Some days it is really fun and interesting being the admin and author of a blog. I am slightly compulsive about checking my traffic stats throughout the day. This lets me know how many people have visited my blog, and if they got there by clicking some link somewhere else out on the Internet. It also lets me know what search terms were used to find my blog. While most of these are pretty predictable, every now and then I see something really silly that led to my site. But it also gives me a keen insight into what people are interested in, and oddly enough, the search term that keeps leading to me over and over again is “how to get the most brains in zombie dice” and “strategies for zombie dice”. I’ve reviewed Zombie Dice in the past, but I have never delved into the math involved with getting the most brains, and at your request, I’m going to dig into that today.

There are two major pieces of information that you should look at when playing zombie dice: What are the conditions in the cup, and what are your odds of rolling a shotgun? If you break each one of those down, your decision becomes much easier. The first thing to consider is what is the likelihood of rolling a brain, shotgun, or feet on each color of die.

**Odds of rolling each
symbol **

**on a single die**

Brain | Shotgun | Feet | |
---|---|---|---|

Red |
16.7% | 50% | 33.4% |

Yellow |
33.4% | 33.4% | 33.4% |

Green |
50% | 16.7% | 33.4% |

As you can see, your gut instincts for what symbols are likely to be rolled are largely justified. It makes sense that you are so much more likely to roll a shotgun on a red die and a brain on a green die. It’s just common sense. However, what is more interesting is the likelihood of rolling one or more shotguns on a single roll of three dice.

**Odds of rolling 1 shotgun on a single roll of three dice,
depending on the colors of the dice rolled. **

R & R | R & Y | R & G | Y & Y | Y & G | G & G | |
---|---|---|---|---|---|---|

Red |
50% | |||||

Yellow |
44% | 38.9% | 33.3% | |||

Green |
38.9% | 33.3% | 27.8% | 27.8% | 22.2% | 16.6% |

**Odds of rolling 2 shotguns on a single roll of three dice,
depending on the colors of the dice rolled. **

R & R | R & Y | R & G | Y & Y | Y & G | G & G | |
---|---|---|---|---|---|---|

Red |
25% | |||||

Yellow |
19.5% | 15.12% | 11.1% | |||

Green |
15.12% | 11.1% | 7.7% | 7.7% | 4.9% | 2.7% |

**Odds of rolling 3 shotguns on a single roll of three dice,
depending on the colors of the dice rolled. **

R & R | R & Y | R & G | Y & Y | Y & G | G & G | |
---|---|---|---|---|---|---|

Red |
12.5% | |||||

Yellow |
8.6 % | 5.8% | 3.69% | |||

Green |
5.8% | 3.69% | 2.1% | 2.1% | 1.09% | 0.45% |

These statistics reveal some really interesting things. Firstly that your odds are never any worse than 50/50. That’s better than any casino in the country, and thus means that the odds of you rolling a shotgun is stacked in your favor, and the odds of you rolling multiples in a single roll drops drastically. But what I find equally fascinating is when the same percentage appears in the table in two different spots. For example, the odds of you rolling a single shotgun blast is the same regardless of if you are rolling all yellow dice, or if you are rolling a red, a yellow, and a green. This can be easily expressed as:

** R + G = 2Y**.

If you apply some simple integers into those variables it becomes easy to figure your odds in much the same way that an individual counts cards. For example lets substitute in the value 5 for the R and 1 for the G. 5 + 1 = 2Y. That would force Y to be 3, a value nicely in the middle. From this we can make some conclusions based upon dice combinations:

If we have three red dice, the combined value is 15, and is our worst odds possible.

If we have three green dice, the combined value is 3, and is our best odds possible.

If we have three yellow dice, the combined value is 9, and is the middle of the pack.

If we have one red (5), one yellow (3), and one green (1) the combined value is 9: the same odds as having three yellow dice. So you know the odds of rolling a shotgun is the same on both sets of dice.

What does this mean? It means that you can quickly access the amount of risk a roll will have, and if it is above or below the threshold of risk you would like to maintain. For instance, if you would like to maintain that you only roll when you have a 25% chance of getting a shotgun, you would want to keep the value of your roll to four or less (aka 3 greens (3), or a yellow(2) and two greens(2). However 2 yellows and a green would put you over your desired threshold of risk.)

But that is assuming that we all know exactly the hand we are going to draw out of the cup each time we decide to roll, and of course we don’t (although you might have an idea based upon how many feet you had rolled previously). These odds can also be evaluated based upon a simple count of the point system (because points work much better than coming up with the exact percentage when dealing with that many variables). The game starts out with 33 points in the cup (3 red (5 points each), 4 yellow (3 points each), and 6 green (1 point each)), and the percentage chance of drawing at least one red die is 23.07% when starting. As you draw dice out of the cup your point value in the cup will change, by subtracting the value drawn each time. For example, if you draw a red, a yellow, and a green die on your first turn the point value of your cub will drop from 33 to 24, by subtracting 9 which was the value associated with the dice you just drew. There are now only 10 dice left in the cup, and the percentage chance that you will draw another red die on your next draw is 20%. Notice that by drawing a red, and making the point value of the cup lower, the likelihood of you drawing another red has decreased.

But this leaves one big gaping whole in our plans to access risk, we have to access that risk in the context of how many dice are left. This can easily be expressed as:

**( R+G+Y)/T = Ri**

**or
**

**(Reds + Greens + Yellows)/Total number of Dice in cup = Risk**

So what does this really mean? It gives a weight to each die left in the cup that is between 1 and 5. The higher this number the more risk associated with taking it. For example, in the picture above there are 5 greens, 2 reds, and 1 yellow die in the cup. This totals to 18 points. Given that there are 8 dice in the cup, we can access the risk of each die to be 2.25. This is saying that each dice is less risky than a yellow, but nearly twice the risk of a green.

18/8 = 2.25

Now I know what you are thinking, “But I’m not going to be able to figure out 2.25 in my head, on the fly”. Well, maybe not, but you can easily determine that 8 goes into 18 twice, but not three times, so you know the risk is between 2 and 3. Less risky than yellow, but twice as risky as a green. In retrospect, 2.25 is actually a slightly better risk assessment than you initially start each round with.

33/13 = 2.53

When you apply this to our previous evaluation of risk on a per roll basis, that means that your first roll of the game is always going to have a risk level of 7.6 (3 times the risk assessment of a single die, or 3 x 2.53). Slightly less then the our “middle of the pack” number of 9, but much higher then our ideal risk of 3.

So, that is a whole lot of math and logic, and possibly way more than the folks who search for “how to get the most brains” wanted to know. However, I think it’s a very interesting look at a very simplistic game. So now, go have fun eating brains, and good luck with not getting shot gunned.

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The Green odds in your first table are incorrect; the symbol distribution is 3B 1S 2F (50 16.7 33.3), not the 4B 1S 1F (66.7 16.7 16.7) that you’ve listed. Is that a simple typo, or does it affect the calculations?

Your math is very very badly off for most of this.

Right off the bat, you have the odds table wrong for colors

ALL dice have 2 feet.

Greens have 3 brains, 1 shot. Reds are 1 brain, 3 shot. Yellows are 2/2/2

That’s not the only problem though. You say that that the odds of rolling one shotgun on 3 reds is 50%. It’s not. It’s 37.5%

You have the same odds, 37.5%, of getting 2 shotguns on 3 reds.

Getting 3 shotguns is 12.5%

Thus the odds of getting at least one shotgun on 3 reds is in fact 87.5%

Also, Red+Green does NOT equal 2 yellows. Green+Yellow+Red does not have the same odds as 3 yellows.

http://anydice.com/program/b8f

This shows odds of getting shotguns (ignoring other results), and as you can see, they are clearly not equal.

Rev. Bob,

Thank you for pointing this out. I didn’t have my copy of Zombie Dice in front of me when I wrote this article, and instead looked up what the dice had on them (which is oddly not printed in many places) the source that I got those from had the greens listed incorrect, and so that is what I used, never thinking that those numbers might be wrong. The good news though? The number of shotguns on the dice were correct, which is the value we use for all future calculations.

I have corrected that first table, so that it now shows the percentages accurately. Thank you for pointing this out!

Palmer,

A red die has 3 shotguns, 2 feet, and 1 brain. So the likely hood of rolling a shotgun is 50/50. When you have three red dice, you have 9 possible sides with shotguns, and 18 possible sides all total. This is still 50/50. However, if you want to figure the likelihood of rolling more then one shotgun at a time you multiply the probability of it happening on one die by the probability of it happening on the second die. This gives you the probability of it happening on two dice: .5 X .5 = .25, thus the likelihood of you rolling two shotguns on three red dice is 25%.

Check out this link for how to figure probability, which was the same means I used for these calculations:

http://easycalculation.com/statistics/learn-multiple-event-probability.php

While I completely agree that I had the feet and brains messed up on the first table for green dice (which I explained above in my comment back to Rev. Bob), the rest is spot on, because it deals with the probability of being shotgunned, which had the proper figures to start.

On a final note, the Red + Green + Yellow = 3 Yellows. These two values are equivalent when we are discussing them in terms of “likelihood you will roll a shotgun”. This is because of the following:

3 Shotguns (red) + 1 Shotgun (green) + 2 Shotguns (yellow) = 6 Shotguns

2 Shotguns (yellow) + 2 Shotguns (yellow) + 2 Shotguns (yellow) = 6 Shotguns

So either way you go you are dealing with a 6/18 chance (or 33.4% chance) that you will roll at least one shotgun.

[…] Blog Carnival, so we thought we’d link to this awesome and thorough statistical analysis on getting more brains in Zombie Dice by One Gamer’s Opinion. A must read for owners of the […]

I love this game. It’s one of my go-to, always take it with me, games. Even the kids like it.

Palmer is correct, your math is pretty much all wrong.

Specifically, the probability of rolling shotguns when rolling 3 RED dice is:

Rolling 0 Shotguns: 1/8

Rolling Exactly 1 Shotgun: 3/8

Rolling Exactly 2 Shotguns: 3/8

Rolling Exactly 3 Shotguns: 1/8

Rolling at least 1 Shotgun: 7/8

Rolling at least 2 Shotguns: 4/8

Those are correct. For similar reasons, basically all of your percentages are wrong.

What you said about multiplying probabilities together to find the probability of getting TWO shotguns is not completely the wrong idea, but it’s a little more complicated than that. .5 x .5 = .25 is the probability of rolling two shotguns when rolling two red dice. For obvious reasons, the probability increases when you roll a third die.

I’d be happy to explain all of the probability to you if you’d like, because this is a really interesting article at the core.

I’m probably going to get laughed off the page for saying this, but I was playing this game with my wife last night and rolled 13 brains. In a row. I had a zero score after rolling variations of brains, then 3 shotguns for a few turns (shouldn’t have pushed my luck), but on turn 3 it all came together and I rolled 3 brains, then 3 more, then 3 again, then 3 AGAIN, then put them all back in, took three dice out of the jar and rolled them one at a time. The first one was a brain, for a total of 13.

Needless to say, I stopped.

Jason

Wow, that is awesome. I actually had someone offer to buy my dinner at a con one time if I could do that, but alas, it didn’t happen. Still, kudos on an awesome feat!

This thread is pretty old, but I wanted to add something to the discussion.

I’m currently working on a dice game myself and writing scripts to work out all sorts of math on dice combinations.

I love Zombie Dice, and the push-your-luck dice odds that come with it. I think what the OP and others might be missing is the distinction between “exactly” X shotguns, and “at least” X shotguns, because there is a huge difference.

I’m not a mathematician (didn’t even graduate college!), but your odds of rolling *at least* one Blast do not get lower as you add more dice – they get higher. The odds for rolling *exactly* one Blast Here’s an example of how you can look at the Red zombie dice. 50/50 chance a single dice will roll a Blast. That means, in ANY given roll with a red die in it, you will always have that 50/50 chance of rolling the blast from that red die. Keep in mind we are talking about a single throw.

In terms of Blasts, that Red die is like flipping a coin. If I flipped two coins at once, (using a Yes/No output) the possible permutations are Y+Y, Y+N, N+Y, and N+N. You’ll notice out of the 4 permutations, 3 have a Yes in their output. 3/4 chance you will roll *at least* one Yes … The math from wizardofodds.com is “1 – (1/X)^N” where X is the number of sides, and N is the number of times rolled. That leaves “1 – (1/2)^2 = 1 – 1/4 = 3/4”. This basically calculates the number of NOT-Blasts and removes that from 1. There is 1/4 chance that both coins will ‘roll’ Yes’s – that’s exponential, as we know from many two-dice games … What are our chances of rolling a 2 with 2d6? (1/6)^2 = 1/36th chance. So, Chances of rolling exactly two Y with 2d2 – (1/2)^2 = 1/4. But, what are our chances of rolling exactly one Yes? “N*(1/X)^Y*(1/X)^Y” … Woah, what? You’re saying “Multiply the number of dice rolled by the chance that exactly Y roll what you don’t want, and multiplied by the exact ones that you DO want”. That makes “2*(1/2)^1*(1/2)^1 = 2 * 1/2 * 1/2 = 1/2” … this gets more clear with more dice.

For 3 red dice, permutations are Y+Y+Y, Y+Y+N, Y+N+Y, Y+N+N, N+Y+Y, N+Y+N, N+N+Y, N+N+N. What Teddy meant about Eigths above, is there are 8 permutations of “Blast” or “No Blast” in the Red dice specifically, as I just wrote out. If you have 3 red dice, the chances you will roll *at least* one Blast is 7/8, represented by “1 – (1/2)^3 = 1 – 1/8 = 7/8”. The chances you will roll *exactly* one Blast is 3/8 … “3 * (1/2)^2 * (1/2) = 3 * 1/4 * 1/2 = 3/8”. This works in reverse for rolling brains on the Green dice. 87.5% chance you’ll roll at least one Brain. 37.5% chance you’ll roll exactly one brain.

In the case of all Yellow (2S/2F/2B) or all Green (1S/2F/3B) the odds change, because you’re not a coin-flip anymore, but the math does not.. But, since in a single roll, Brains and Feet are “Not Blasts”. Green means 1/6 Blasts, and Yellow means 1/3. What are the chances of rolling *Exactly* 3 of the same thing on 3 Yellow? “3*(1/3)^3″ = 3/27 = 1/9”. Exactly 1 Blast? “3 * (2/3)^2 * (1/3) = 12/27 = 44.4%” … Chances of rolling at least 1 Blast? “1 – (2/3)^3 = 19/27 = 70.3%”

Now, is that the same as rolling 1 Red, 1 Yellow, and 1 Green? At least 1 Blast: “1 – (1/2)*(2/3)*(5/6) = 26/36 = 72.2%” … Nope, isn’t the same as 3xY.

The calculation to see what the odds are for rolling exactly 1 Blast on 3 different-odd dice is probably beyond my comprehension, as it probably has something to do with adding together all three different computations for each different dice rolling a Blast and the other two not. Here’s a stab…

Chance that only Green will roll Blast: (2/3)*(1/2) * (1/6) = 2/36 = 5.6%

Chance that only Yellow will roll Blast: (5/6)*(1/2) * (1/3) = 5/36 = 13.9%

Chance that only Red will roll Blast: (5/6)*(2/3) * (1/2) = 10/36 = 27.8%

Not sure, but add them all together to get the overall chance that only 1 Blast will roll? 47.3%?

Funny that multiple real errors were pointed out back in 2011 and still aren’t fixed even though the author commented in 2012. Sorry Victoria, most of your numbers or statistics descriptions are still wrong.