amistre64
  • amistre64
spose i wanted to calculate the probability of getting 2 of a kind in a 5 card poker hand ....
Mathematics
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schrodinger
  • schrodinger
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amistre64
  • amistre64
would it be a good idea to do this: first card can be any card: 1/13 next card assume the pair: 3/51 next 3 cards different: 48/50, 47/49, 46/48
amistre64
  • amistre64
then multiply that by the number of ways to get 2 of a kind in a hand of 5?
amistre64
  • amistre64
ive gotta run to class, so ill read this later if yall want to argue amongst yourselves :)

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anonymous
  • anonymous
I looked this up on the internet, maybe you should read it,
anonymous
  • anonymous
The probability of getting two SPECIFIC jacks is the probablity of getting one of them, multiplied by the conditional probability of getting the other one given that the first one was also received, multiplied by the conditional probability of getting no more jacks other than the first two.. The first jack can go to either one of the 5 cards in the hand or to one of the other 47 spots, The probability is 5/52. If the first jack is known to be in the hand, there are only 4 spots left in the hand and only 51 cards unaccounted for, so the conditional probablity of getting the second jack is 4/51. With 50 cards remaining, and two of them jacks, the proability of getting three non-jacks out of the three remaining cards in the hand is (48/50)(47/49)(46/48). Finally, there are 6 distinct combinations of two out of the four jacks, (Think of the 4 sides of a square plus the two diagonals, or just calculate C(4,2).) So, there are 6 different mutually-exclusive ways, each with the above probability product, to get a hand with a pair of jacks. These combined to: p = 6(5/52)(4/51)(48/50)(47/49)(46/48) which is about 4%. If all you wanted was the probability of getting a hand with exactly two jacks (a pair) then that's your answer. If you want a hand that is called "a pair of jacks", then you have to rule out a two-pair hand or a full house. (3 or 4 of a kind in jacks have already been eliminated, as have been straights and flushes.) To do this, modify the product above to: p = 6(5/52)(4/51)(48/50)(44/49)(40/48) because after picking the first non jack (48 choices out of the 50 remaining cards), you cant pick that card in another suit, so the number of choices for the 2nd non-jack character has been reduced to 44 cards (52 cards - 4 jacks - 4 cards of the 3rd card rank) out of the 49 cards unaccounted for. The final product is now 40/48 for the same reason since the final card can't be jack (-4 cards), can't match the third card (-4 more cards) for (52-4-4-4=40) desirable cards out of the 48 remaining. That's about 3 and a quarter percent
dan815
  • dan815
@amistre64 you are not accounting for the 3 or 4 of a kind case
amistre64
  • amistre64
at the moment, i am just considering getting a hand that has ONLY 2 of a kind; since 2 of a kind is valued differently as a hand then 3 of a kind, or 4 of a kind ... even a full house has 2 of a kind built into it.
amistre64
  • amistre64
i had consider a possible binomial setup like this; which gives me about 4%, which is what my original idea turned up as well \[\large \binom 52~(\frac1{13})^2~(\frac{12}{13})^3\]
dan815
  • dan815
i dunno maybe im doing it wrong lol
dan815
  • dan815
the probability of getting atleast a 2 pair is way too high for my stuff
dan815
  • dan815
like 60-70%
dan815
  • dan815
lemme think about another method then
dan815
  • dan815
4-5% for 2 pair is about right though
amistre64
  • amistre64
"at least" 2 of a kind is around 50% if i trust the internet :)
dan815
  • dan815
|dw:1377701051312:dw|
dan815
  • dan815
doesnt seem that convincing to me what do you think of this method?
dan815
  • dan815
i looked at one case where its A 2 3 4 5 in order so 4^5/(52!/47!) * 13C5 * 5! <---- = number of ways to get = doubles
dan815
  • dan815
so 1 - that = 0.49 chance of getting doubles
amistre64
  • amistre64
i was thinking more along the lines of: 1 draw: 1/13 2 draw: 48/52 3 draw: 45/51 4 draw: 42/50 5 draw: 39/48 for the chances to get none of a kind, times how many ways there are to arrange that .. (13 5) ?
amistre64
  • amistre64
got my denoms off a bit
dan815
  • dan815
shouldnt your 3rd draw be 52-8 / 50 = 44/50 and 4th = 52-12/49 =40/49 5th = 36/48
dan815
  • dan815
1-(((1/13)*(48/51)*(44/50)*(40/49)* (36/48)*(13Choose5))/5!)
dan815
  • dan815
is approximately 60% acceptable?
dan815
  • dan815
http://www.wolframalpha.com/input/?i=1-%28%28%281%2F13%29*%2848%2F51%29*%2844%2F50%29*%2840%2F49%29*+%2836%2F48%29*%2813Choose5%29%29%2F5%21%29
amistre64
  • amistre64
ive got to focus on the paying job at the moment, but these are fun to read :) good effort and ideas tho, i appreciate them.
dan815
  • dan815
sure fun question :)

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