 Show The denominator is 52! i believe. The number of ways to arrange 52 cards. The numerator I'm not 100% on. If we consider all four aces to be one fixed element then we have 48 non-ace cards and one unit of 4 aces stuck together. Then we have 49! as the numerator. So I got 49!/52!. Is this on the right track. There is no answer in the back of the book for this problem. Thanks! level 1 You're close. Also the aces can appear in any of 4! orders, so 49!*4!/52! level 2 · 3 yr. agoCollege Instructor/M.S. Mathematics I got the same answer however I thought of it as: (4/52) * (3/51) * (2/50) * (1/49) is the probability of the 4 aces being the first 4 cards but there are 49 ways the aces could be together (1 - 4, 2 - 5, 3 -6, ......., 49 - 52). level 1 · 3 yr. ago · edited 3 yr. agoNew User 49(4!)(48!)/52! There are 49 ways to place the set of aces in the deck. Edit: 1.8x10-4 level 1 You almost got it, but don't forget that the access are different so they have an order too, so that would multiply by 4! $\begingroup$ A deck of 52 cards is shuffled thoroughly. What is the probability that the four aces are all next to each other? What I tried To choose the first ace I took $ \left( \begin{array}{c} 13 \\ 4 \\ \end{array} \right)/\left( \begin{array}{c} 52 \\ 4 \\ \end{array} \right)$ Is my working correct. Could anyone explain. Thanks
peterh 2,6196 gold badges21 silver badges37 bronze badges asked Aug 24, 2015 at 16:46
$\endgroup$ 2 $\begingroup$ We have $52!$ different shuffles. The aces have $4!$ and the others have $48!$ different permutations. Now, we insert the aces somewhere into the permutations of the others as 1 card, we have $49$ places to to that (before the first... after the last), thus the answer is $$\frac{49\cdot48!\cdot4!}{52!}=\frac{24}{50\cdot51\cdot52}$$ answered Aug 24, 2015 at 16:53
Ákos SomogyiÁkos Somogyi 1,7278 silver badges17 bronze badges $\endgroup$ 1 $\begingroup$ Possible positions for a stack of 4 aces in the {interstices + ends} of the 48 other cards = 49 against total possible positions for aces of ${52\choose 4}$ $$\text{Probability} = \frac{49}{52\choose 4} = \frac{1}{5525}$$ answered Aug 24, 2015 at 18:32
true blue aniltrue blue anil 31.6k4 gold badges25 silver badges48 bronze badges $\endgroup$ $\begingroup$ For the whole deck there are 52! permutations. In how many of these the four aces stick together: answered Aug 24, 2015 at 17:02
Shashi DwivediShashi Dwivedi 3011 gold badge2 silver badges8 bronze badges $\endgroup$ What is the probability you draw 4 aces from a deck of 52 cards?Explanation: There are 4 aces in the 52-card deck so the probability of dealing an ace is 4/52 = 1/13. In a 5-card hand, each card is equally likely to be an ace with probability 1/13. So together, the expected number of aces in a 5-card hand is 5 * 1/13 = 5/13.
What are the odds of 4 aces in a row?The number of outcomes that have four aces in a row is 4! Save this answer. Show activity on this post. The correct answer to the question posed is: The probability is 1.
What is the probability to get 4 ace in a deck of cards?The chance of drawing one of the four aces from a standard deck of 52 cards is 4/52; but the chance of drawing a second ace is only 3/51, because after we drew the first ace, there were only three aces among the remaining 51 cards.
How many ways can you draw 4 aces from a deck of well shuffled cards?Answer and Explanation: Therefore, there are 270,725 ways to draw 4 aces from a deck of cards.
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