How do you do an equation using a definite and indefinite loop?? How many cards must you draw from a deck of 52 cards to have at least a 50% chance of getting two aces I have the answer! I like sharing equations with others. If you have any please share. Is there someone out there in equation land that can help me?Here is my original equation exactly as I put it into the equation solver:
(Probability in n draws)
P=12x(N-1)x(FACT(#)/FACT(#-(N-2)))/(FACT(##)/FACT(##-N))x100 I have tested it and it works perfectly For using 52 cards and drawing e.g. two aces this is the equation formula: Probability in n draws = 12(n-1)*P(48,n-2)/P(52,n) Therefore(12 is a constant) (-1 is a constant) & (-2 is a constant) (N=number of cards drawn from the deck without replacement) (#=48) (##=52) Therefore P N # ## are the variables
If N=2 P=.45 next N=3 P=.87 next N=4 P=1.25 add up all the P's=2.75 Now using a loop variable I want to add N=2 to N=30 the answer should be 80.2 (Please teach me how to create a definite and indefinite loop its on page 25 of Technical Applications HP-27S & HP-19B my problem is I don't understand the equation example. But, If you use MY equation example I will be in a position to understand.)
I know that by using the GET function e.g. P=G(P)+2x(N-1)x(FACT(#)/FACT(#-(N-2)))/(FACT(##)/FACT(##-N))x100 All I have to do is keep entering N=2 then N=3 etc. My objective is to learn how to use the syntax of the equation solver. Please help if you can. Thanks.
By the way My orginial equation can be slightly modified for using 104 cards (2 decks)The equation formula:Probability in n draws = 2(n-1) x
P(102,n-2)/P(104,n)e.g.Probability in n draws to get e.g. two queens of spades. Probability in n draws = 2(n-1)*P(102,n-2)/P(104,n) the only constant that needs changing is 12 to a 2.Therefore P N # ## are the same variables in this case (#=102) (##=104)
I'm using a HP-17BII
Regards,
Michael
How do you do an equation using a definite and indefinite loop?? How many cards must you draw from a
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