52 card deck. There are 5 people. Person 1, 2, 3, 4, 5.
First person pulls a card off the top
2nd person pulls a card off the top
3rd person pulls a card off the top
4th person pulls a card off the top
5th person pulls a card off the top
What is the probability of the 4th and 5th person pulling the 2 highest cards.
In before 3.50%
In b4 “you’re dumb you can’t do simple math
Thank you


05222020, 05:15 AM #1
Can someone calculate the probability for me srs
New York

05222020, 05:19 AM #2
 Join Date: May 2013
 Location: New York, New York, United States
 Posts: 10,540
 Rep Power: 95052
Probability is not simple! .... though this question is relatively straightforward, I think.
Probability that the first card is NOT one of the top two: 50/52
Probability that the second card is not: 49/51
Third: 48/50
The fourth person pulling one of the highest two cards is: 2/49
and the fifth person pulling the remaining one is: 1/48
Now multiply those all together, getting about 0.00075."The right to be heard does not automatically include the right to be taken seriously."
Hubert Humphrey
Training Log: http://forum.bodybuilding.com/showthread.php?t=170707741&p=1427864821#post1427864821

05222020, 05:21 AM #3
No idea, but the probability surrounding a deck of cards always boggled my mind. There are 52! (factorial) ways to arrange a standard 52 card deck. That means there are 8.06e+67 or 80,658,175,170,943,878,571,660,636,856,403,766,975 ,289,505,440,883,277,824,000,000,000,000 different ways to arrange a deck of cards. That's roughly the number of atoms in our galaxy. That means every time you shuffle a deck of cards, it's almost a certainty that it is a completely unique arrangement that has never happened before, for anyone, ever.
"Say that there exists 10 Billion people on every planet, 1 Billion planets in every solar system, 200 Billion solar systems in every galaxy, and 500 Billion galaxies in the universe. If every single person on every planet has been shuffling decks of cards completely at random at 1 Million shuffles per second since the BEGINNING OF TIME, every possible deck combination would still yet to have been “shuffled”."
https://blog.priyanshrastogi.com/cra...y7afe34c8dcb3**Doesn't have a sig crew**

05222020, 05:21 AM #4
As in picking the two highest cards in the deck being the king and queen of hearts or picking two cards higher than 1, 2, 3? Are you ranking by suits?
If the latter it’s 1/5 x 1/4 = 0.2x0.25 = 5% or 1/20.
I don’t think it’s more complex than that for what you’re asking. There’s random probability based on shuffle but each person has a 20% chance of picking the highest card at the start. Once the first is gone there are only 4 in the equation which is the Monty hall problem.Last edited by Spazzzzy; 05222020 at 05:30 AM.

05222020, 05:22 AM #5

05222020, 05:24 AM #6

05222020, 05:25 AM #7

05222020, 05:29 AM #8

05222020, 05:31 AM #9
This
Unless the question is asking the two highest cards among the 5 cards that were pulled by the group. In which case I have no f*cking idea. Good luck solving that one.
Edit:
Brah above seems to be claiming to have solved it the other way I mentioned but didn't describe how he solved it so skeptical hippoProfessional Misc Rustler
"But the people we saved, they're our legacy. And they'll remember us and then I guess we'll eventually fade away too. But that's fine. Cause we left the world better than we found it, ya know."
Sam Winchester

05222020, 05:33 AM #10
 Join Date: May 2013
 Location: New York, New York, United States
 Posts: 10,540
 Rep Power: 95052
Nonono in that case it's actually even easier (I think)you just track the number of possible arrangements overall, set that as the denominator, and then track the number of favorable arrangements as the numerator.
So like there are 52*51*50*49*48 on bottom (ways to pull five cards) and then on the top you'd do 3*2*1*2*1.
could be wrong, haven't had second coffee yet"The right to be heard does not automatically include the right to be taken seriously."
Hubert Humphrey
Training Log: http://forum.bodybuilding.com/showthread.php?t=170707741&p=1427864821#post1427864821

05222020, 05:34 AM #11

05222020, 05:34 AM #12

05222020, 05:34 AM #13

05222020, 05:43 AM #14
What’s the probability that 4 and 5 both pick higher than 1, 2, 3
1/5 x 1/4 = 5%. It’s no more complex than that if it’s a randomised draw. Every 20 times you pull the draw they will have the highest two cards.
It’s not every 1333 times you pull the draw they will have the highest two cards. Miscers trying to sound smart again...

05222020, 05:50 AM #15

05222020, 05:50 AM #16

05222020, 05:52 AM #17
But isn't the probability they'll pull the highest two cards influenced by the cards the first 3 draw?
I get what you're saying, if you have 5 people 1 of the 5 will have the highest and 1 of the next 4 will have the second highest. It's possible I'm over thinking it but when we specify it has to be the 4th and 5th person who have the two highest cards I feel like there should be more to it than just 1/5 * 1/4.
For instance, if person 1, 2 & 3 all pull 2's from the deck then the probability person 4 & 5 will have the two highest cards shoots up to nearly 100%. In other words you've have to calculate all potential scenarios? Maybe? LOL
Like I said, maybe overthinking thisProfessional Misc Rustler
"But the people we saved, they're our legacy. And they'll remember us and then I guess we'll eventually fade away too. But that's fine. Cause we left the world better than we found it, ya know."
Sam Winchester

05222020, 06:02 AM #18
Not very well it seems
It doesn’t matter how big the draw size is. Card value could go up to 10,000,000,000,000. 1/5 chance of any person picking the highest card in the draw and a 1/4 chance for the next person = 1/20 chance of it happening.
No more complex than that. It is unlikely to happen but it will happen frequent enough.

05222020, 06:04 AM #19

05222020, 06:05 AM #20

05222020, 06:08 AM #21
According to my maths its 50%. They either pull the 2 highest cards or they don't.
not srs, awaiting miscmathematician because this is way more complicated of a math question than the previous answers assume.
at least when calculating poker odds you don't take into consideration that someone might have the card that you are waiting for in the deck already. So when 2 people are playing and you're waiting for your flush on river, you are taking 52 minus opponents 2 cards minus your 2 cards minus the 4 cards on deck = 44.
13 suits in the deck but there's already 2 on the table and 2 in your hand so 9/44 is your odds of hitting your flush on river.

06212020, 05:58 AM #22

06212020, 06:08 AM #23
Top of my head with no mathematical reference / memory of my classes.
Because you want specifically person 4 and person 5 to have the highest two...
There is a 2/52 (1/26) chance that person 4 will receive the highest, or second highest, card.
Once person 4 is done, now there's only 51 cards in the deck from which the highest is derived, and person 5 will then have a 1/51 chance of getting the remaining 1/2 highest cards.
So with the above logic, it'd be (2/52) * (1/51) chance. Or .00075.
The reason I'm ignoring persons 13 (which would reduce the size of the deck upon drawing), is we're also banking on them to NOT get the highest cards. Unfortunately I believe to calculate this probably, this in itself requires its own addition to the formula...
It might look like something like this:
Person 1: 50/52 chance he doesn't get the 2 highest
Person 2: 49/51 chance he doesn't get the 2 highest
Person 3: 48/50 chance he doesn't get the 2 highest
Person 4: 2/49 chance he gets one of the 2 highest
Person 5: 1/48 chance he gets 1/2 of the 2 highest after persons 14 have gone
I think this is how I'd do it, and just multiply the fractions all together to get the result. If I'm wrong then oh well lol.
EDIT: One more thing is....we're assuming there's 52 cards that have 52 different values lol. If this is a normal deck of cards, any Aces are likely the highest value (of which there are 4 per deck). So if that's true, it completely changes the formula above lol. IE their chance to pull an ace at any point in time is 4/52, and to NOT pull an ace would be 48/52. And since we're saying "2 highest value cards", that might include King right? So the odds of getting either a king or ace is 8/52, and to not pull those is 44/52.
In other words, depending on the specifics / interpretations of what is meant by "2 highest value cards in a 52 card deck", would really change the requirements of the formula.Last edited by Visel; 06212020 at 06:24 AM.
Bookmarks