## Monday, March 15, 2010

### Avoid the Flooding Today

Player A has one more coin than player B.  Both players throw all of their coins simultaneously and observe the number that come up heads.  Assuming all the coins are fair, what is the probability that A obtains more heads than B?

1. 50%

If you go case by case, you can see the pattern.

If B has 0 coins and A has 1:
A has a 1/2 chance of getting a head, and therefore having more heads than B.

If B has 1 and A has 2:

A has a 1/4 chance of having 2 heads (which will always beat B), and 1/2 chance of having 1 head (which will beat B if he throws tails).
B has a 1/2 chance of throwing tails.
So putting that together, the chances of A having more heads than B:

1/4 + (1/2 * 1/2) = 1/2

If B has 2 coins and A has 3:

1/8 chance A throws 3 heads
3/8 chance A throws 2 heads
3/8 chance A throws 1 head

1/4 chance B throws no heads
1/2 chance B throws 1 head

So the chances of A throwing more heads than B =

1/8 + (3/8 * 3/4) + (3/8 * 1/4)

4/32 + 9/32 + 3/32 = 1/2

2. I'm not surprised to see you got it, Andy. The chances are 50/50, although the answer is not intuitive at all.

3. Gee, probability is never intuitive to me but that was one of the few problems that was. The other coins don't enter into the solution, only PlayerA's extra coin, which has 50% chance of landing heads.

4. Thank you for the interesting puzzle, which generated quite a bit of thought in my maths department, as well as three solutions of various levels of complexity.

I don't think I can include pictures on the answers here, so see

http://joningram.org/blog/2010/03/one-more-coin-three-solutions/

for the low level one, but the high-level argument is very nice:

Let p be the probability that A throws more heads than B.

1 - p = P(A throws the same or fewer heads than B) = P(A throws more tails than B) = P(A throws more heads than B) = p.

So 2p = 1, and the probability of A throwing more heads than B is exactly 1/2.

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