probability question (1 Viewer)

[mathsbrain]

Hi all,

can someone please help and try and explain how to find the probability of obtaining one pair when being dealt a 5 card hand from a standard pack of 52 cards?

 

[jeffwu95]

i got 0.52 might of made a mistake though

i think i forgot to remove some of the possibilities zzz
OK i got 0.47

 

[nightweaver066]

There are 52 cards where we select 5.
Total combinations: 52C5

There are 13 different cards (4 of each) and you want to select one of them. Then you want to select 2 cards from the 4 cards.
Selecting the pair: 13C1 x 4C2

Now to select the 3 other cards, we need to consider the other cards. From these there are 4 choices from each.
Other cards: 12C3 x 4 x 4 x 4 = 12C3 x 4^3

So Probability =

 

[jeffwu95]

dont you also have to minus the probability that there is 2 pairs so (aa,bb,c)

 

[nightweaver066]

dont you also have to minus the probability that there is 2 pairs so (aa,bb,c)

I don't think so. I dealt with the other 3 cards by selecting 3 different cards (hence the 12C3)

 

[jeffwu95]

well i thought it was 4C2 times 13 times 48C3 (u choose 3 other card other then the pair) - 4C2 times 4C2 times 13C2 (as u have to choose 2 pairs out of 13) times (52-8)
divide by 52C5

 

[jeffwu95]

ok i realise what i did wrong haha

 

[mathsbrain]

There are 52 cards where we select 5.
Total combinations: 52C5

There are 13 different cards (4 of each) and you want to select one of them. Then you want to select 2 cards from the 4 cards.
Selecting the pair: 13C1 x 4C2

Now to select the 3 other cards, we need to consider the other cards. From these there are 4 choices from each.
Other cards: 12C3 x 4 x 4 x 4 = 12C3 x 4^3

So Probability =

Your method makes total sense, but i have an alternative here which i am in a slight confusion:

I have the same denominator being 52C5.

For numerator, i have 52C1 to choose the first card, then i have 1 choice out of 3 to complete the pair. However, i have to divide by 2! due to the fact that the first 2 cards since order is not important (not sure why). Then for the next 3 cards, i do 48C1 * 44C1 * 40C1 /3!
It's the divide by 3! that i don't fully understand. i get that this gets rid of the ordering problem, but WHEN do you have to do this. Like looking at this question at the start, i would have thought to divide by 5! overall, instead of 2! and 3!. Like how do you tell? any tricks?

 

[Shadowdude]

Your method makes total sense, but i have an alternative here which i am in a slight confusion:

I have the same denominator being 52C5.

For numerator, i have 52C1 to choose the first card, then i have 1 choice out of 3 to complete the pair. However, i have to divide by 2! due to the fact that the first 2 cards since order is not important (not sure why). Then for the next 3 cards, i do 48C1 * 44C1 * 40C1 /3!
It's the divide by 3! that i don't fully understand. i get that this gets rid of the ordering problem, but WHEN do you have to do this. Like looking at this question at the start, i would have thought to divide by 5! overall, instead of 2! and 3!. Like how do you tell? any tricks?

where did you get these numbers from


If you set out your working logically, you'll figure it out much faster.

 

[mathsbrain]

once you chose 52C1 * 3C1 /2!, you have a pair, and you DONT want another pair from then on. you have 48 cards to choose from (since you dont want any of the first 2 cards and the 2 cards that has the same number), so 48C1, the next one cant be the same number, so choose one from 44, ie 44C1, and 40C1. But why divide by 3! ?

 

[mathsbrain]

mathman help?

 

[braintic]

Your method makes total sense, but i have an alternative here which i am in a slight confusion:

I have the same denominator being 52C5.

For numerator, i have 52C1 to choose the first card, then i have 1 choice out of 3 to complete the pair. However, i have to divide by 2! due to the fact that the first 2 cards since order is not important (not sure why). Then for the next 3 cards, i do 48C1 * 44C1 * 40C1 /3!
It's the divide by 3! that i don't fully understand. i get that this gets rid of the ordering problem, but WHEN do you have to do this. Like looking at this question at the start, i would have thought to divide by 5! overall, instead of 2! and 3!. Like how do you tell? any tricks?

The 13C1 and 4C2 give no order to the solution - they give you a pair somewhere in your hand.
The 48x44x40 picks three cards in order, so you must divide by 3! to select just one of those orderings.

Note that the first part of the calculation could have been done using ordered selections:
Pick a card, any card: 52 ways
Now pick another card with the same denomination: 3 ways
But you are double-counting, as 2C followed by 2D is the same as 2D followed by 2C, so you need to divide by 2!
The calculation: 52x3 / 2! which is the same as 13C1 x 4C2.

Back to the last 3 cards: if you start with a pair of 2s, then pick 3 more then you are hex-counting, ie. there are 3!=6 ways of coming up with the same 3 cards.

Perhaps what confused you was the nCr notation - it should really have been nPr - but you know that nP1 = nC1.