Cards Probability Question
From a well shuffled deck of 52 playing cards(no jokers) ,a person draws out cards one by one until he picks all 13 hearts in the deck.
On which turn is he most likely to pick the last heart? Don't give me 52 as the answer bcos the question also includes the probability that he reaches that turn to pick the 13th heart. 
Can't any gurus answer this question

I would venture that the final heart is most likely to be drawn in either 49th, 50th, 51st, or 52nd position (each occurring with equal probability).
In an infinite serious of trials, probability will ensure that after 12 hearts are drawn, 12 of each of the remaining suites (clubs, spades, diamonds) are drawn as well. This gives us 48 cards out and 4 remaining. It is not possible to determine whether or not the case heart will necessarily be drawn in the next card but at this point there is a 25% chance that it will be drawn in 49th, 25% for 50th, 25% for 51st, and 25% for 52nd. Without hard math, common sense would suggest that the final heart will most often be found in this 4card grouping (remaining 4) when compared with similiar groupings occurring before 48 cards are selected. Any thoughts? 
Quote:
arunbg, what have you tried? If you show some work, you are much more likely to get responses. By the way what do you mean by "Don't give me 52 as the answer bcos the question also includes the probability that he reaches that turn to pick the 13th heart."? Specifically the bold part, I don't see what you're getting at. 
I do think that HokieBalla34 is on to something. Doesn't it seem reasonable to assume that over many draws the probability is that the hearts come off in a regular pattern and are evenly distributed?
If we wind up with one heart in the last four cards we have: 1/4 to get it the first time, but 3/4 of the time we miss giving: 3/4x1/3=1/4 on the second draw, and so forth: 1/4 + 3/4(1/3)+1/2(1/2) + 1/4(1)= 1 or the total probability in the four draws. In fact, as a general case, if we have one heart among n cards, we have 1/n chance of picking it up immediately, and (n1)/n chances of not, but we might get it on the second try: [tex]\frac{n1}{n}*\frac{1}{n1}=1/n[/tex] and so on down the line, [tex]\frac{n2}{n}*\frac{1}{n2}=1/n [/tex] etc... So that the probability is the same, 1/n, for each of the n possible cases. Or as Laplace would see it, probability consists of sorting things into equally probable cases. For cards, that could consist of all permutations. If we have 1 heart in n cards, then there would be n permutations which consist of putting the one heart in one of the n choices. 
Quote:
As for the other replies regarding an infinite no of trials, I am compelled to think that there might be a lead there but can someone give me concrete mathematical backing for this? Thanks guys 
Quote:
Let's call P(m) the probability your mth draw is the 13th heart, so P(1)=P(2)=...=P(12)=0. We can work P(m) out directly in the other cases. In order to draw the last heart at pick m, at m1 you must have picked 12 hearts. Find the probability of picking 12 hearts with m1 picks (this is a little simpler as we don't care where in these m1 picks these 12 happen). Given 12 hearts in hand after m1 picks, you can find the probability the next pick is a heart. there's absolutely no reason to consider an "infinite" number of trials or some kind of limit as there are a finite number of outcomes. 
Allow me to posit that the opposite question might be easier to study.

Quote:
I was thinking about maximising the obtained probability function for nth turn to get the most probable turn. what do u think. 
I think if you find what I called P(m) it's not hard to find where it's max occurs, you could always just calculate P(m) for m=13, 14, ... 52 if you can't see any other way of maximaizing P(m). What did you get for P(m)?

There is a way to get an exact answer without regard for probability. Just construct the rectangular row. Say in the case of n=6, looking for two hearts, we proceed:
123456 xx xsx xssx xsssx xssssx sxx sxsx sxssx sxsssx ssxx ssxsx....ect. (An x represents a heart and a blank representing a space, and where I had to put some s's for space, since it condensed it otherwise.) In this particular case we will have to continue for 6C2 = 15 rows within our 6 columns. So, therefore you can do 15 lines, say using Excel, and determine the total matrix for all outcomes based on an equal probability for each case. Then just add up the cases. 
Enumerating all possible outcomes is not reccomended. You would have to look at all possible ways to distrubute 13 hearts in a deck of 52, i.e. 52C13=635,013,559,600, which will take up alot of space.

Once you get into these things, there is an easy way to see the diagram. (It does help to do some drawing.) We are going to have 52 slots available for 13 hearts. We put one heart in the left most, or first, slot. Then we distribute 51C12 rows of hearts all over the remaining necessary spaces.
To get the next case, we leave the first slot blank, put a heart in the second place, and distribute 50C12 hearts over the remaining cases. We continue on down until we have 12 hearts to put in 12 slots, or 12C12=1. Now we turn the situation around and call the first slot the last. Now lets look at the actual situation, which seems to work out: [tex]\frac{\sum xC12, x=12...51}{52C13} =1[/tex] This gives the final ratio in the 52 slot as the highest at: [tex] \frac{51C12}{52C13}=1/4.[/tex] 
Quote:
I do not believe that you can dissect which position the 13th heart is the most likely to occur in. Instead, I believe that it would only be possible to determine the GROUP of cards it most likely appears inand my conclusion is that it is most likely to occur in the group that consists of positions 49,50,51,52. 
I'll make my hint more explicit!
It's much easier (but still not trivial) to deal from the bottom of the deck, and ask when you find the first heart. Quote:
It's a perfectly well defined question, with a perfectly straightforward (if tedious) solution. The only interesting part is to find a good shortcut to speed things up, or a good bit of theory to promote understanding. And yes, the 13th heart is more likely to be at the bottom of the deck than at any other particular position. (Of course, it's more likely not to be at the bottom of the deck than it is to be at the very bottom) 
Hurkyl: It's much easier (but still not trivial) to deal from the bottom of the deck, and ask when you find the first heart.
What we could have done is constructed a computerized robot that would endlessly shuffle and deal. Then we could search its memory banks and determine the hearts in the 52nd place. BUT, we could have had the robot deal from right to left and then considered the first card as the last. BETTER YET, we could ditch the robot, and not check anything. Just ask yourself what is the heart probability for the first card to be dealt? 
Quote:
nCr here means the usual number of ways to select r objects from n. 
Quote:
So for P(n) we can get a series like P(n)= 39/52 * 38/51 * 37/50 ..... (n1 terms) * 13/(52n+1) where n <= 40 Here P(n) refers to picking up the first heart when the deck is dealt from the bottom. Note that the last term becomes 1 when when n=40 ,ie if he reaches the 40 th turn he is sure to pick the first heart. But the overall probability is the least. I think the maximum comes when n=1 where P(n) = 1/4 but I am not sure. I'll try working it out in C++. Thanks for the feedback guys 
All times are GMT 5. The time now is 04:47 PM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums