Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail
From: "Carl G." <carlgnews@microprizes.com>
Newsgroups: rec.puzzles
Subject: Re: Roll a Penny Game
Date: Wed, 4 Jun 2025 11:52:08 -0700
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On 6/4/2025 1:53 AM, David Entwistle wrote:
> You have been given the job of designing the "roll a penny" game for your
> school's summer fair.
> 
> Background: The player is given a small 'shoot' down which they roll a
> coin on to a large board marked with a square grid of lines. The coin
> rolls across the board and when it stops, and falls flat, it will lie
> either, entirely within a square - not touching a line, or only partly
> within a square and across a line. If it is entirely within a square the
> player wins. If it is across a line the player loses.
> 
> a) Your first effort should provide the player with an equal probability
> of winning and losing. If the coin has a radius of one unit, what is the
> side length of the square that gives an equal win / lose probability? You
> can ignore any element of player skill and any complications associated
> with line thickness etc.
> 
> b) The school vicar, who knows more about such things than he likes to
> admit, points out that the game would be more interesting if the prize
> could be made larger. The headmaster, responsible for school finances,
> wants the game to at least break even. You are asked to modify your design
> such that the player only has a 1 in 10 chance of winning on each roll. If
> the coin has a radius of one unit, what is the side length of the square
> that gives an 1 in 10 win, 9 in 10 lose, probability? You can still ignore
> all complications.
> 
> I remember playing the game at primary school in the 1960s and working out
> the probability at secondary school, when introduced to the subject, some
> fifty odd years ago.
> 
Spoiler?
Spoiler?
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When the game is won, the center of the penny will fall on a square 
region within each grid square. If the side of the grid square is a, the 
side of the inner square is a-2, since a penny has a radius of 1.  The 
ratio of this inner square's area to the grid square is the probability 
of winning ("P"), (a-2)^2 = P a^2.  This results in the quadratic equation:
(1-P)a^2 - 4a + 4 = 0
Solving for a,
a = (2 + 2 sqrt(P))/(1-P)

For a probability of 0.5 (win half the time), a = 6.83 (approx.)

For a probability of 0.1 (win 1/10 of the time), a = 2.92 (approx.)

-- 
Carl G.


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