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?
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 0.45585003
10 0.41344538
11 0.37406963
12 0.33757503
13 0.30381753
14 0.27265676
15 0.24395604
16 0.21758242
17 0.19340659
18 0.17130298
19 0.15114969
20 0.13282852
21 0.11622495
22 0.10122818
23 0.08773109
24 0.07563025
25 0.06482593
26 0.05522209
27 0.04672638
28 0.03925016
29 0.03270847
30 0.02702004
31 0.02210730
32 0.01789639
33 0.01431711
34 0.01130298
35 0.00879121
36 0.00672269
37 0.00504202
38 0.00369748
39 0.00264106
40 0.00182842
41 0.00121895
42 0.00077569
43 0.00046542
44 0.00025856
45 0.00012928
46 0.00005541
47 0.00001847
48 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 20060817 11:40:25 