Random Cards

gondorff

I shuffle a deck of cards and take three at random.

Can someone show me mathematically what is the exact probability that one of these three cards is an ace, two or a three? (or any other three arbitrarily named cards). Disregard suits.

All you have to do is divide the number of favourable outcomes by the number of possible outcomes, thus, the probability for any event P(E) is Favourable/Possible.

So, for an Ace:

4 (four Aces)/52 (52 cards) = 1/13.

But if you want to do it for three cards, you recalculate each probability and multiply, so for a 2:

4 (number of twos)/51 (now there's only 51 cards in the deck, as you've already taken one out) = 4/51.

And, just say, a three:

4 (number of threes)/50 (same reason as above) = 2/25.

Now the total probability is the three probabilities multiplied, so it's 1/13*4/51*2/25 = 8/16575.

So the probability of pulling out the three cards that you want to pull out is 16575 in 8.

(I think I'm right, but, I could be completely wrong with this, I haven't done probability yet, but I'm sure one of the maths guru's here will correct me).

I think I may have misread. Did you mean what is the probability of pulling out an Ace, two and a three? Or pulling out one of the three?

If it's one of the three, 12/52 I think.

gondorff

Ace or two or three.

This is how it was explained to me:

Probability for first card: 12/52
second card: 12/52
third card: 12/52


Add'em up: 36/52

An odds on chance! I think it's pretty close but I want to see a conclusive proof.

gondorff

gondorff wrote: »

Ace or two or three.

This is how it was explained to me:

Probability for first card: 12/52
second card: 12/52
third card: 12/52


Add'em up: 36/52

An odds on chance! I think it's pretty close but I want to see a conclusive proof.

The probability for the first card is 12/52 alright, but now for the second card it's 11/51 as you've only 51 cards to choose from, and you've already one of your favourable choices gone; and for the third it's 10/50 for the same reason. I'm actually not sure whether to add or multiply them now that you say it, but I think my odd's are right.

Webmonkey

I would have thought it was:

P(ace) = 4/52
P(one) = 4/52
P(two) = 4/52

Probability of getting one of these events is

4/52 + 4/52 + 4/52 = 12/52 = 3/13 = 23%

I think -JammyDodger- you are getting mixed up with conditional probability in that there are 3 consecutive goes (and cards not replaced, memory based), where probability first card is an ace, then the second card a one and third card a two. Then you multiply giving proability of 0.9%.

But this is a once off, memoryless event.

It being years since i've done it but this would be my understanding. Could be totally wrong too

LeixlipRed

The OP said he wanted to pick three cards though. It's a bit ambiguous. Maybe he'll come back and clear it up for us.

gondorff

I take three cards from a shuffled deck. Without looking at any of them, the probability of the first being A,2 or 3 is 12/52.

I still haven't looked at it. Now the probability of the second card being A,2 or 3 is still 12/52 and the same for the third card.

So surely the total chances of any of these three cards being A,2, or 3 is:

Roughly 12/52 + 12/52 + 12/52 = 36/52 no?

Not even something close to this?

Webmonkey

Oh i see, then -JammyDodger- is more correct, but I'd imagine you multiple then instead add.

Grudaire

gondorff wrote: »

I take three cards from a shuffled deck. Without looking at any of them, the probability of the first being A,2 or 3 is 12/52.

I still haven't looked at it. Now the probability of the second card being A,2 or 3 is still 12/52 and the same for the third card.

So surely the total chances of any of these three cards being A,2, or 3 is:

Roughly 12/52 + 12/52 + 12/52 = 36/52 no?

Not even something close to this?

don't think that works, cause what if you pick 6 cards, you get 72/52 as your probability.

think like this:

1 - 40/52 *39/51*38/50

ie. 1 - Prob of not getting any of the cards at all

that right?

RoundTower

you cant add up the probabilities like this. The reason you cant is because you count some situations twice. For example suppose you picked an ace, a two and a five. You'd have counted that situation twice in your 12/52 + 12/52 + 12/52.

Adding the probabilities up is a good way to get an approximate answer if the probabilities are very small, for example if you wanted to figure the chance of getting at least one ace the chances would be close to 3/52.

The simplest correct way to solve this problem is to figure out the chances of not getting any ace, two or three on any of the cards. There are 40 "big" cards in the deck so the chance of getting no A23 on the first card is 40/52, for the others 39/51 and 38/50. Multiply these together you get 59280/132600. Nowe we needed to get the chance of at least one A23, so take this number away from 1 and get 73320/132600 which is 47/85 or about 55%. So still an odds on chance, but closer than you thought.

RoundTower

well cliste got it first, but he's definitely right, and no one else is, unless they were trying to answer a different question.

gondorff

RoundTower wrote: »

if you wanted to figure the chance of getting at least one ace the chances would be close to 3/52

Ok except for this (typo?), yourself and Cliste have cleared this up for me.

Cheers!

RoundTower

gondorff wrote: »

Ok except for this (typo?), yourself and Cliste have cleared this up for me.

Cheers!

yeah an unusual typo where all the letters were replaced with the wrong ones

i meant either 12/52, or "the chance of getting at least one ace of spades would be exactly 3/52"

Adrock-aka

Hi guys,

Have been reading this post and was wondering what the percentage probability is of shuffling a deck of cards, dealing the first four from the top and them being all the same card. I.e shuffle for a minute and deal four Jacks from the top of the deck only.

Is it...

4/52 * 3/51 * 2/50 * 1/49

=> 24/6,497,400

=> 1/270,725

=> .0000037

= .00037%?

Not sure if thats correct!

Cheers!
Adrock

LeixlipRed

Looks good to me.

Adrock-aka

success!

Lol, thanks.

Pherekydes

Adrock-aka wrote: »

Hi guys,

Have been reading this post and was wondering what the percentage probability is of shuffling a deck of cards, dealing the first four from the top and them being all the same card.

If the question is, "What is the probability of dealing 4 cards of the same value off the top of the deck?" the answer is 0.00005 approx.

Why?

If you name the card in advance then yours is the correct answer. If you don't name the card in advance then the first card is 52/52.

LeixlipRed

Should have made that distinction myself. Lucky no one relys on just me on here

gondorff

RoundTower wrote: »

yeah an unusual typo where all the letters were replaced with the wrong ones

Wow!

What are the odds of that?!

LeixlipRed

60% of the time, it works every time I believe. Just like sex panther.

Adrock-aka

The reason I ask is because I got a phone call at 1:45 am last Tuesday morning from my little brother saying he had been shuffling a deck while watching a film for 20 mins or so and dealt himself 4 cards... JJJJ. Aparently he needed to know the odds of that happening THAT VERY MINUTE! And it just couldnt wait until sociable hours.

Even though i had work the next day I was obviously impressed with the feat he had accomplished. Took him years to get that good!!

Adrock-aka

Pherekydes wrote: »

If the question is, "What is the probability of dealing 4 cards of the same value off the top of the deck?" the answer is 0.00005 approx.

Why?

If you name the card in advance then yours is the correct answer. If you don't name the card in advance then the first card is 52/52.

I see what you mean, the fact is he wasnt trying to get anything especially not JJJJ so its 52/52 * 3/51 * 2/50 * 1/49 = 1/20,825.

TBH I dont think he was even going for the 2nd or 3rd (it was late and he didnt know what was going on....)

52/52 * 51/51 * 50/50 *1/49 = 1/49

To be really honest, me must have thought it was impossible to get the fourth and didnt go for it so...

52/52 *51/51 * 50/50 * 49/49 = 1/1

Only joking!
Thanks for the reply. I get you.

If he had turned the four over at once would it be the 4/52 * 3/51 * 2/50 *1/49???

Is it only when you intend to pick a jack it turns from 52/52 to the 4/52, 3/52 etc etc???

LeixlipRed

If he decided beforehand that he wants to draw 4 Jacks then the probability of doing so is 4/52 * 3/51 * 2/50 * 1/49 = whatever.

If he decides that he'll draw a card at random and then try draw three more of the same then the first probability is 1 as it does not matter what card he draws. So total probability is

52/52 * 3/51 * 2/50 * 1/49 = whatever.

Adrock-aka

LeixlipRed wrote: »

If he decided beforehand that he wants to draw 4 Jacks then the probability of doing so is 4/52 * 3/51 * 2/50 * 1/49 = whatever.

If he decides that he'll draw a card at random and then try draw three more of the same then the first probability is 1 as it does not matter what card he draws. So total probability is

52/52 * 3/51 * 2/50 * 1/49 = whatever.

Gotcha, thanks

kincsem

I think Roundtower's approach is correct but I think there is a slight miscalculation in his numbers.

The chance of the first card not being an ace, duece, trey is 40/52.
The chance of the second card also not being an ace, duece, trey is 40/51.
The chance of the third card not being an ace, duece, trey is 40/50.

The chance in my opinion is 1 - (40/52 x 40/51 x 40/50) or 51.73%

Just guessing here. The poker forum knows.

hivizman

The probability of the first card not being A, 2, or 3 is 40/52, but then you've drawn one of the "big cards", so there are only 39 left. Hence the probability of the second card not being A, 2, or 3, conditional on the first card not being A, 2, or 3, is 39/51. If both cards are "big cards", then there are only 38 "big cards" left in a pack of 50, so the overall probability of drawing three cards none of which is an A, 2 or 3 is:

40/52 x 39/51 x 38/50 =0.4471.

Hence the probability of at least one A, 2 or 3 is 1 - 0.4471 = 0.5529 or 55.29%.

So I think that Roundtower is correct here.

kincsem

I was wrong. Sorry.

My bad maths might merit an infraction or banning.

Roundtower was correct with (1-(40/52 x 39/51 x 38/50) = 55.29%

Looking at the opening question I saw that it said one card but did not say “one card only” so I assume two cards, or even three cards of ace; deuce; trey are acceptable.

That got me thinking what are the odds of those –

One only … 40/52 x 39/51 x 12/50 x 3 = 42.35%
Two only … 40/52 x 12/51 x 11/50 x 3 = 11.95%
All three ….12/52 x 11/51 x 10/50 x 1 = 1.00%

And these add to 55.30%

Am I wrong again?

hivizman

kincsem wrote: »

That got me thinking what are the odds of those –

One only … 40/52 x 39/51 x 12/50 x 3 = 42.35%
Two only … 40/52 x 12/51 x 11/50 x 3 = 11.95%
All three ….12/52 x 11/51 x 10/50 x 1 = 1.00%

And these add to 55.30%

Am I wrong again?

No, that's what I make them as well. The difference between 55.30% and 55.29% is simply rounding.

By the way, these are the probabilities of getting one, two or three cards out of the 12 aces, twos and threes - they are not the probabilities of getting a pair of aces, twos or threes, or three aces, three twos or three threes. Of the 1320 acceptable combinations for "all three", only 12 combinations give a "three of a kind".

lisbon_lions

I would have approached the initial question from a combination viewpoint.
The order in which the A,2,3 appears is not important, just that we get them.
So, C(n,r) = C(52,12)
= 52!/12!*(52-12)!
[edit: ignore the above its way off what we are trying to calculate]

However, after rereading the question - you ask what is the probability that one of these cards is an ace,two or three.

In that case,
P(1st card success AND 2nd card Not And 3rd card Not) OR P(1st card Not AND 2nd card success AND 3rd card not) OR P(1st card Not AND 2nd card Not AND 3rd card success)

=(12/52 * 40/51 * 39/50) + (40/52 * 12/51 * 39/50) + (40/52 * 39/51 * 12/50)

(.23*.784*.78) + (.77*.23*.78) + (.77*.765*.24)

= .14 + .138 + .141
= .42 [rounded up]...

gondorff

I'll word this so there can be no dispute as to what I mean.. (hopefully)

Take 3 cards at random from a deck of 52 cards.

What are the odds that at least one of these three cards is an ace, deuce or a three?


Now I know that if I take one card then the odds are 12/52 obviously. But surely if I take three cards my chances are almost tripled?

Example:

Take 3 cards and place them face down on the table. The odds for the first card are 12/52, and in my opinion, the odds for the second card (without looking at the first card), are still 12/52, and the odds for the third card (without looking at cards one and two), are yet still 12/52.

Where do we go from here? I know it's odds on but I need to be shown a conclusive answer (with proof) for the question above typed in blue.

Pherekydes

gondorff wrote: »

Take 3 cards at random from a deck of 52 cards.

What are the odds that at least one of these three cards is an ace, deuce or a three?


Now I know that if I take one card then the odds are 12/52 obviously. But surely if I take three cards my chances are almost tripled?

In that case, the probability (odds) are 1 less (the probability of not drawing any one of those), which equals

1 - (40/52 * 39/51 * 38/50)

You cannot simply treble the probability of drawing one, as people are inclined to do with probabilities of this nature. To see why, change the question to, "What is the probability of drawing at least one of the five lowest cards in 3 goes?"

The number of cards to choose from becomes 20, and the probability of drawing one in one go is 20/52. 3 times this is 60/52, which is impossible.

MathsManiac

Pherekydes wrote: »

In that case, the probability (odds) are...

At the risk of starting a needless tangent, it's possibly worth pointing out that the "odds" are not the same thing as the probability. The odds of an event E are: P(not E)/P(E).

Pherekydes

MathsManiac wrote: »

At the risk of starting a needless tangent, it's possibly worth pointing out...

Not a needless tangent at all. We all know a lot of people (not mathematicians) use the terms interchangeably.

The probability of getting an ace with one draw of a card is 1/13, while the odds are 12/1, which is P(not E)/P(E).

RoundTower

Adrock-aka wrote: »

The reason I ask is because I got a phone call at 1:45 am last Tuesday morning from my little brother saying he had been shuffling a deck while watching a film for 20 mins or so and dealt himself 4 cards... JJJJ. Aparently he needed to know the odds of that happening THAT VERY MINUTE! And it just couldnt wait until sociable hours.

Even though i had work the next day I was obviously impressed with the feat he had accomplished. Took him years to get that good!!

This is a difficult one, it all depends. Is your brother this guy?

https://www.youtube.com/watch?v=GErpOl3KG_w

gondorff

OK I'm happy with the explanation that Cliste gave. This is good for calculating probability for one hit.
Any fool can calculate the probability for three hits but..

What is the probability of hitting exactly twice?

Do you calculate for

HIT HIT MISS
HIT MISS HIT
MISS HIT HIT

and take an average of the three?

MathsManiac

gondorff wrote: »

OK I'm happy with the explanation that Cliste gave. This is good for calculating probability for one hit.
Any fool can calculate the probability for three hits but..

What is the probability of hitting exactly twice?

Do you calculate for

HIT HIT MISS
HIT MISS HIT
MISS HIT HIT

and take an average of the three?

No. You add them (which only works when they are "mutually exclusive" events - i.e. they don't overlap).

Since all three have the same probability, It's three times any one of them.

gondorff

Please expand. What is the percentage probability of hitting twice?

MathsManiac

The total number of possible outcomes when you choose 3 cards, in a specific order, from a deck of cards is 52*51*50.

Now, how many of these outcomes are you interested in? I'm assuming you're still talking about "hit" meaning getting an ace, two or three.

The total number of ways of getting HIT, HIT, MISS is: 12*11*40
The total number of ways of getting HIT, MISS, HIT is: 12*40*11
The total number of ways of getting MISS, HIT, HIT is: 40*12*11.

So the total number of outcomes of interest is the sum of these three.

When outcomes are equally likely, then the probability is (number of outcomes of interest)/(total number of possible outcomes).

So, the answer is 3(12*11*40) / (52*51*50), which is about 0.12, (i.e., 12%).

MathsManiac

I should also have mentioned, by the way, that since the order the cards are drawn doesn't matter, you could do it with "nCr"s instead:

Total number of (unordered) selections of 3 cards from 52 is 52C3 = 22100.
Number of ways to get 2 of the 12 you want and 1 of the 40 you don't want is (12C2)*(40C1) = 66*40 = 2640.

Probability = (of interest)/total = 2640/22100 = 132/1105 ~= 12%