# A Puzzle based on probability

(Based on true story) A friend of mine and myself were drinking and we wanted to decide who will pay for the next round of drink. We decided to toss a coin so as to ensure a fair chance (1/2 prob.) for each of us. However since none of us had a coin with us, we decided on the following game that seemed to imitate the same probability numbers.

"We have a box containing a number of matchsticks (n) (not know to either of us beforehand). I ask a third friend to grab a bunch of matchsticks(k) and throw away. So now we are left with some matchsticks in the box (n-k).Now one of us calls whether the number left is even or odd.If I call even and the number is even, I win or else I lose."

My questions

1. Is my probability of winning 50% ?
2. If my opponent knows beforehand whether the number n is even or odd, does it impact my chances in the game?
3. If my opponent knows beforehand whether the number k is even or odd, does it impact my chances in the game?

I have a feeling 1 is true. However I think 2 and 3 may not always go in my favour. (In the real game, we borrowed the matchbox from an unknown patron and asked someone else to throw the matchsticks away)

• I believe if $n$ is even, then there's a slightly higher change that removing $k$ matchsticks will be even if you include the possibility that $k=n$ and $k=0$, where slightly higher should be $1/(n+1)$. – Ian Coley Jan 30 '14 at 21:53

Humans are bad random generators in general. The problem is that a human picking $k$ out of $n$ follows a biased distribution. Even if we asssume that the number $k$ is picked really randomly, it is a good guess that i twill not happen according to some uniform distribution but something like a binomial distribution. This introduces a bias - though a bias that is hardly predictable for a one-time experiment (note that due to individual preferences your $k$ may be distributed with a mean of $\frac n2$ and your opponent's $k$ may be distributed with a mean of $\frac n3$; likewise due to different exposure to alcoholic beverages the standard deviations may vary. Nevertheless, if we knew these two distributions, we'd quite surely obtain probabiliteis slightly off the intended $50:50$ chance.
If both $n$ and $k$ are at least moderately large, the chances will be very close to $50\%$ At the small end I would predict a bias for odd numbers, because if there is one match to start there will probably be none thrown away and if there are two there will probably be one thrown away. Once $n \gt 10$ I would expect the end effects to be negligible. For 2 and 3, I would say neither helps your opponent as long as $n$ is large enough. This is obviously a real life answer, not a mathematical one.