Siddharth and Alok have 10 coins and 9 coins respectively. They are playing a game in which they throw up all their coins simultaneously and observe the number of coins which land heads up. The person with more heads wins. If there are no trick coins, what is the probability that Siddharth wins?
Additionally, calculate the probabity that Siddharth loses?

It seems that the answer to part 1 will be 1/2 whenever Siddharth has one more coin than Alok, and (unlike part 2) can be found without binomial theory.

Let Alok and Siddharth throw n and n + 1 coins respectively, and remove any one of Siddharth’s coins from our initial consideration - let’s say the last one to land.
Let p be the probability that both groups of n coins contain the same number of heads. Then, by symmetry, the probability that one of them is winning (or losing) on the basis of the two groups of n coins is (1 - p)/2.

Siddharth’s remaining coin will determine the final result. If he is winning after n coins, he will go on to win. If he is drawing after n coins, a further head will give him a final win.

P(Siddharth wins) = P(winning after n coins) + P(drawing after n coins) * 1/2

                        = (1 - p)/2  +  p/2

                        =  1/2

Posted by Harry on 2010-11-09 20:40:44