I was playing ?Zombies!!! The Director?s Cut?. You move your counter round a map of a town, trying not to be killed by zombies. The winner is the first to make it to the helipad. You start with 3 ?heart? tokens, and 3 ?bullet? tokens. You can collect more of these tokens during the game. You can hold any number of bullets, but can hold a maximum of 5 hearts.If you land on the same square as a zombie, you need to try to kill it, by rolling a 4, 5 or 6 on a dice. If you fail, you can either spend a heart token, or you can spend bullet tokens. If you spend a heart token, you roll the dice again. If you fail again you can spend another token?Each bullet token adds one to the dice roll. So if I rolled a 2, I could spend 2 bullet tokens to make 4, and kill the zombie. If you run out of tokens and still haven?t killed the zombie, you go back to the start.Which is better, a bullet or a heart token?You collect tokens by landing on a square with a token in it. If a square has both a token and a zombie, should I try to collect the token (I have to fight the zombie first to collect the token)?How many zombies can I expect to kill before being sent back to the start?

- Traden4Alpha
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Interesting. By one way of looking at it it takes on average, 2 hearts to kill a zombie but only 1.5 bullets. And if one has rolled a 3, a bullet seems like a good investment (assuming one gets to decide how many bullets to use AFTER the roll).You kill 1/2 a zombie with every zombie-square you enter and every time you use a heart.... more later....

- Traden4Alpha
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If one has 0 hearts and 0 bullets left (the [0,0] case), one will, on average, survive 1 turn and kill 1 zombie (z[0,0] = 1). In the [1,0] case, the average number of zombies killed, z[1,0], is:z[1,0] = 0.5*z[0,0] + 0.5*(1+z[1,0]), which implies z[1,0] = 2more generally,z[N,0] = 0.5*z[N-1,0] + 0.5*(1+z[N,0]), which implies each heart kills an average of 1 zombie The M bullets case is more complicated.... At the very least it seems it takes an average of 1.5 bullets to kill a zombie above and beyond the normal zombie fatality patterns.And the [N,M] case.... Hmmm.... Well that's the point of your question!My current VERY ROUGH guess (probably a lower bound) is that one would survive 6 turns starting with [3,3] and not including any tokens picked up along the way.

- Traden4Alpha
**Posts:**23951**Joined:**

The case of one bullet:z[0,1] = (0)/3 + (1+z[0,0])/6 + (1+z[0,1])/2z[0,1] = 5/3Two bullets:z[0,2] = (0)/6 + (1+z[0,0])/6 + (1+z[0,1])/6 + (1+z[0,2])/2z[0,2] = 23/9more later.....

For the general case, For simplicity,we can write z[-1,N]=0 and z[M,-1]=z[M,-2]=z[M,-3]=-1And for M>=0 and N>=0, z[M,N]=1+1/3*Max(z[M-1,N],1+z[M,N-1])+1/3*Max(z[M-1,N],1+z[M,N-2])+1/3*Max(z[M-1,N],1+z[M,N-3]).1 corresponds to dice=4,5,6, second term to dice=3, third dice=2 and last dice=1.We can see:* for dice= 3, 1+z[M,N-1] is always > z[M-1,N] so use a bullet if you roll 3.* for dice= 1, 1+z[M,N-3] is always < z[M-1,N] so use a heart if you roll 1.* for dice=2, it depends. It's better to use an heart if the number of heart is >= number of bullet-x(number of bullet), x starts from 0 and increases slowly (x(5)=1 and x(10)=2).z[3,3]=8.18.Need probabilities to meet a token, a zombie or both to solve the general case (with the rules of 5 hearts)

Last edited by Vanubis1 on September 16th, 2015, 10:00 pm, edited 1 time in total.

Nice analysis! No zombies eating your brains!I had the dice = 1 & 3 solutions, but was stuck with the dice = 2. I'd been expecting to find a neat solution that x hearts could be thought as equivalent to a bullet. I hadn't considered that 'it depends'.In the rules, as an alternative to reaching the helipad first, you can win by killing 25 zombies. For a follow-up question, how many heart & bullets would you need to expect to win by killing zombies (max 5 hearts, assume you don't collect more tokens)?

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