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Odds

mummymonkey

Did you spill my pint?
mummymonkey
Joined
Dec 3, 2002
Messages
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Location
Scotland
There are five cards; three black and two red. If you pick three, what are the odds of picking both red cards?

I worked it out at 30% by scribbling down all the combinations and picking the first three cards each time. (Don't know if that's right or not.) Presumably there is a rather better way of doing it.

Also would the same question with ten cards (six and four) give odds of 15% for picking all four reds with six picks?

And yes, counting was never my favourite subject as school. :blush:
 
Dear monkey,

I figure it's a 10% chance. My logic is this:

Suppose out of the five cards available, you draw one. There is a 40% chance of getting a red one.

Out of the four cards remaining, however, you only have a 25% chance of drawing a red card.

25% of 40% is 10%.

Cpl Ferro
 
CplFerro said:
Dear monkey,

I figure it's a 10% chance. My logic is this:

Suppose out of the five cards available, you draw one. There is a 40% chance of getting a red one.

Out of the four cards remaining, however, you only have a 25% chance of drawing a red card.

25% of 40% is 10%.

Cpl Ferro
Click to expand...

That can't be right, because it's out of three tries. you can't use a bernuli trial here either. you might have to count all the probability of all the possible ways to pick both cards.

I actually get a 30% chance.

you could get two reds by either getting a black then two reds, getting two reds and a black, or by getting a red, then a black, then another red.

for each way, I got a 10% chance, added up gives 30%

EDIT: whoops, yeah. didn't see that you also got 30%. anyway, I agree. and AFAIK, that's the only way to get the answer here. there's easier ways to get answers like this if you have independent events, but I don't know any for dependent events.
 
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Edited 'cause I didn't read the question properly.

There are 10 combinations of 2 cards, only one of which is red,red. So that answer is 10%. In addition there are 3 ways to choose 2 items out of 3. So it's 10% * 3 = 30%. That equals 7 to 3 odds against.

Type in the google search box the following. 5 choose 2. Then 3 choose 2

Google knows.;)
 
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Lable the cards as follows ...

B1 B2 B3
R1 R2

How many ways can one pick 3 cards?

5 x 4 x 3 = 60 permutations.

How many ways include 2 red cards?

R1 R2 B1
R1 R2 B2
R1 R2 B3

R2 R1 B1
R2 R1 B2
R2 R1 B3

R1 B1 R2
R1 B2 R2
R1 B3 R2

R2 B1 R1
R2 B2 R1
R2 B3 R1

B1 R1 R2
B2 R1 R2
B3 R1 R2

B1 R2 R1
B2 R2 R1
B3 R2 R1

18 ways with 2 reds out of 60 = 30%
 
CplFerro said:
Dear monkey,

I figure it's a 10% chance. My logic is this:

Suppose out of the five cards available, you draw one. There is a 40% chance of getting a red one.

Out of the four cards remaining, however, you only have a 25% chance of drawing a red card.

25% of 40% is 10%.

Cpl Ferro
Click to expand...

I believe you determined the probability of getting 2 red cards from picking only 2 cards.
 
ways of selecting 3 cards from 5 = 5!/3!2! = 10
ways of selecting 1 black card from 3 = 3!/2!1! = 3
ways of selecting 2 red cards from 2 = 1

so to select 2 red and 1 black you have 3x1 ways out of 10 = 30%

so far so good


for 4 red cards and 6 black cards

6 cards from 10 = 10!/6!4! = 10.9.8.7/4.3.2 = 210
2 black cards from 6 = 6!/4!2! = 15
4 red cards from 4 = 1

so 15 ways out of 210 is 1/14 = 7.14%

I hope that's correct
 
Thanks all. I did it the same way JustThinking did it.
Max, I can't follow your post, what do the exclamation marks signify?
 
Thanks for the correction, my eyes fail me sometimes.

In re-response to this:

There are five cards; three black and two red. If you pick three, what are the odds of picking both red cards?

When drawing the first card, the odds of getting red are 40%.

When drawing the second and third cards, the combined odds of getting red are 50%.

40% X 50% = 20%.

Cpl Ferro
 
MaxHarcore is correct, here's why.

If you have M black cards and N red cards, and pick K of them, what are your chances of getting exactly C black cards?

Answer: [choose(M,C)*choose(N,K-C)]/choose(M+N,K)

choose(a,b) should be the number of ways to choose b things from a things. That ends up being a!/(b! * (a-b)!). See <a href="http://en.wikipedia.org/wiki/Binomial_coefficient">wikipedia</a> for an explanation.

For M=3, N=2, K=3, and C=1, that's c(3,1)*c(2,2)/c(5,3)=3/10.

For M=6, N=4, K=6, and C=2, that's c(6,2)*c(4,4)/c(10,6)=15/210=1/14.
 
mummymonkey said:
Thanks all. I did it the same way JustThinking did it.
Max, I can't follow your post, what do the exclamation marks signify?
Click to expand...
Exclaimation marks signify "factorial". Basically, n-factorial is equal to the product of the integers from 1 to n. So ...
3! = 3 x 2 x 1 = 6
6! = 6 x 5 x 4 x 3 x 2 x1 = 720
The reason factorial occurs so often in counting things is the nature of the beast. How many ways can a standard deck be ordered. Well, the first card can be any one of 52 cards. After selecting the first card, the second can be any one of the remaining 51 cards, the third any one of the remaing 50, the fourth ...

As a result, the standard deck of cards can be ordered in 52 x 51 x 50 x ... X 3 X 2 X 1 ways, or simply 52!.

So in the original problem, you have 5 cards. So if you put them in random order, you would have 5! possible shuffled decks.
5! = 5x4x3x2x1=120
When you wrote down as you mentioned in your very first post that you wrote down all the combinations, how many do you get? 120 I expect.

Walt
 
mummymonkey said:
There are five cards; three black and two red. If you pick three, what are the odds of picking both red cards?

I worked it out at 30% by scribbling down all the combinations and picking the first three cards each time. (Don't know if that's right or not.) Presumably there is a rather better way of doing it.

Also would the same question with ten cards (six and four) give odds of 15% for picking all four reds with six picks?

And yes, counting was never my favourite subject as school. :blush:
Click to expand...

Or, the equivalent. What's the probability of picking 2 black cards since that's what remains.

The first card has a 60% probability of being black. Given that, the second card has a 50% probability of black being picked. Overall probability .6*.5 or 30%.
 
n! represents the number of orders in which one can arrange n objects.

So, the number of different ways to arrange those five cards is 5!=120. Now, how many of those "different" ways are actually the same? Well, once you've picked out 3 cards, there are 3!=6 different ways that you can rearrange them without really chaging anything. You can also rearrange the other 2 cards 2!=2 different ways. So you have 5!/(3!*2!)=10 truly different ways to pick three cards (there's a special term for this: the number of ways to choose k items from a list of n is equal to n!/((n-k)!*k!) and is called n choose k or nCk), and three of them have two red cards. So the probability is 3/10=30%.

mummymonkey said:
Also would the same question with ten cards (six and four) give odds of 15% for picking all four reds with six picks?
Click to expand...
Let's just do this in general. Let's say you have b black cards, r red cards, you pick p cards (so you have l cards left). There are (p+l)!/p!l! total possibilities. Of those, b!/l!(b-l)! have nothing but black cards left.
So the probability is [b!/l!(b-l)!]/[(p+l)!/p!l!]=b!p!/(b-l)!(l+p)!

If b=3, p=3, l=2, then this is 3!3!/1!5!=6*6/120=3/10
If b=6 p=6 l=4 then it is 6!6!/2!10!=6!/(2*10*9*8*7)=1/14
 
Just thinking said:
Correct .. but then how does one get 0! = 1? ;)
Click to expand...
There are no positive integers less than or equal to zero. Therefore, there should be no multiplication. 1 represents no multiplication; if you multiply a number by 1, that's the same as doing no multiplication.
 
Just thinking said:
Correct .. but then how does one get 0! = 1? ;)
Click to expand...
Because a foolish consistency is the hobgoblin of little minds.

Someone said, n! = n * (n-1)!, which is obviously true in most cases.
Then they said 1! = 1 * 0!
Therefore 0! = 1!/1 = 1

Actually I think I learned the real reason long ago in math, had something to do with the gamma function if I recall correctly. But if you can make 10 to the power 0 = 1, why not make 0! = 1. Just to annoy people.
 
Walter Wayne said:
Actually I think I learned the real reason long ago in math, had something to do with the gamma function if I recall correctly.
Click to expand...

gamma(n) = integral tn-1*e-t dt, 0 to oo

Integrating the gamma function, one gets:

gamma(n) = (n-1)!

and

gamma(1) = 1

but from above gamma(1) = (1-1)! = 0! = 1
 
Walter Wayne said:
Actually I think I learned the real reason long ago in math, had something to do with the gamma function if I recall correctly. But if you can make 10 to the power 0 = 1, why not make 0! = 1. Just to annoy people.
Click to expand...
The real reason is that it is convenient. By setting 0! = 1, you can use it in legions of formulae, without need of a special case. An example from basic calculus is Taylor's expansion

[latex]
\displaystyle
f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n
[/latex]

If 0! were 0, we would have to write a lengthier formula. The same thing happens everywhere. It also fits with the identification Gamma[n+1] = n! (for positive integers), as T'ai Chi pointed out.
 
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