If you are dealt 3 cards, what are the chances that there is at least one jack queen or king in the 3 cards?

Question

How do you figure this out?

Answer 1

LOL!  Wow!  THAT was a long-winded answer, wasn't it!

ZCT, with his original answer, was almost dead on.  He just missed one minor point.

You add the odds of each card, but only if the previous card(s) missed.

Odds of first card being a face card - 12/52 = 23.1%

Odds of second card being a face card if the first wasn't - 12/51 * 76.9% = 18.1%

Odds of third card being a face card if the first two weren't - 12/50 * 58.8% = 14.1%

So the odds of getting at least one ... 23.1% + 18.1% + 14.1% = 55.3%.

--------------------------------------...

It's the same answer DarkLord got, because it's the same calculation, only expressed differently (and with fewer words.)

It's the answer ZCT would've got if he'd remembered to account for misses.

It's the answer Sandy would've gotten, if he'd remembered to reduce his count of available non-face cards with each draw.

It's not the answer Bill would've gotten, but hey, at least he tried ....

Answer 2

I think your all wrong and that you need to calculate the probibility of getting NONE of those cards by multipliing

40/52 *40/51 *40/50

then DEDUCT IT FROM 1 for all remaining possibilites which include having anywhere form one to three of them.

so you would get, 51.8%

Answer 3

First off, great question!

I have a mathematics degree, so bear with me as i get a little geeky :)

Unfortunately, EVERYONE ABOVE IS WRONG :(   but.. the guy just above got close.

Off the bat - odds are 55.29 %... but please read the whole answer ? I think you'll enjoy.

this is long-ish, but hopefully will help you with similar question in the future.

I will list some basics of probability for all the readers.. you may already know these, but some others may not.  I hope you don't mind :)  

I think it will help with the logic of HOW you get the answer, rather than just giving you the straight up probability value....again, useful for figuring out other card combination percentages.

My assumption is that we are dealing with a single deck of cards..not typical in a casino, but it works for us here asa starting point.

The way you want to think about this is as follows:

You want AT LEAST one face card in your 3-card hand.  That is the same as saying, you don't want ANY hand that has only non-face cards.

1) Probabilities (called P from now on) are expressed as numbers between zero and one - a P = 1 means certainty of 100% and a P = 0 means 0 %.

2) We commonly use the FREQUENCY METHOD of P-value determination.  Basically it means that you take the total number of ways a SPECIFIC outcome can happen (call it S), and the total number of ways EVERY outcome can happen (call that E), including your specific outcome.... and then you divide them like this,

S/E ---> = the probability of outcome S happening out of all possible outcomes E.  

In our case, E = 52 (total number of cards) ... S is the specific outcome-  what we want to determine for a 3 card draw.

E is made up of particular states, all of which combine to form what we then call the "total space E" over which outcomes like S can occur.

If there are several events/states that are similar/equivalent to the outcome we've called S, we might call them S1, S2, S3, etc.  and then we would call all of those events together part of the "space S".  

For example, a 3 card hand with a Jack is the same as a 3 card hand with Queen...IF...all we care about is having a face card in the mix.

The key concept is that the "space S" exists INSIDE of the "space E".

it helps to think of circles to visualize this... "space E" is a big old circle with a smaller circle called "space S" inside of it.  

All the points inside the circle "space S" are those similar events (or states) we talked about above (Jack vs Queen)!

And, all the area outside of the circle "space S" are other events... Like, say, all the 3-card hands with NO FACE cards whatsoever.

3) If a P-value for a sequence of events, like dealing cards one at a time, is independent from the previous round (i.e., each card dealt), then the probability of the sequence of events is the product (multiplication) of the individual probabilities of each event.

that is, for event 1, 2, and 3.... Probability of event 1 , called P(1) for short, and P(2), AND P(3) altogether (called P(1,2,3)) is given by,

P(1,2,3) = P(1) * P(2) * P(3)  ------  equation #1

4) Finally, the P value  of every possible outcome always sums up to  a total of 1.   That is the same as saying that, out of all possible things that could happen... the probability that SOMETHING (anything) happens is 100%...pretty much common sense.  This helps the P values make real-life sense too :)  

Another way to think of that is that the total area of the circle "space E" described above is set up to be exactly equal to 1, and "space S" has some area that is a fraction of that total area.

Phew! okay let's move on....

Now, think about the two possible scenarios based on your question:

a) you have a 3 card hand with 1 or more face cards in it.

b) you have a 3 card hand with no face cards in it.

Those two events above are events that can happen and so the P value for event a)  and for event b), denoted as P(a) and P(b), must add up to a total of 1.

that is,

P(a) + P(b) = 1     -----------  (equation 2)

Notice that P(a) is exactly what you asked.

so, rearranging the equation #2,

P(a) = 1 - P(b)        -----------  (equation #3)

it turns out that P(b) is very easy to figure out!

you have 52 cards in the standard deck, after removing the joker, and this is the total "space" we have to deal with for any  first deal of a card.   We'll call this Ei, where the "i" refers to 1,2, or 3 in the dealing of our 3 card hand.

the total "space of cards to choose from is 52 on the first deal, 51 on deal 2, and 50 on deal 3.  

Why is that you ask ? Remember, you are losing a card from the deck with each deal!

so....

E1 = 52, E2 =51, E3 = 50

12 of those cards in E1 are face cards.... and 40 are not.  We'll call the P-values for drawing from the 40 card deck  P(Si) ... where the "i" refers to 1,2, or 3 in the dealing of our 3 card hand.  

Notice that the S1, S2, and S3 values change just like the E values ( S1 = 40, S2 = 39, and S3 = 38)

-- remember again, you lose a non-face card with each deal.

Similarly, S1, S2, and S3 are our "events" from equation #1 listed above, such that:

P( S1, S2, S3) = P(S1) * P(S2) * P(S3)  ----- (equation #4)

therefore, from the frequency definition:

P(S1) = S1/E1 = 40/52

P(S2) = S2/E2 = 39/51

P(S3) = S3/E3 = 38/50

and from Equation #4,

P( S1, S2, S3) = 40/52 * 39/51 * 38/50 = P(b)  (equation #5)

which comes out to a value of approximately = 0.4471

YES! we have just determined the P-value for drawing a 3 card hand that has NO FACE CARDS at all.

using equation #3 and our result from equation #5,

P(a) = 1 - 0.4471 = 0.5529

Since odds are typically stated as percentages, we just multiply the answer by 100 %, like so

0.5529 *100 % = 55.29%

THEREFORE , YOU HAVE BETTER THAN 50% CHANCE OF GETTING AT LEAST ONE FACE CARD IN THIS 3 CARD DEAL.

Please take note that the *wording* of your question greatly determines the correct answer.  That is because in EVERY probability question, the amount of information is what determines the overall probability of a given result.

For example, if you had specified a PARTICULAR face card, the P-value would be dramatically reduced.

OR... if you had specified the ORDER IN WHICH THE FACE CARD was dealt... that would also lower the odds.  This all goes back to the idea of P-value "spaces"

.... clearly, getting ANY face card is more liley than getting, say a jack

... and that is more likely than getting the jack of hearts exactly

... which is more likely than getting the jack of hearts exactly on the second deal

... and so on and so on

But...all of those events fall within the realm of getting AT LEAST 1 face card.... Neat huh?

Okay!  I hope this was both informative and fun as an exercise.  I know it was long, but if you apply this same line of thinking, you can solve multitude of card related probability problems.

Just to re-iterate:  your chances are 55.29%

Good luck in your card games, and may the gods of the shuffle be at your side !! :)

take care,

- mtt

Answer 4

There are 52 cards in a deck with no jokers.  Of those there are four jacks, four queens, and four kings.

So if someone gives you a card from a randomly shuffled deck, you have a 12:52 chance of getting a picture card.  Or 12/52=0.23 which is 23%.

If you get to do this three times, you have three chances to get one of those cards:

12/52+12/51+12/50=0.71 which is 71%

So by my math if you are given three random cards, there is a 71% chance that at least one of them will be a face card.

That's how I would figure it, but I could be wrong.

Answer 5

ztc's method doesnt work because you can keep adding probabilities and get a probability greater than 1

this is how you do it

there are 52 cards in a deck

let's do it one card at a time

there are 4 jacks, 4 queens, and 4 kings in the deck

so there are 40 other cards

the odds that the first card isnt a jack queen or king is 40/52

assuming that happens there are 51 cards left

so the odds of the next one not being a jack queen or king either is 40/51

and the odds of the 3rd one is 40/50

so the odds of all 3 events happening is

40/52 * 40/51 * 40/50 = 64000/132600 = 320/663

or as a decimal -- .482655

as a percent - 48.2655%

Answer 6

i'm not sure ... but my guess would be 36 out of 52 (or, simplified: 9 out of 13). i got this because there are 4 of each face card in a deck of 52, so that would mean the face cards make up 12 out of 52. so ... if you draw only one card from the deck, your chances of getting a face card are 12 out of 52. if you are dealt 3 cards, i would assume that your chances of getting at least 1 face card would be 3 times higher. so 12 times 3 is 36 ... and that gives you the 36 out of 52.