SOLUTION: We had to make a probability game in Gr 12 Data Management. We now have to find out the probability of winning and the expected return. Mine was called Multiples. Here are the rule

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Question 65521: We had to make a probability game in Gr 12 Data Management. We now have to find out the probability of winning and the expected return. Mine was called Multiples. Here are the rules:
· It costs $2 to play.
· ace = 1, jack, queen, king = 10.
· Roll the die.
· If you roll a 1 you win automatically (because all numbers are multiples of 1)
· If you roll any other number then pick the number of cards that corresponds to the number you rolled. (Example: if you roll a 4, choose 4 cards from the deck)
· If all the cards add up to a multiple of the number you rolled then you win your bet back plus $1.
· If all the cards add up to a multiple of the number you rolled, plus the number you rolled is one or more of the cards you selected, you win your bet back plus $4. (Example: if you roll a 4, there would have to be at least one 4 selected out of the 4 cards)
· If your cards do not add to a multiple of the number you rolled, you lose.
I don't get the probability Or the expected return. there must be a million cases because there are so many different sums of 2, 3, 4, 5, 6 you can get with a deck of 52 cards. Please help.:
Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!
.SEE THE GUIDE LINE ANSWER BELOW.
I AM CALLING IT AS GUIDELINE ANSWER AS IT IS LONG PROCEDURE IN A "PROJECT WORK" TYPE OF ASSIGNMENT
AND YOU SHOULD BE ABLE TO CARRY FORWARD WITH THIS SYSTEMATIC METHOD OF APPROACH
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We had to make a probability game in Gr 12 Data Management. We now have to find out the probability of winning and the expected return. Mine was called Multiples. Here are the rules:
· It costs $2 to play.
OK
· Roll the die.
OK
CASE 1
· If you roll a 1 you win automatically (because all numbers are multiples of 1)
OK·
PROBABILITY = 1/6......EXPECTED TAKING = ???NOT GIVEN..LET ME TAKE IT AS 6.
IF IT IS DIFFERENT , YOU CAN CORRECT EASILY.
HENCE EXPECTED VALUE 1 = EV1 = 6*(1/6)=1.....................1
CASE 2
If you roll any other number then pick the number of cards that corresponds to the number you rolled. (Example: if you roll a 4, choose 4 cards from the deck)
OK·
THERE ARE 5 ALTERNATIVES HERE GIVING ROLLED NUMBER OF 2,3,4,5,6 EACH WITH AN EQUAL PROBABILITY OF 1/6..FIND EVS FOR EACH ALTERNATIVE AS SHOWN BELOW.
ALTERNATIVE 1
ROLLED NUMBER =2
TOTAL COMBINATIONS OF TAKING 2 CARDS OUT OF 52..=52C2
RANGE OF TOTALS FROM 2 CARDS =2 TO 20
FOR SUCCESS TOTAL SHOULD EQUAL....2 OR 11 OR 20.....3 VARIANTS
VARIANT 1
TOTAL = 2
POSSIBLE COMBINATION...2 ACES...TAKING 2 ACES OUT OF 4 ACES=4C2
PROBABILITY = 4C2/52C2
...OR....
VARIANT 2
TOTAL =11
POSSIBLE COMBINATIONS = 1+10,2+9,3+8,4+7,5+6....
1+10.......FROM 4 ACES & 10/J/Q/KINGS TOTALING 16 CARDS = 4C1*16C1
PROBABILITY = 4C1*16C1/52C2
SIMILARLY DO FOR 2+9,3+8,4+7,5+6
VARIANT 3
TOTAL =20
POSSIBLE COMBINATIONS 10+10
2 TENS/J/Q/K ..NUMBER OF SUCH CARDS=16...TAKING 2 OUT OF 16 CARDS= 16C2
PROBABILITY = 16C2/52C2
HENCE TOTAL PROBABILITY UNDER THIS ALTERNATIVE 1=
(1/52C2)[4C2+4C1*16C1+.........+16C2]
EXPECTED VALUE = (2+1)*(1/6)*(1/52C2)[4C2+4C1*16C1+............+16C2]..........2
1/6 IS FOR THE PROBABILITY OF OCCURRENCE OF THIS ALTERNATIVE
If all the cards add up to a multiple of the number you rolled then you win your bet back plus $1.
OK
ADD UP ALL EVS OBTAINED ABOVE FOR ALL THE VARIANTS UNDER EACH ALTERNATIVE AND ALSO THE TOTAL OF ALTERNATIVES.
If all the cards add up to a multiple of the number you rolled, you win your bet back plus $4.
"I meant to say that, PLUS, if your number is included in those cards (ie. you roll a 4, there are 4 cards, but a 4 is one of the cards)"
OK
THIS IS ALREADY INCLUDED AS A PART OF THE ABOVE CALCULATION.HENCE TAKE THOSE OUT SEPARATELY MULTIPLY WITH THE ADDITIONAL RETURN YOU GET (4-1=3 EXTRA)
TO GET EV UNDER THIS CASE.
If your cards do not add to a multiple of the number you rolled, you lose.
OK
THIS PROBABILITY IS 1- TOTAL YOU GOT ABOVE WITH EV OF ZERO.
=============================================================
NOW ADD UP ALL THE EVS YOU GOT ABOVE TO GET THE TOTAL EV!!
MERRY X'MAS AND HAPPY NEW YEAR WITH THIS PROJECT!!!AND MANY MORE!!!!
YOU THINK I AM CRUEL? ANY WAY MY BEST DEFENSE OR HOPE IS THAT YOU WONT OPEN TILL THE DAWN OF THE DAY AFTER NEW YEAR!!!!!!
I don't get the probability. there must be a million cases because there are so many different sums of 2, 3, 4, 5, 6 you can get with a deck of 52 cards. Please help.
NORMALLY WE COUNT AN ACE AS 1,JACK/QUEEN/KING AS TEN AND THE REST AS PER THE NUMBERS ON THEM.IS THIS THE PRACTICE OR ANY OTHER COUNT IS THERE?
SURE..BUT FIRST CLARIFY PLEASE.YOU CAN MAIL THE SITE ACKNOWLEDGING THIS RESPONSE AND ALSO POST REVISED QUESTION IN THE SITE.
AT THE OUTSET THERE ARE LIMITED POSSIBILITIES ONLY..FOR EXAMPLE WITH
2 CARDS, SUM POSSIBLE IS 2 TO 20....SUCCESS CASES ARE THOSE TOTALING 2,11,20 ONLY
3 CARDS.....SUM....3 TO 30 ...SUCCESS...3,12,21,30...ETC....

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