## March 24, 2007

### The cost of Safeway’s Score & Win contest.

I’ve often wondered what the odds of winning the Score & Win contest's \$1,000,000 grand prize. Every game you see the ad and know that there is almost no chance of actually winning it (especially with the Canucks who can’t even score 5 goals combined). I can’t imagine what it’s like to know you’re the one who has a chance at a million though. Presumably they’ve got some accountant type calculating the odds and making sure they are able to pay award with the advertising if such an unlikely even occurs. So I’m just going to provide you with the accountant’s math so you all know the chances of winning the big prize.

First I looked at 2002-2003 to 2005-2006 data to get this list:
Probability of scoring exactly 1 goal given you scored at least 1 goal
= 0.89279 = P(G=1|G>0)
Probability of scoring
exactly 2 goals given you scored at least 1 goal
= 0.09583 = P(G=2|G>0)
Probability of scoring
exactly 3 goals given you scored at least 1 goal
= 0.01043 = P(G=3|G>0)
Probability of scoring
exactly
4 goals given you scored at least 1 goal
= 0.000953 = P(G=4|G>0)

Using the assumption that goal scoring is Poisson, which is proved to be reasonable by Ryder, implies P(G=X) = exp(-k)*k^x/(x!).

Here’s where it might get confusing. [P(A|B) = P(A and B)/P(B)]
P(G=1|G>0) = P(G=1)/P(G>0)
P(G=2|G>0) = P(G=2)/P(G>0)
P(G=1|G>0)/P(G=2|G>0) = P(G=1)/P(G=2)
P(G=1)/P(G=2) = [exp(-k)*k^1/(1!)]/[exp(-k)*k^2/(2!)] = 0.89279/0.09583 = 9.3

Solve for k: [with computer]
k = 0.215
You can do the same for 2->3 and get K = 0.32 and 3->4 gets K = 0.274
Since I consider 1->2 to be the most accurate I went with 0.215

The probability of an average player getting 5 goals is:
exp(-0.215)*0.215^5/(5!) = 0.000003
The probability that a given player doesn’t score 5 is 99.99969%, so the probability that none of the 18 possible scorers gets 5 is:
(0.9999969)^18 = 99.994%

Or the probability that a player does score 5 in a game is: 0.00005577

Since the prize is an annuity distributed over 20 years it’s only worth about \$680,000 when it is won (@ 4% discounting).
Multiply the costs by the odds of winning in a given game:
\$680,000*0.00005577 = \$38/game
Works out to one 5 goal game every 7-8 seasons. [Does anyone have a list of 5-goal games?]

There are about 2 million possible shoppers (exaggeration) and Safeway has a 30% market share, so there are about 600,000 entrants (assuming they all shop equally often). Making the final value to the individual viewer:
\$0.00006/game or 0.006 cents/game.

The Puck Stops Here said...

A summary of 5 or more goal games (since you asked):

7 goals: Joe Malone Quebec Bulldogs in 1920

6 goals: Newsy Lalonde Montreal in 1920, Joe Malone Quebec in 1920, Corb Denneny Toronto St Pats in 1921, Cy Denneny Ottawa Senators in 1921, Syd Howe Detroit in 1944, Red Berenson St Louis in 1968, Darryl Sittle Toronto in 1976

5 goals: accomplished 45 times in NHL history (I wont list them all). It happened 4 times in the 1917/18 season. Joe Malone did it 3 of the times. It happened 7 times in the 1920's. 3 times in the 1924/25 season. Babe Dye was the only player to accomplish this twice in the 20's. It only happened twice in the 30's and not after 1932. It happened 3 times in the 40's (two during the WWII years). It happened once in the 50's and once in the 60's. It happened 3 times in the 70's but not until 1977. In the 1980's it happened 15! times. Wayne Gretzky was the first to do it twice since Babe Dye (in fact Gretzky did it 4 times). Bryan Trottier also did it twice - once in the 70's. It happened 7 more times in the 1990's. Mario Lemieux did it 3 times (the first time in the 80's). It has not happened since Sergei Fedorov in 1996.

Anonymous said...

thoes odds make me wanna shop at buck or two