([a,b,c,d,x], p)*probstate(p[0],p[1],p[2],p[3],p[4]) for p in ps])
    return probstatemat[a]**[c][d][x]  

total = 0.0
for i in range(5):
    total += probstate(0,0,0,0,i)
print(total)

total = 0.0
for i in range(6):
    total += probstate(0,0,0,0,i)
print(total)

```

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