What are the odds?
December 9, 2009 3:59 PM   Subscribe

I'm rolling 5 dice. I win if I roll two 1's, one 2, or one 3. What are my odds of winning, expressed as a percentage?

This question is actually about the board game Battlelore, but I've simplified it above for ease of discussion.

In the game, you roll dice to score hits. Instead of numbers on the dice, there are three banner color icons (red, green, and blue), a "lore" icon, a retreat flag icon, and a bonus strike icon.

In combat, if you roll the banner color that matches the color of the target, you score a hit. In most circumstances, you also get a hit if you roll the bonus strike icon, and sometimes if you roll the lore icon or the retreat flag icon, just depending on what cards are in play and where the pieces are on the board.

I'd put together a spreadsheet for determining the odds of almost any given die roll in Battlelore. So, for example, if you want to know the odds of scoring two hits if you're rolling four dice which hit on bonus strikes and blue banners, you'd look at cell D31 and see that you have 40.74% chance of that happening.

But there's this one case from a recent game that's giving me trouble: I need to score one hit, I'm rolling 5 dice, and I score hits on the blue banner, the retreat flag, and on bonus strikes. BUT: The target ignores the first bonus strike.

No problem, I thought: That would be the odds of scoring one hit on on two die faces with five dice (cell D8, 86.83%) plus the odds of scoring two hits on one die face with five dice (cell C32, or 19.62%).

Unfortunately, those results add up to more than 100%, so I know that's wrong. The odds are high, probably over 90%, but there's still the chance that I'll roll all green and red banners or something.

How do I handle this oddly specific situation?
posted by JDHarper to Grab Bag (3 answers total)
 
Can you think of it in terms of subtraction? I think your odds are

1 - (chance of rolling 1 yellow, without retreats or blues) - (no retreats, no blues, no yellows).

Anything else should count as a hit.
posted by zamboni at 4:15 PM on December 9, 2009


You have a 5 Dice, and they are all identical. We call their sides:

R G B L F S

Which stand for (R)ed, (G)reen, (B)lue, (L)ore, Retreat (F)lag, and Bonus (S)trike.

You want to calculate the probability of scoring more than 1 of (B, F), or more than 2 of (S). There are a couple ways to do this nicely, I'll let somebody else explain the Bayesian probability technique, I'll give you the way to do it using the basic rules of probability you might learn in an introductory class.

The first thing to realize is that the 5 Dice rolls are completely independent. So if we just consider all the ways to roll the 5 Dice with 6 Faces, we have 6^5 = 7,776 possible outcomes. One way to calculate your probability of success is to count how many of the outcomes will succeed. Alternately, we are in a situation where we suspect there to be far fewer outcomes where we will fail, so perhaps we should try counting those.

We fail for every die roll that doesn't contain a B or F, and if it contains at most one S.

We separate the two cases, one S, and no S. Lets deal with the no S case, because it's easier.

5 Dice that can roll R, G, L have 3 independent choices each, so we have 3^5 = 243 outcomes with no S.

If there is one S, the simplest way to solve this is to count the ways this can be done assuming the first die is the S, then multiplying by 5 (because our search space differentiates between the different dice being S). The odds for the remaining 4 dice are 3^4 = 81, multiplied by 5 = 405.

Our solution comes from subtracting the two 'fails' from our search space and dividing by the search space to get a percentage: (7,776 - 243 - 405)/7,776, or about 91.7%.

On preview, zamboni is correct (and more succint).
posted by onalark at 4:22 PM on December 9, 2009


Awesome. Thanks for the help!
posted by JDHarper at 4:43 PM on December 9, 2009


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