Another Probability Question

[jay47]

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A 52 card deck is laid on a table. What is the probability of drawing a king in 4 tries? One card is removed after each attempt. Success is defined as drawing AT LEAST 1 king in the four tries.

My logic on this would be:

Prob: 1st Draw= 4/52

2nd Draw= 4/51

3rd Draw= 4/50

4th Draw= 4/49

Therefore, is the answer, (4/52)+(4/51)+(4/50)+(4/49)= 0.316987103 ??

 

[hausee]

A 52 card deck is laid on a table. What is the probability of drawing a king in 4 tries? One card is removed after each attempt. Success is defined as drawing AT LEAST 1 king in the four tries.

My logic on this would be:

Prob: 1st Draw= 4/52

2nd Draw= 4/51

3rd Draw= 4/50

4th Draw= 4/49

Therefore, is the answer, (4/52)+(4/51)+(4/50)+(4/49)= 0.316987103 ??

after your first draw, you only have 3 kings and total 51 cards left....

Anyway, here is a hint: what is posibility you will not get a King....

 

[system7]

not 100% sure but i think this how i'd do it:

(probability of drawing a king) x (probability of not drawing one)

1st Draw= 4/52 x 51/52

2nd Draw= 4/51 x 50/51

3rd Draw= 4/50 x 49/50

4th Draw= 4/49 x 48/49

(4/52)(51/52)+(4/51)(50/51)+(4/50)(49/50)+(4/49)(48/49) = 0.3107

 

[orthosm2021]

jay47,

you should be multiplying the 4 probability factions (not adding them up).
And as hausee has mentioned, the number of total cards left should decrease by 1 after each draw. so the answer should be: (4/52)*(3/51)*(2/50)*(1/49)


If you want to approach this as a combination problem. Im sure you know the formula for combination), you can also view the problem like this. You are making a 4 card set from a 52 card deck. So there are C(52,4)=52*51*50*49/(4*3*2*1)=270720 ways you can make this 4 card set. Out of these 270725 ways, you only want the set where you pick all 4 cards from the 4 kings and none from the 48 non-king cards. Common sense/intuition tells us that there is obviously only 1 way to pick all four kings from the available 4 kings and pick 0 cards from the 48 non-king cards. You just pick all of the four kings...and thats it. (*doesnt matter in what order you pick the four kings since this is a combination, not permutation)
*** C(n,k) denotes number of k-combinations from a given set of n elements. C(n,k)=n!/[k!*(n-k)!]****


so you just divide 1 by the total number of ways 270720 or in factorials, 52*51*50*49/(4*3*2*1) and get 1/270720 or 4*3*2*1/(52*51*50*49) -> same answer!



If you are a math person and likes equations,,
probability of drawing 4 kings in 4 draws from 52 card deck
= C(4,4)*C(48,0)/C(52,4)=4*3*2*1/(52*51*50*49)


What the above equation says is
first term: C(4,4)=1 : #ways you can draw all of your 4 cards from the 4 kings. (only 1 way from intuition!)
second term: C(48,0)=1: #ways you can draw 0 card from 48 non-king cards. (only 1 way from intuition!)
third term: C(52,4)=270725: total #ways you can draw any 4 cards from a 52-card deck.

So you multiply the first two terms and divide the product by the third term.

This is essentially the exact same operation as the one you use, except its more formula oriented and less brain thinking once you get use to the operation. You can also use this formula as a way of checking your answer. Also this formula is useful if you get a bit twisted problems like.... probability of getting 3 kings...and 1 non-king.
you simply do C(4,3)*C(48,1)/C(52,4)

hope this helps!

 

[Streetwolf]

after your first draw, you only have 3 kings and total 51 cards left....

Anyway, here is a hint: what is posibility you will not get a King....

Thank you! At least someone here knows what they are doing.

Nice try above me explaining combinations and such but what you did was not what the question was asking for.

Allow me to repeat that statement in the quote above:

Anyway, here is a hint: what is possibility you will not get a King....

Good, now here's the answer if you really can't figure it out.







You want the probability of drawing AT LEAST ONE king. Whenever you have a problem that asks for the probability of at least one, SOLVE FOR THE PROBABILITY OF NONE! Then subtract that answer from 1 (or 100%). *WHY?*

The probability of 0 kings in 4 draws is (48/52)(47/51)(46/50)(45/49) = 0.719.

So there's a 71.9% chance you won't draw a king after 4 draws.

So what does the other 28.1% represent? The chance that you WILL draw at least one king within 4 draws.

That's the answer you want.

So either 0.281 or 28.1%, whatever they list.

 

[orthosm2021]

Thank you! At least someone here knows what they are doing.

Nice try above me explaining combinations and such but what you did was not what the question was asking for.

Allow me to repeat that statement in the quote above:

Anyway, here is a hint: what is possibility you will not get a King....

Good, now here's the answer if you really can't figure it out.







You want the probability of drawing AT LEAST ONE king. Whenever you have a problem that asks for the probability of at least one, SOLVE FOR THE PROBABILITY OF NONE! Then subtract that answer from 1 (or 100%). *WHY?*

The probability of 0 kings in 4 draws is (48/52)(47/51)(46/50)(45/49) = 0.719.

So there's a 71.9% chance you won't draw a king after 4 draws.

So what does the other 28.1% represent? The chance that you WILL draw at least one king within 4 draws.

That's the answer you want.

So either 0.281 or 28.1%, whatever they list.

*** ignore my post below. I read the question wrong. (my apologies to streetwolf. man i really cant read. What says below is for finding the probability of getting all 4 kings in 4 draws. Im just gonna leave it here in case someone indeed need to find the probability of finding all 4 kings in 4 draws or do similar problems*


Hey,
before you say someone else's answer wrong, check your own solution first. And think rationally when you do. and, my answer IS correct.

So why your answer is wrong:
first of all, think about your number rationally. 28.1% (your answer) is way too high to be realistic. Have you ever played card games? If your answer is indeed true, you get a four kings in your hands more than 1/4 times you draw 4 cards from a full deck. Doesnt sound so realistic. If it was, why would a four-of-a-kind be so rare in poker?

I did not use your method on purpose b/c that way is actually more complicated than my, or jay's way. Here is what you are supposed to do if you would like to find the answer by finding the probability of NOT getting any king in 4 draws:
If you just subtract from 1 the probability of not getting any king (0 king in other words)in 4 draws, you end up with the combined probability you will get 1 king, 2 kings, 3 kings, or 4 kings. This is why your answer is way off -- your answer was only quarter way done.

So what you should do is,
on top of subtracting the probability of not getting 4 kings, you should also subtract the probabilities of not getting 3 kings, 2 kings, and 1 king. After you do this, THEN you are finally done.

So in mathematical expression ,
1-C(4,0)C(48,4)/C(52,48)-C(4,1)C(48,3)/C(52,48)-C(4,2)C(48,2)/C(52,48)
-C(4,3)C(48,1)/C(52,48) = 1-270724/270725=1/270725.
Here is the correct answer, again same as the answer from my approach. The number of ways you can draw 3 kings, 2 kings, 1 kings, or 0 king is 270724... and the total number of you can draw any 4 cards is 270725. Not surprisingly, 270724 is just 1 short from 270725.... and 1/270725 is the correct answer I got above.


alright gluck studying.

 

[jay47]

Hey,
before you say someone else's answer wrong, check your own solution first. And think rationally when you do. and, my answer IS correct.

So why your answer is wrong:
first of all, think about your number rationally. 28.1% (your answer) is way too high to be realistic. Have you ever played card games? If your answer is indeed true, you get a four kings in your hands more than 1/4 times you draw 4 cards from a full deck. Doesnt sound so realistic. If it was, why would a four-of-a-kind be so rare in poker?

I did not use your method on purpose b/c that way is actually more complicated than my, or jay's way. Here is what you are supposed to do if you would like to find the answer by finding the probability of NOT getting any king in 4 draws:
If you just subtract from 1 the probability of not getting any king (0 king in other words)in 4 draws, you end up with the combined probability you will get 1 king, 2 kings, 3 kings, or 4 kings. This is why your answer is way off -- your answer was only quarter way done.

So what you should do is,
on top of subtracting the probability of not getting 4 kings, you should also subtract the probabilities of not getting 3 kings, 2 kings, and 1 king. After you do this, THEN you are finally done.

So in mathematical expression ,
1-C(4,0)C(48,4)/C(52,48)-C(4,1)C(48,3)/C(52,48)-C(4,2)C(48,2)/C(52,48)
-C(4,3)C(48,1)/C(52,48) = 1-270724/270725=1/270725.
Here is the correct answer, again same as the answer from my approach. The number of ways you can draw 3 kings, 2 kings, 1 kings, or 0 king is 270724... and the total number of you can draw any 4 cards is 270725. Not surprisingly, 270724 is just 1 short from 270725.... and 1/270725 is the correct answer I got above.


alright gluck studying.

Not to disrespect you, but shouldn't the answer be at least greater than 1/13 (or 7.69%)? I don't think that you quite understand the situation that is being presented. Probability of getting at LEAST one king out of 4 draws.

The reason I say this is because 1/13 is the chance of drawing a king in ONE try out of the deck. If I have a chance at drawing 1 king, or more than one king in 4 tries, logic would tell me that the answer would have to be greater than 1/13. I don't think that your answer makes logical sense. I believe that when you said "if you just subtract from 1 the probability of not getting any king (0 king in other words)in 4 draws, you end up with the combined probability you will get 1 king, 2 kings, 3 kings, or 4 kings." you are simply supporting Streetwolf's answer.

This is the reason I posted this question here, it seems so simple at first, but I think it is a lot more complicated than we think.

 

[TexasOMFS]

Seeunmonk is correct. The fallacy in Streetwolf's logic here is that indeed, when subtracting only the probability for drawing 0 kings from 1, one fails to recognize the probabilities of drawing 4, 3, or 2 kings. You MUST remember to account for all combinations.

 

[TexasOMFS]

In addition, what seeunmonk laid out for you above is the hypergeometric model, I believe. It's a model commonly used for drawing without replacement. And anyways I wouldn't argue with someone from MIT about statistics...they tend to know what they're doing 😛

 

[jay47]

In addition, what seeunmonk laid out for you above is the hypergeometric model, I believe. It's a model commonly used for drawing without replacement. And anyways I wouldn't argue with someone from MIT about statistics...they tend to know what they're doing 😛

I'm not doubting the seeunmok is a talented mathematician, he did get a 30 on the math, but I am doubting that he understands the question. I am only trying to get at least one king. Maybe I should explain the question more clearly.

I have a deck of cards, assuming they are randomly shuffled, I am going to draw 4 cards in a row off of the top (or hell, from the middle, or from the bottom, it doesn't matter!). What is the probability that I will get at least one king? This implies that I can get one king, or two kings, or three kings, or four kings during the four draws.

Now I KNOW for a fact after playing poker myself for several years that there is not a 1/270,720 chance (a one in a quarter million chance, lol seriously) that I will get a king (or two kings, or three kings, etc...). It does not make sense. I don't know what he is calculating, but it definitely isn't the probability that I will draw at least one king.

 

[Streetwolf]

Seriously people?

The question asks for the probability AT LEAST ONE CARD IS A KING.

That means we consider these cases:

1 out of 4 is a king
2 out of 4 are kings
3 out of 4 are kings
4 out of 4 are kings

Read again closely, you'll see.

If you want only the case of 4 out of 4 kings then I'd straight up do (4C4)/(52C4) which is what you did.

 

[orthosm2021]

man. I did read the question wrong. I was calculating getting ALL 4 kings in 4 draws.
( see? I have terrible reading skills haha)

ok if its is getting AT LEAST 1 KING in 4 draws, streetwolf has completely right approach.

and fyi, i'm not a he, im a female.

 

[jay47]

Hey,
before you say someone else's answer wrong, check your own solution first. And think rationally when you do. and, my answer IS correct.

So why your answer is wrong:
first of all, think about your number rationally. 28.1% (your answer) is way too high to be realistic. Have you ever played card games? If your answer is indeed true, you get a four kings in your hands more than 1/4 times you draw 4 cards from a full deck. Doesnt sound so realistic. If it was, why would a four-of-a-kind be so rare in poker?

I did not use your method on purpose b/c that way is actually more complicated than my, or jay's way. Here is what you are supposed to do if you would like to find the answer by finding the probability of NOT getting any king in 4 draws:
If you just subtract from 1 the probability of not getting any king (0 king in other words)in 4 draws, you end up with the combined probability you will get 1 king, 2 kings, 3 kings, or 4 kings. This is why your answer is way off -- your answer was only quarter way done.

So what you should do is,
on top of subtracting the probability of not getting 4 kings, you should also subtract the probabilities of not getting 3 kings, 2 kings, and 1 king. After you do this, THEN you are finally done.

So in mathematical expression ,
1-C(4,0)C(48,4)/C(52,48)-C(4,1)C(48,3)/C(52,48)-C(4,2)C(48,2)/C(52,48)
-C(4,3)C(48,1)/C(52,48) = 1-270724/270725=1/270725.
Here is the correct answer, again same as the answer from my approach. The number of ways you can draw 3 kings, 2 kings, 1 kings, or 0 king is 270724... and the total number of you can draw any 4 cards is 270725. Not surprisingly, 270724 is just 1 short from 270725.... and 1/270725 is the correct answer I got above.


alright gluck studying.

Ahaa! I see what he is answering, but I am not asking that question. He is finding the probability of drawing 4 kings from a deck of cards. I am not asking about drawing 4 kings, I am asking about drawing at LEAST 1 king. Therefore all I have to draw is 1 king out of the four tries for it to be successful.

 

[jay47]

man. I did read the question wrong. I was calculating getting ALL 4 kings in 4 draws.

ok if its is getting AT LEAST 1 KING in 4 draws, streetwolf has the right approach.

and fyi, i'm not a he, im a female.

Lol, I just saw what you were answering too. Thanks for the excellent explanation though, I appreciate it.

 

[TexasOMFS]

Ah yes, I stand corrected.
My apologies, I read the problem as drawing ONLY one king in 4 tries, for which Seeunmonk's answer would've been wrong anyways.

 

[orthosm2021]

yup. my answer was for finding prob of getting 4 kings in 4 draws. lol. man, i just can't read....

And FYI, for TexasOMFS's understanding of the question (finding ONLY 1 king in 4 tries), you quickly fix the equation in my quote to below:
1-C(4,4)C(48,0)/C(52,48)-C(4,3)C(48,1)/C(52,48)-C(4,2)C(48,2)/C(52,48)
-C(4,0)C(48,4)/C(52,48)

These card probability questions become very simple once you get use to this operation. 🙂

alright gnite

 

[Streetwolf]

To find 1 king in 4:

(4C1)(48C3) / (52C4)

 

[orthosm2021]

To find 1 king in 4:

(4C1)(48C3) / (52C4)

yup. this is indeed easier (in fact the best) approach for TEXASOMFS's question. I was just showing how to do it in the reverse way.
For problems like getting only 1 card out of 4 trial, finding probability of NOT getting something something and then subtracting from 1 is a lot more complex than doing it this way as he did here. i mentioned this above. (in the context of the original question, however, subtracting from 1 is the best approach. So hausee and streetwolf's approach is indeed the best one for the original question)

 

[jay47]

To find 1 king in 4:

(4C1)(48C3) / (52C4)

Yeah, now you guys are making sense.....I was really going crazy for about 10 minutes because I knew that it couldn't be right....This makes perfect sense in conjunction with what Seeunmok was saying earlier.
Thanks.

 

[hausee]

Yeah, now you guys are making sense.....I was really going crazy for about 10 minutes because I knew that it couldn't be right....This makes perfect sense in conjunction with what Seeunmok was saying earlier.
Thanks.

🙄 didnt realize the post can go this far. Should've just post the solution ....

 

[jay47]

🙄 didnt realize the post can go this far. Should've just post the solution ....

I didn't have an answer, that is why I was asking. It came up at a family reunion and I blanked out.

There are some posts on SDN that are dozens of pages long, this one is relatively short.