anyone good at probability / maths?

[zombie no.one]

little problem I'm trying to work out

if a standard deck of 52 cards is randomly shuffled, and then dealt out to 4 people (13 cards each), what are the odds that one person will end up with a 'straight'

i.e.

A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K

(of any suit... don't all have to be the same suit)

[Alex]

I hope this helps.

https://math.stackexchange.com/questions/999613/card-game-probability-13-card-hand

[zombie no.one]

haha, way over my head but thanks anyway... those clearly math-y guys don't even seem to agree on how to answer that OP's question. think this might be a more complex problem than I hoped.

[Trevor]

little problem I'm trying to work out

if a standard deck of 52 cards is randomly shuffled, and then dealt out to 4 people (13 cards each), what are the odds that one person will end up with a 'straight'

i.e.

A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K

(of any suit... don't all have to be the same suit)

No, sorry: I was so bad at math at school I thought 2 + 2 = 22 

[zombie no.one]

No, sorry: I was so bad at math at school I thought 2 + 2 = 22 

well I'd say there's definitely an argument that it does... depends how you look at the question really.

I mean, we take it for granted that 10 ÷ 2 = 5... but where in the question does it state that the 10 has to be divided equally?

[Rev. Powell]

This may be both an approximation and cheating, but...

https://en.wikipedia.org/wiki/Poker_probability

Odds of a straight are 0.3925%. But that excludes straight flushes so you need to add them in: 0.00139% and 0.000154% for a total probability of 0.394044%.

Assuming the chances of any player drawing a straight are absolutely independent of the others, that's 4 chances to draw a straight at 0.394044% so multiply the probability four times for = 1.576176%.

Now, if your example specifies that one and only one person draws a straight, it's a little bit less than that because you have to remove the unlikely multiple straight scenarios. I think 1.5% chance is a fine approximation, though.

[zombie no.one]

 thanks Rev 

actually I could've specified at least one person gets a straight...

1.5% seems a little high but then again I don't know, and I trust your working more than my instincts

an interesting thing to note is the odds of 3 people getting dealt a straight are exactly the same as 4 getting a straight, because if 3 have one, the only other remaining combo of cards is another straight, which the 4th person must have.

[Rev. Powell]

thanks Rev 


an interesting thing to note is the odds of 3 people getting dealt a straight are exactly the same as 4 getting a straight, because if 3 have one, the only other remaining combo of cards is another straight, which the 4th person must have.

Not at all.

Person 1: 2C-3D-4H-5S-6C
Person 2: 2D-3H-4S-5C-6D
Person 3: 2H-3S-4C-5D-6H
Person 4: 7C-7H-9D-KD-AD

for example.

[zombie no.one]

I'm talking about a straight of all 13 cards...from Ace all the way to King

if 3 people get dealt that configuration, the 4th must also have it

- to be clear, this is what I meant in my original question... what are the odds of one (or more) people getting dealt a straight of all 13 cards from Ace to King

[Rev. Powell]

I'm talking about a straight of all 13 cards...from Ace all the way to King

if 3 people get dealt that configuration, the 4th must also have it

- to be clear, this is what I meant in my original question... what are the odds of one (or more) people getting dealt a straight of all 13 cards from Ace to King

Oops! I totally misread your question---I thought we were talking about standard 5-card poker hands.

I probably could have solved your problem years ago when I was in college, I aced statistics. But today I'd have to do some research to remember the proper formulas.

[Rev. Powell]

I believe the number of combinations of 13 cards dealt to four people from a 52 card deck is

52!/(52-13)13!=635,013,559,600

https://www.calculator.net/permutation-and-combination-calculator.html?cnv=52&crv=13&x=Calculate

Now, we just need to find out how many of those are 13 card straights (a much thornier question), and we're golden!

Maybe tomorrow...

[zombie no.one]

haha, no worries - I actually didn't word my question that well, reading it back... that looks more like the kind of number I'd expect

I aced statistics. 

I see what you did there 

I'm not terrible at maths but I have a certain 'cut off point' where it becomes gobbledygook

[lester1/2jr]

I got a 440 on my math SAT so no not at all

[Rev. Powell]

Possible number of straights might be simply 13!, or 6,227,020,800.

In that case 6,227,020,800/635,013,559,600=0.0098061225714967, or about a %0.9 chance (9 in a thousand hands) for any one player to get a straight. If chances of any player getting a straight are independent (not sure if they are or not) then multiply by 4 chances for a %0.4 for any player to get a straight.

[zombie no.one]

ok nice so it's about somewhere between a 1/100 and 2/100 chance of getting dealt the full 13 card straight?