Someone asked me this today and I am stumped . A player has been dealt aces of any suit combination in a 2 card holdem hand. He/She is holding Aces in their hand. What are the chances the player will be dealt aces of any combination in the immediate next five deals. Cards are being randomly shuffled after each deal.
Asked
Active
Viewed 41 times
0
-
5Not following. if the cards are put back in the deck, which is then properly shuffled, then past hands have nothing to do with the probabilities going forward. Is that what you are asking? – lulu Mar 14 '17 at 13:54
-
Are you keeping those two aces and getting 5 new cards on top of them dealt to you to make a hand of 7 cards? – Heavenly96 Mar 14 '17 at 16:39
2 Answers
0
The chance of getting two aces on a given hand is $\frac 4{52} \cdot \frac 3{51}=\frac 1{221}=p$ and the chance of not getting aces on a given hand is $1-p$. The chance you don't get two aces on any of the next five hands is then $(1-p)^5$ and the chance you get them at least once is $1-(1-p)^5$. How can this hand affect any of the next five if the deck is shuffled?
Ross Millikan
- 374,822
-
I realize that if the deck is shuffled that each is an independent event. If I am following the answers correctly this makes this problem more straightforward ? You should decrease or multiply the probability of a single occurrence by the power of (# of recurrences) after the first one ? – I-Script-Alot Apr 12 '17 at 18:58
-
Yes. If you have independent events, the probability that something happens in each of them is multiplied. The probability of flipping $n$ heads in a row is $\frac 1{2^n}$ – Ross Millikan Apr 12 '17 at 19:42
0
Prior hand does not effect next 5
Chance of AA is 4 / 52 * 3 / 51 = 1/221
Any of the next 5 is same as 1 - none of the next 5 = 1 - (1 - 1/221) ^ 5 = 2.242% = 1 / 44
Exactly 1 in the next 5 I don't know how to do that
paparazzo
- 330