According to David Sklansky the odds of drawing 3 aces in a game of 7 card stud is 1/425, where as by my calculations it would be 1/5525 (4/52x3/51x2/50). I see that 5525/13 is equal to 425, but i do not see where the 13 comes into play. Could someone please explain?
Hmm.
I know for certain your number is wrong, at least the way you have phrased it. The odds of drawing three aces in three straight draws from a freshly shuffled deck are indeed 1 in 5525, as you have calculated, but the odds of getting three aces in SEVEN CARDS - it’s 7-card stud, right? - are obviously a lot better.
Well, I’m not going to work it out, but I think you’re ignoring the fact that you have several chances to draw that first ace. You’re also ignoring the fact that YOUR cards are not drawn consecutively.
The first point is relevant. The 2nd is not in the least.
Horsecrap. How can the fact that 3 other players are also drawing cards NOT be relevant???
Despite the uncivilness of your 2nd post, I will reply anyway. But, please, this is elementary probability and not GD.
Let’s say Nametag is the 2nd to be dealt of 4 players. This means that after the cards are shuffled and cut, Nametag gets cards 2, 6, 10, 14, 18, 22, and 26 in the deck.
I sneak in and right “Nametag” on those cards. Take the deck to China for a month, broadcast the deck on “Larry King Live”, write a rap song about the deck giving the cards out in order, mix the deck up a couple dozen times and finally return the deck to the poker table.
Now re-arrange the deck anyway you want. Put the marked cards on the top, the bottom, whatever. Have the dealer deal out the cards anyway he pleases. Face up, going in any direction, from any part of the deck. One rule: just the ones with “Nametag” on them are dealt to Nametag. Note that Nametag gets the exact same cards as before.
Notice how it’s the order of the cards in the deck after the shuffle that matters. The order of being dealt after that fact is all that counts. Put Nametag’s card on the top and deal those first.
Note that the cards being initially 2,6, etc. in the deck is irrelevant. Any 7 cards anywhere in the deck, dealt in any order will do. The only reason poker players insist on a fixed order of dealing is to discourage obvious cheating.
I strongly suggest that those with limited knowledge of elem. prob. keep the tone of their posts mild.
I’m not going to multiple this out but the method is to determine the number of possible hands where 3(and only 3) aces are present(numerator) and divide by the total number of combinations of 7 card hands(denominator)
The general combination formula is:
n!/((X!(n-X)!)
The numerator would be the product of combinations of three (and only 3)aces(n=4,X=3) times combinations of the four other cards(n=48,X=4).
You would divide all that mess by the general combination formula
where n=52,X=7.
Apologies if this comes twice.
I see aahala’s posted method while I wrote this, but here’s another way.
I don’t have a calculator handy, but the solution to the OP is that this is a hypergeometric distribution (sampling without replacement) with a population N = 52 and successes k = 4, with a sample size n = 7.
The probability P(x) = [ C(k,x) * C(N-k, n-x) ] / C(N,n)
where C means ‘combination’, i.e. C(a,b) = a!/( b!(a-b)! )
In this case, we want P(x=3).
I have now compared the odds of drawing three aces in a 7 card hand and a post concerning it appearing on this board within 5 minutes of hitting the submit buttom. I believe the former is a better bet.
The chance of drawing at least 3 consecutive aces out of a run of 7 cards from a 52-card deck is 5 times the figure you give, because your 3-card run can start in any of the first five positions. The correct figure is 1 chance in 1105.
The chances of drawing 3 aces in any position within a seven card run are 35 times your figure, since the three aces can appear in any of 35 slots. (7x6x5/3x2). This figure would be about 1 in 158.
I don’t know anything about seven card stud, so it may be that the figure cited by Sklansky doesn’t correspond to either of these situations.
It does. Drawing 3 aces and drawing 3 consecutive aces are different.
No, ftg is right ( see his second post). Unless we are told something about these other cards( e.g., there are four other players and none of their face-up cards is an Ace), they don’t affect the probability. The probability that the first three cards are Aces is the same as the probability that the first, fifth and ninth are all Aces.
This is how I figure it. The number of sets of seven cards in which exactly 3 are aces is:
(number of choices for what the three aces are) X (number of choices for what the four non-ace cards are).
Letting C(n,d) = n! / (d!*(n-d)!), the above number is C(4,3) * C(48,4). Since there are a total of C(52,7) possible hands of 7 cards, the odds of getting one in which exactly three of the cards are aces is
C(4,3) * C(48,4) / C(52,7)
If I’ve computed correctly, this works out to be
4 * 48! * 7! * 45! / (4! * 44! * 52!) = 4 * 7 * 6 * 5 * 45 / ( 52 * 51 * 50 * 49) = 37800 / 6497400 = 0.0058177117
This differs from 1/425 = 0.0023529412. I don’t know if I made a mistake, or if I should have not counted those hands in which the non-ace cards contain two-pair or other valuable hands.
Right. If you don’t know what cards have been dealt, then you can only assume that the individual cards have been dealt in accord with the probabilities so the original probabilities should still be maintained.
The deal is just a method of continuing the shuffle. I seems to me that the probability wouldn’t be changed if player 1 were dealt all of his cards, then player 2 all of his, etc.
Correct me if I’m wrong, but the actual actual figure would be ever-so-slightly lower than this one, because hands which have all four aces in a row are counted twice.
If you do not require 3 consecutive aces, yes.
Indeed, you are correct. This lowers the chances from 1 in 1,105 to 1 in 1,123.
I know it’s a little late to mention this, but I noticed that I’d read too quickly on preview – my method and aahala’s are of course exactly the same.
By way of atonement, I calculated the odds of getting only a 3A hand.
Using the calculation of 3A hands(shown correctly by Tyrell McAllister), there are 778,320 7-card hands with exactly 3 aces. (probability : ~1/172)
Of these,
[ul]
12 are non-A four of a kind – kind of trivial, but 12 ranks * 1 way to get dealt 4 of a kind;
1485 are flushes – C (12 cards in a suit,4 cards) * 3 suits (only 3 since one A completes it);
81 are A-high straights – 3[sup]4[/sup] (4 cards, 3 suits);
but 3 of those are royal flushes, which we already counted, so don’t count them twice;
77832 are non-A pairs – C(4,2)*C(47,2) (ways to make a pair * combinations of remaining cards);
of which 2376 are two-pair hands, so strike them off – C(12,2)*C(4,2)*C(4,2) (Combinations of 2 pairs * ways to make a pair from 4 suits for each pair)
[/ul]
leaving 706,041 7-card hands which have 3A as the best possible hand. Dividing by the total number of 7-card hands (C(52,7)=133784560), we end up with 0.00527744, or ~1/189.
Hopefully I haven’t made a mistake there. However, I still can’t see where Mr. Sklansky’s numbers came from, other than that 1/425 is the odds of being dealt three of a kind in three consecutive cards (as already pointed out).
Well I made at least two mistakes there.
My calculation of the 77832 pair hands included the two four-of-a-kinds with the other ace. So two less hands from that.
In addition, the numbers don’t add up (I’d made a mistake with the two-pair hands and then added instead of subtracting. )
Taking that into account, I get 701,291 hands, yielding 0.00524194, or ~1/191.
At least until another mistake is found.
I take it you may have thought we had included cases of 4 aces in the hand in our method. We didn’t. There are four combinations of three cards and our result is the same as Tyrell’s.
Whether the OP meant to include the cases of 4 aces may be subject to some interpretation, but the three of us thought it excluded such cases.