Akio has one more coin than Bansi. They throw all of their coins and count the number of heads. If all the coins are fair, what is the probability that Akio obtains more heads than Bansi?

Sunday, April 15, 2012

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### 3 comments

### Credit

This puzzle is taken from:

- Hobson, Nick. "Solution to puzzle 18: One extra coin." Nick's Mathematical Miscellany. 9 June 2004. 21 Apr. 2012 <http://www.qbyte.org/puzzles/p018s.html>.

## John Grint solved this puzzle:

The probability is \(\frac{1}{2}\).

Let \(H\) be the event that Akio obtains more heads than Bansi, and let \(T\) be the event that Akio obtains more tails than Bansi.

The four possible combinations of outcomes are \((H \text{ and } T)\), \((H \text{ and } \lnot T)\), \((\lnot H \text{ and } T)\) and \((\lnot H \text{ and } \lnot T)\). But at least one of the events \(H\) or \(T\) must occur since Akio has more coins than Bansi, hence \(P(\lnot H \text{ and } \lnot T) = 0\). We cannot have \((H \text{ and } T)\) since this would require Akio to have at least 2 more coins than Bansi, hence \(P(H \text{ and } T) = 0\). Therefore the only two possible outcomes are \((H \text{ and } \lnot T)\) or \((\lnot H \text{ and } T)\), i.e. the events \(H\) and \(T\) are mutually exclusive and precisely one of them must occur.

Since all the coins are fair, the probabilities of the two events \(H\) and \(T\) are equal, by symmetry. Therefore \(P(H) = P(T)\), and we have \(P(H) + P(T) = 1\).

Therefore \(P(H) = P(T) = \frac{1}{2}\), i.e. the probability that Akio obtains more heads than Bansi is \(\frac{1}{2}\).

## Ed Murphy solved this puzzle:

Label Akio's coins \(A_1, A_2, \dots, A_n, A_{n + 1}\). Label Bansi's coins \(B_1, B_2, \dots, B_n\).

For all the coins except \(A_{n + 1}\), there are \(2^{2n}\) ways for them to land, all equally probable, and these can be divided into three sets:

Each way in (1) is the mirror image of a way in (3), so those sets have the same size.

For each way in (1), Akio has more heads than Bansi overall, regardless of how coin \(A_{n+1}\) lands.

For each way in (2), Akio has a \(\frac{1}{2}\) probability of having more heads than Bansi overall, depending on how coin \(A_{n+1}\) lands.

For each way in (3), Akio does not have more heads than Bansi overall, regardless of coin \(A_{n+1}\) lands. At most, he might have the same number of heads as Bansi overall.

Let \(x = \frac{m}{2^{2n}}\) where \(m\) is the number of ways in set (1). Then the overall probability of Akio obtaining more heads than Bansi is \(1 \cdot x + \frac{1 - 2x}{2} + 0 \cdot x = \frac{1}{2}\).

## Christian Bau solved this puzzle:

Akio throws \(n + 1\) coins, Bansi throws \(n\) coins. We put Akio's last coin aside, comparing Akio's \(n\) coins with Bansi's \(n\) coins. If the probability that Akio has more heads is \(p\), then the probability that Bansi has more heads is also \(p\). The probability that both have the same number of heads is \(1 - 2p\).

Looking at Akio's \(n + 1\) coins and Bansi's \(n\) coins, Akio has more heads if he either had more heads with the first \(n\) coins, or if he had the same number of heads as Bansi's with the first \(n\) coins and the \((n + 1)\)th coin is heads. The probability for this is \(p + \frac{1 - 2p}{2} = p + (0.5 - p) = 0.5\).