The promotion comsists of 7 high hands a day for 7 days for a total of 49 hands.
Every winner draws a card from the deck. At the end of the promo for the week, there are 3 cards left.
The A is worth 5k.
What are the chances of that card not being drawn?

Part 2: what are the chances of tne card not being drawn going into the last day with 7 drawings left for a total of 10 remaining cards.
Thank you!

Sorry! Forgot the 2nd part: The A was not drawn the 1st week.
So now they put another A in the deck & there is a total of 53 cards & we go another week for another 49 drawings.

What are the chances of neither A being drawn going into the last day after 42 drawings & 11 cards left?

Thanks again!

.

C(51,49)/C(52,49)=(51!*49!/2!)*(3!/52!*49!) =3/52=0.0577

Don't understand all that! Why not just take the 3 cards left & divide by the total in a deck & you come up with .0577.
That doesn't mean the A is going to be one of the 3 cards.

for a 'non-math guy' by a 'non-math guy'


simplify the problem to be able to do it by 'manual' counting, then extrapolate a formula for that 'manual count' and see if its the same formula as statmanhals.



we have 6 pieces of fruit, 5 oranges and 1 apple, if you pick pieces of fruit until you have 4 pieces, how often is the apple not picked?

picking 1 at a time or picking all 4 at once makes no difference, either way you wind up with 4 pieces from which you will look to see if an apple is present.

to differentiate the oranges from one another they are labeled A,B,C,D,E the apple is labeled X.

all the ways you can pick 4 fruits

ABCD
ABCE
ABCX
ABDE
ABDX
ABEX
ACDE
ACDX
ACEX
ADEX
BCDE
BCDX
BCEX
BDEX
CDEX

the ones that don't include an X are the ones where the apple wasn't picked

ABCD
ABCE
ABDE
ACDE
BCDE

the apple is not present in 5 of the 15.

the chance the apple won't be present is 5/15
5 out of a possible 15 or .33



making this manual approach into a formula:

to arrive at the 15 using combinations is;
how many ways from a set of 6 can we choose groups of 4;
which is written C(6,4) which can be read as 6 choose 4
(6 choose 4 can be googled for a result)

to arrive at the 5 ;
how many ways from a set of 5 can we choose groups of 4;
is 5 choose 4 =C(5,4)

its a set of 5 and not 6 because we removed the apple(x).
because we are looking for the ways the groups of 4 can be picked that doesn't include an apple.

to determine the probability;
C(5,4)/C(6,4) = .33

extrapolate to the problem presented with a set of 52 cards and a selection of 49 to determine how often 1 specific card is not selected
rather then a set of 6 fruit and a selection of 4 to see how often 1 is not selected;

C(51,49)/C(52,49)