Math puzzle: coin tossing carnival game

Over the holiday weekend, I went to a county fair and noticed a lot of carnival games.

It got me thinking about a probability question arising from one of the games. Here is the puzzle:

In a carnival game, you are to toss a coin on a table top marked with a grid of squares. You win if the coin lands without touching any lines–that is, the coin lands entirely inside one of the squares.

If the squares measure 1.5 inches per side, and the coin has a diameter of 1 inch, what is the chance you will win?

Can you solve it?

Bonus: find a formula if the squares measure S inches per side and the coin measures D inches in diameter.

As usual, leave a comment on how long it took you to solve the puzzle. I will have posted a solution in the comments on Thursday.



Share this post:

| More

Previous post:

Next post:



  • http://stochasticseeker.wordpress.com/ Stochastic Seeker

    I wish I could draw a picture, but basically what I did was considered the center of the coin as a single point and derived the set of points (that are the center of the coin) such that coin rests inside the square.

    Conveniently, these points form a square and dividing the area of the square formed by the points and the unit squares of the grid I got:

    A 1/18 chance.

    I’ll try out the general solution in a bit. It took me roughly about 3-4 mins.

  • http://stochasticseeker.wordpress.com/ Stochastic Seeker

    Okay, just plugging in unknown variables into my process yielded a ((S-D)/S)^2 chance of winning (assuming that D S then the chance of winning is obviously 0.

  • Ben Sauer

    @Stochastic, if the general solution is ((S-D)/S)^2 (which is what it seems it should be) then the first answer needs to be 1/9th. (Which is what I think it needs to be).

    Took me about 2 minutes and the approach I had was to take it in one dimension. The okay part is S-D and the total part is S. And then I squared it to get to two dimensions.

    But perhaps I’m missing something?

  • Matt

    I got 37.78%. It took me 15 min.

  • Frank

    ~30 seconds, graphed the set of points Seeker describes, but ended up with…spoiler alert from here on…1/9. Another couple minutes…in general, the coin can’t land near the borders, which have total area 2SD-D^2 (edges minus corners), so the formula should be 1-(2SD-D^2)/S^2. Seeker’s solution, finding the sides of the inner square (after removing the edges) is more straightforward.

  • Benjamin

    It took me about four minutes to come up with 1/9. The circle’s only a circle for illustration. Without loss of generality, the left-most point of the circle must fall within 0 and .5 inches of the left hand side of a square. There’s a 1/3 chance of that. The top-most point must fall within 0 and .5 inches of the top side. There’s a 1/3 chance of that.
    Generally, I suggest the probability is (1-D/S)^2.

  • James

    I thought about it while walking the dog around the block and got the same answer, 1/9, as Ben and the same general solution, (S-D)^2/S^2, as Seeker.

    Seeker had wished to draw a picture: I thought about it as walking a dog on a long leash (length d/2) in a square (side length S) dog park; and finding all points where I can stand and she can’t cross the boundary.

    This puzzle reminded me of Buffon’s needle tossed onto parallel lines http://en.wikipedia.org/wiki/Buffon%27s_needle
    and now I’m thinking about tossing a Buffon needle onto a grid, or maybe tossing a square coin onto the same grid.

  • http://www.lomont.org Chris Lomont

    I also get ((S-D)/S)^2 when Dinfinity and the limit converges to the above.

    Or more sloppily look at the inner square in each given square has side length S-D where the center, if landing there, misses touching the edge. The ratio of inner squares to big squares is also ((S-D)/S)^2.

    Took about 5 minutes to solve, 5 minutes to write up.

  • http://www.vpsgraphics.com V Paul Smith Jr

    I got the same thing in about 30 seconds
    ((S-D)/S)^2
    But by the time I could respond, it’d already been posted several times.
    So, to contribute something new:
    ((S-D)/(S+L))^2
    …with L being the line thickness.

  • skal

    1 minute finding (1-D/S)^2, 3 minutes convincing me it was right and i didn’t miss anything obvious, and trying not to think of the Buffon’s needle analogy.

    now, my contrib: it’s (1 – D/S)^n in dimension n :)

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Thanks all, the correct answer is 1/9. The idea is the circle’s center has to lie in a specific area for the circle not to intersect the grid. Then the probability is calculated as the ratio of the area that the center can lie versus the total area of the square.

    Here is a diagram that can help.

    The small blue square has a side length of S – 2 *(D/2) = S – D.

    The general formula I got is [(S-D)/S]^2.

  • http://billmill.org Bill Mill

    I got 1/9 in about 15 minutes by considering only the top-left point of the bounding box of the coin. If it’s in the range 0 < x < .5 and 0 < y < .5, the coin is fully in the square in which the top-left point resides; else it's in at least two squares. .5/(3/2) = 1/3, so square it to get 1/9.

Previous post:

Next post:

This site is for recreational and educational purposes only (privacy policy)

Powered by WordPress