High school probability query
Albert
A hand of 5 cards is dealt from a well shuffled pack of 52 playing
cards. Find the probability that the hand will contain the ace of
spades, given that there is at least one ace.
The probability of the hand containing at least one ace is 1 minus the
probability of the hand containing no aces:
1 - (48 * 47 * 46 * 45 * 44) / (52 * 51 * 50 * 49 * 48) = 18 472 / 54 145.
This is so far correct, for the teacher revealed the value above after
everyone had attempted the above question.
The probability of the hand containing the ace of spades _and_
containing at least one ace is the same as the probability of the hand
containing the ace of spades, is it not? I calculate this probability as
1 - (51 * 50 * 49 * 48 * 47) / (52 * 51 * 50 * 49 * 48).
The answer to my question should therefore be
(1 - (51 * 50 * 49 * 48 * 47) / (52 * 51 * 50 * 49 * 48)) / (18 472 / 54
145) = 20 825 / 73 888, shouldn't it?
Thanks In Advance (TIA),
Albert
Achava Nakhash, the Loving Snake
The number of hands, the first card of which is the ace of spades, is
51*50*49*48. The number of hands withthe second, or third, or fourth,
or fifth card beomg the ace of spades is obviusly identical, so the
number of hands containing the ace of spades must be 5*51*50*49*48.
Thus the probability that a hand contains the ace of spades is
5*51*50*49*48 / 52*51*50*49*48 = 5/52. You are correct that the
intersection of the set of hands containing the ace of spades and the
set of hands containing at least one spade is the same is the set of
hands containing the ace of spades and so the probabilites are
equal.so the result you are looking for is (5/52)/(18472/54145), where
I have trusted your arithmetic (your reasoning is certainly fine) for
the denominator fraction. I believe your assessment of the
probability that a hand contains the ace of spades is quite wrong.
HTH,
Achava
William Elliot
> Am I correct?
>
> A hand of 5 cards is dealt from a well shuffled pack of 52 playing cards.
> Find the probability that the hand will contain the ace of spades, given that
> there is at least one ace.
>
> The probability of the hand containing at least one ace is 1 minus the
> probability of the hand containing no aces:
>
> 1 - (48 * 47 * 46 * 45 * 44) / (52 * 51 * 50 * 49 * 48) = 18 472 / 54 145.
>
> This is so far correct, for the teacher revealed the value above after
> everyone had attempted the above question.
>
> The probability of the hand containing the ace of spades _and_ containing at
> least one ace is the same as the probability of the hand containing the ace
> of spades, is it not? I calculate this probability as
>
> 1 - (51 * 50 * 49 * 48 * 47) / (52 * 51 * 50 * 49 * 48).
>
The probably of a hand not containing the Ace of Spades is
51_C_5 / 52_C_5 = 51! * 5! * 47! / 52! * 5! * 46! = 47/52.
The probablity of a hand containing the Ace of Spaces is
1 - 47/52 = 5/52 = .0962
The probablity of a hand not containing an ace is
48_C_5 / 52_C_5 = 48! * 5! * 47! / 52! * 5! * 43!
= 47 * 46 * 45 * 44 / 52 * 51 * 50 * 49
= 47 * 46 * 9 * 44 / 52 * 51 * 10 * 49
= 47 * 46 * 9 * 11 / 13 * 51 * 10 * 49
= 47 * 23 * 9 * 11 / 13 * 51 * 5 * 49
= 47 * 23 * 3 * 11 / 13 * 17 * 5 * 49
= 35673 / 54145
The probablity of a hand containing an ace is
1 - 35673 / 54145 = 18472 / 54145
p(Ace of Spaces) = p(Ace of spades | an ace) * p(an ace)
p(Ace of spades | an ace) = 5/52 / 18472/54145 = .0328
> The answer to my question should therefore be
> (1 - (51 * 50 * 49 * 48 * 47) / (52 * 51 * 50 * 49 * 48)) / (18 472 / 54 145)
> = 20 825 / 73 888, shouldn't it?
>
Does 18 472 mean 18,472? Then we agree.
Are you sure this is high school stuff?
The Qurqirish Dragon
Considering that basic probability is often introduced in Algebra I
(high school algebra, not college algebra), I wouldn't be surprised if
this was a high-school problem, or even an elementary / middle school
problem for an advanced class.
Ray Vickson
He computed the probability that the hand does not contain the ace of
spades, then subtracted that from 1. His computation is OK: there are
51 cards that are not the ace of spades, then after choosing the first
one, there are 50 left that are not the ace of spades, etc.
R.G. Vickson
>
> HTH,
> Achava
Achava Nakhash, the Loving Snake
Right you (and he) are. His method is not even more complicated than
mine except the part where he doesn't see the huge amount of
cancellation that comes up front. Mea culpa. That being said,
finding techniques to simplify results, the object being not to have a
pretty answer but to have an answer that gives more insight into what
is happening, is a useful skill to develop.
Regards,
Achava
Achava Nakhash, the Loving Snake
>
I concur. My daughter took this in high school, and she had problems
every bit as complicated as this.
Regards,
Achava