Answer.
before you calculate an exact answer it's good to make a rough estimate first - this gives an indicator that things haven't gone astray during the number-crunching
as a 13 card hand will have on average one ace in it hands with no aces in them have to be more common than hands with two or more aces in them
you can expect to see one ace, no ace, two aces, three aces ( - seldom), or four aces (-rare), in that order of probability - this infers that you will see a lot more hands with one ace than with two aces in them - maybe twice as many? - if this is true then when lolly ignores the hands with no aces there might be only one in three hands containing two or more aces - after the arithmetic the exact answer turns out to be 37% - a very bad bet to take at even money!
in the second case, the hand consists of the ace of spades plus 12 of the other remaing cards - there are 3 aces in these 51 cards so any card dealt has a 3/51 chance of being an ace - with 12 cards the average number of aces will be
12 x 3/51 = 2/3
obviously there will actually be zero, one, two, or three aces in the 12, with no or one ace being the most common - to make up the average of 2/3 of an ace then one ace will have to be more likely than none - the chance of there being at least one more ace in the 13 cards therefore seems to be between 1/2 and 2/3
in fact, it turns out to be 56% - so you should take the offered evens bet and slowly remove all of lolly's money
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