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Probability

A hand of 8 playing cards is selected at random from a standard pack of 52. Calculate the probability that the hand contains no royalty (J,Q,K)

I know this should easy but I just cant seem to get my head around it
Original post by Amyherdman
A hand of 8 playing cards is selected at random from a standard pack of 52. Calculate the probability that the hand contains no royalty (J,Q,K)

I know this should easy but I just cant seem to get my head around it


Well first of all, how many Jacks, Queens, and Kings is there in a single pack of cards?

Then, consider the probability of NOT picking any of those cards out of the pack without replacement.
Reply 2
Avatar for Amyherdman
Amyherdman
OP
8
Original post by RDKGames
Well first of all, how many Jacks, Queens, and Kings is there in a single pack of cards?

Then, consider the probability of NOT picking any of those cards out of the pack without replacement.


So I did 12/52 * 12/51 * .... * 12/46 * 12/45 and that gave 1.417x10^-5 which seemed far too small
Original post by Amyherdman
So I did 12/52 * 12/51 * .... * 12/46 * 12/45 and that gave 1.417x10^-5 which seemed far too small


12/52 represents the probability of picking one of the cards you're not supposed to have. The further ones are conditional probabilities given that the previous pick(s) weren't any of the cards you're not supposed to have.
Basically, this is not the probability you're looking for.
Reply 4
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Amyherdman
OP
8
Original post by RDKGames
12/52 represents the probability of picking one of the cards you're not supposed to have. The further ones are conditional probabilities given that the previous pick(s) weren't any of the cards you're not supposed to have.
Basically, this is not the probability you're looking for.


So it wont work changing it to 40/52?
Original post by Amyherdman
So it wont work changing it to 40/52?


It would. The probability you're looking for is 4052⋅3951⋅…⋅3345\dfrac{40}{52} \cdot \dfrac{39}{51} \cdot \ldots \cdot \dfrac{33}{45}
(edited 6 years ago)
Original post by RDKGames
It would. The probability you're looking for is 4052⋅3951⋅…⋅3244\dfrac{40}{52} \cdot \dfrac{39}{51} \cdot \ldots \cdot \dfrac{32}{44}


You just have to assume that with this one he is taking them out one at a time instead of 8 in one go. I started contemplating this too myself and confused myself too much haha
Reply 7
Avatar for Amyherdman
Amyherdman
OP
8
Original post by Y11_Maths
You just have to assume that with this one he is taking them out one at a time instead of 8 in one go. I started contemplating this too myself and confused myself too much haha


Yeah, thats whats happening with me, I just keep confusing myself
Original post by Amyherdman
Yeah, thats whats happening with me, I just keep confusing myself


As an alternative method:

Number of possible hands where there are no picture cards is 40C8

Number of possible hands where there are no restrictions is 52C8.

So, probability of there being no picture cards is 40C8/52C8, which will work out to the same value you had.
Original post by RDKGames
It would. The probability you're looking for is 4052⋅3951⋅…⋅3244\dfrac{40}{52} \cdot \dfrac{39}{51} \cdot \ldots \cdot \dfrac{32}{44}


it will be till 33/45 since 40/52 is counted as one too...
Original post by brainmaster
it will be till 33/45 since 40/52 is counted as one too...


Yes this is correct
Original post by brainmaster
it will be till 33/45 since 40/52 is counted as one too...


Yeah, meant to write that and didn't notice. :smile:
Reply 12
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Amyherdman
OP
8
Original post by ghostwalker
As an alternative method:

Number of possible hands where there are no picture cards is 40C8

Number of possible hands where there are no restrictions is 52C8.

So, probability of there being no picture cards is 40C8/52C8, which will work out to the same value you had.


this is the way i was trying to do it but kept making some sort of mistake. Thank you

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