In an event organized by my friend DJ Shon at Mani Square during the Durga pujas there was a particular game event that required contestants to choose the Joker from a pack of cards. The game was announced and a group of 5 college goers surrounded the game table. There were 5 cards laid down on the table and being the generous person Shon is, he guaranteed that one of the participants would surely pick a Joker. However, he is stringent enough to allow at most 1 Joker in the pack of 5 cards laid down at the table. The cards are, of course, laid face down on an opaque table with nobody except the dealer knowing which card is the Joker. Of course, I assume the dealer is not biased!
So, one by one, the participants pick a card, and the lucky participant wins! Not quite very exciting since one of the participants always wins. Initially I felt that going first would increase chances of winning the prize but then felt it was not possible. Breaking this down mathematically, let us take a look.
For a pack of N cards the probability of the first player picking the only Joker becomes 1 / N. For the next player to win we need the first player to lose also, so his probability of winning becomes:
( 1 - 1/N ) * { 1 / (N-1) } = (N-1)/N * 1/(N-1) = 1 / N.
Pretty startling! The second participant has exactly the same chances to win as the first! This continues as a chain and we find all N participants have the same chances of winning and ordering does not matter.
So, this game is actually a fair game.
So, one by one, the participants pick a card, and the lucky participant wins! Not quite very exciting since one of the participants always wins. Initially I felt that going first would increase chances of winning the prize but then felt it was not possible. Breaking this down mathematically, let us take a look.
For a pack of N cards the probability of the first player picking the only Joker becomes 1 / N. For the next player to win we need the first player to lose also, so his probability of winning becomes:
( 1 - 1/N ) * { 1 / (N-1) } = (N-1)/N * 1/(N-1) = 1 / N.
Pretty startling! The second participant has exactly the same chances to win as the first! This continues as a chain and we find all N participants have the same chances of winning and ordering does not matter.
So, this game is actually a fair game.
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