jeffcole2003oct's post gives you part of the answer for the case where
you keep your winnings (and in that case the strategy is obviously to
keep picking until you hit a winner).
I've also examined the situation where you lose all your winnings in
case you hit a dud (in which case the game is mathematically slightly
more interesting as there's an actual element of "strategy").
I haven't done that kind of math for a while, so it'll be good if
someone double-checks my reasoning.
You've got the numbers right for the probability of not losing in the
first n picks:
(12*11*...*(13-n)) / (15*14*...*(16-n))
A more traditional way to reach that number is to count the number of
ways to not lose in the first n picks (which is the number of ways to
pick n out out of 12, C(12,n)) and the total number to make n picks
(which is C(15,n)).
The probability of being a winner after n picks is (12!/n!/(12-n)!) /
(15!/n!/(15-n)!), i.e. (15-n)!/(12-n)!/2730, i.e.
(15-n)*(14-n)*(13-n)/2730.
The maximum number of picks that a player should take is the value of
n that maximizes n*(15-n)*(14-n)*(13-n). Amusingly, both 3 and 4
maximize the value.
The expected value of the game is $100*3960/2730, i.e. approximately $145
The 3-pick strategy has a 132/273 chance of winning $300 (i.e.
approximately 48.3%). The 4-pick strategy has a 99/273 chance of
winning $400 (i.e. approximately 36.2%). The 3-pick strategy obviously
has a lower variance.
If you decide to go for the 3-pick strategy, the expected number of picks is:
1*3/15 + 2*12/15*3/14 + 3*12/15*11/14 i.e. approximately 2.48
If you decide to go for the 4-pick strategy, the expected number of picks is:
1*3/15 + 2*12/15*3/14 + 3*12/15*3/14 + 4*12/15*11/14*10/13 i.e.
appproximately 2.91
JBQ
···
On Nov 18, 2007 3:28 PM, <GRAYTLEEGRAY@aol.com> wrote:
I am trying to grasp a concept for a Game Show type promotion. Maybe someone
could help?
You are selected to play this game. There are a bunch of squares on a board.
Some are duds but most are prizes to start. You keep playing and picking
squares, hitting prizes, until you hit a dud. Pick a winning square to continue.
For an example there are 15 squares to start and 3 are duds. There is a 3/15
chance that the game will be over on the first round. Also a 12/15 chance
that it will continue, but with a 3/14 chance the game will end on the next
round. On the last round there will only be a 1/4 chance to complete the 12 rounds
to win the complete prize of 12 units.
So what is the expected number of rounds to be played in this game? What is
the expected value if each win is equal in value?
In this example I get:
80% chance to win the first round (12/15)
48.35% chance to finish the third round (12/15x11/14x10/13=44/91)
successfully,
down to .22% chance (1/445) of completing the 12 rounds.
So am I on the right path? The question is just what is the average Expected
Value of such a promo if you win one unit for each level? Unit=$100.
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