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Stage 2 of the puzzle is a probability question. Much of what we do in protecting or encrypting data is risk based, with the aim of minimising the probability of an adverse event, so….
Imagine that you have a standard pack of cards. That's a total of 52 cards in four suits with denominations from Ace to King.
If you shuffle the pack and deal out the top 13 cards face up you very obviously would note that each card is different, you've only got one pack.
Now imagine that you have not one but thirteen packs of cards. Shuffle each deck and deal one card from each one.
In bridge, an illegal hand would occur if 2 identical cards form part of the 13 card hand. When taking one card from each of the 13 decks, what is the probability of dealing an illegal hand?
Compute the probability of an illegal hand as a percentage and round down to the nearest integer (0 decimal places). Ditch the percent sign and divide the magic number (from Stage 1) by the result.
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