Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I just got a new board game called Betrayal At The House On The Hill. In the game you can trigger a 'haunt' determined by looking up on a table in which Room an Omen Card is found (after some dice rolling). Let us assume that in each game the haunt chosen is truly random (I doubt this is the case but it seems like a good enough assumption). I think this means I can generalise the chance of starting two identical hunts as a Birthday Problem.

Am I right in thinking that the number of Haunts is, on the wiki page linked above analogous to the days $d$? And games I need to play before I get a match $n(d)$? For reference there are 50 possible haunts, so $d=50$, I think?

Am I correct in assuming that the formula I need to apply is then:

$n(d)=\left\lceil \sqrt{2d\ln2}+\frac{3-2\ln2}{6}+\frac{9-4(\ln2)^2}{72\sqrt{2d\ln2}} \right\rceil$

Or is this 'overkill' (there aren't $10^{18}$ haunts unfortunately)? I Picked this one as the others say things I'm unsure of, i.e.73% of all integers d or a set of integers d with asymptotic density 1. and I don't know what this means in terms of my board game example.

How would I find the probability of drawing two identical haunts within $n$ games of each other? I'm not sure if this is covered in the article above, though I think the section on Near Matches covers haunts that or $k$ numbers away from each other.

Apologies for rambling a bit and bringing this problem to bear on a board game, not a very mathematical or important example. But I was curious, wanted to learn more and I knew maths would work.

Am I right in my use of the maths here? How is it (formula above) changed if I want to look at different probabilities (i.e. not 50% match rate but some arbitrary $p(matching~haunt)$? What about number of games needed to draw the same haunt on two consecutive games?

How should I apply The Birthday Problem to Betrayal At The House On The Hill?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Yes, if you know that $d<10^{18}$, then the formula you give will give the correct answer. But probably even the first formula $\lceil \sqrt{ 2 d \ln 2}\rceil$ will work (what is your exact value of $d$ anyway?). If in doubt, add a few to the result (if you want to make a bet on the original birthday problem at a party, I suggest you do this at parties with at least 25 guests, say, and not just 23). Note that the $n$ obtained is the minimal $n$ such that among $n$ independent choices among $d$ equally probable haunts, it is more likely than not that at least one duplicate occurs. It is as far from assuring you that duplicates occur as you cannot be sure in advance that a coin thrown shows tails.

share|improve this answer
    
Could you expand on that a little? Did I get my n and d the right way round? By independent choices, that would be choices of haunts correct? So that would be the same as games, and there is one haunt per game. I'll edit in more details (value of d). –  Pureferret Sep 2 '12 at 12:27
    
Yes, your interpretation of what $d$ and $n$ refer to is correct. –  Hagen von Eitzen Sep 2 '12 at 12:32
    
After you edited $d=50$ into your question, there is no need to compute $n(d)$ using a complicated formula at all. The table at wikipedia already lists $n(d)=9$ for $43\le d\le 54$. –  Hagen von Eitzen Sep 2 '12 at 12:34
    
I saw the table, and I thought it was accurate, but I'm also interested in a few things the table can't give me, as well as the maths behind it. What if I alter the match probability from 50% to some other p? What if I want to know about consecutive haunts? –  Pureferret Sep 2 '12 at 12:38
1  
With $d=50$ I suggest to simply compute $1-(1-\frac1d)(1-\frac2d)\cdots(1-\frac {n-1}d)$ for growing values of $n$ and see when the value crosses your required match probability $p$; for example, you cross 90% at $n=15$ and 95% at $n=17$. For any game, the probability that the haunt is the same as in the game before is just $\frac1d$. –  Hagen von Eitzen Sep 2 '12 at 12:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.