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 Picking cards (Posted on 2006-08-17)
A player picks random cards from an ordinary card deck, without returning them to the deck. How many cards should he pick so as to get at least one ace with 70% probability?

 No Solution Yet Submitted by atheron Rating: 3.0000 (1 votes)

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 solution -- spoiler | Comment 1 of 7

The puzzle basically asks for how many need be drawn in order to reduce the possiblity of not getting an ace to under 30%.

The first draw has 48/52 probability of not resulting in an ace. If the first did not result in ace, the second draw has 47/51 probability of not resulting in an ace; this has to be multiplied by the first fraction to get the probability of neither having an ace.  Subsequent fractions by which this probability must be multiplied continue to have the numerator and denominator each reduced by one for each draw, so for example, the probability of the first 5 not containing an ace is 48*47*46*45*44/ (52*51*50*49*48).

A table of probabilities of not having received at least one ace, in n draws is:

` 1 0.92307692 2 0.85067873 3 0.78262443 4 0.71873673 5 0.65884200 6 0.60277034 7 0.55035553 8 0.50143504 9 0.4558500310 0.4134453811 0.3740696312 0.3375750313 0.3038175314 0.2726567615 0.2439560416 0.2175824217 0.1934065918 0.1713029819 0.1511496920 0.1328285221 0.1162249522 0.1012281823 0.0877310924 0.0756302525 0.0648259326 0.0552220927 0.0467263828 0.0392501629 0.0327084730 0.0270200431 0.0221073032 0.0178963933 0.0143171134 0.0113029835 0.0087912136 0.0067226937 0.0050420238 0.0036974839 0.0026410640 0.0018284241 0.0012189542 0.0007756943 0.0004654244 0.0002585645 0.0001292846 0.0000554147 0.0000184748 0.00000369`

The first one that shows less than 30% probability of not having drawn an ace is n=14 draws.

 Posted by Charlie on 2006-08-17 11:40:25

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