Between 23,000 and 40,000 points for $46 and a cramped hand

55

Finally, a fun(ish) promo that doesn’t require flights or nights…

On Monday, Deals We Like published a cool idea in the post “Spend $46 and Earn at Least 47,000 IHG Points!”  IHG (Intercontinental Hotels Group) has an upcoming promotion called Priceless Surprises.  Unlike most airline and hotel promotions, this one does not require travel.  The promo offers two ways to earn “Priceless Surprises”: 1) Complete two eligible stays; or 2) Mail in a hand-printed 3” X 5” piece of paper.

Full rules for both options can be found in the IHG and MasterCard®Priceless Surprises® Promotion Official Rules document.

In this post, we’ll look only at the mail-in option…

Number of entries

The official rules state “Each entrant is eligible to receive up to ninety-four (94) Game plays during the Promotion Period, regardless of method of entry.”  So, the idea here is to mail in up to 94 separate entries and rack in the points.  How many points will you get from doing so?  Let’s see…

Chance of winning analysis

Prizes are grouped into Group A, Group B, and Group C.  The Group A prizes are where you’ll find the really good stuff.  Prizes range from $50 prepaid cards to all-expense paid trips to Paris or New York.  The Group A prizes do not have fixed odds of winning.  Instead, they’ve decided how many of each prize will be awarded (3 Paris trips, 2 one million IHG point winners, 525 free night winners, 1050 $50 gift card winners, etc.).  So, your chance of winning will depend upon how many competing entries show up.

Group B and C prizes are IHG points.  The odds of winning each of these is as follows:

Prize Odds
5000 points 1:42.86
2000 points 1:28.57
1000 points 1:14.29
500 points 1:1.18

I (and many others) initially read the odds incorrectly as probabilities.  For example, the 500 point prize odds of winning is listed as 1:1.18.  If you think of this as a probability (1 out of 1.18 entries will win), then it looks like you have a 85% chance of winning this prize.  And, if you add up all of the options, it looks like you have nearly 100% chance of winning something.  In truth, though, 1:1.18 should be read as “for every 1 win, there will be 1.18 losses”.  In other words, the probability of winning 500 points is 1 / (1+1.18) = 46%.

My first draft of this post was based upon the incorrect conversion of odds to probabilities described above.  Luckily, before publishing, I read this comment by JEM in the Deals We Like post:

Hmm… If I remember my middle school math correctly, Odds and Probability have this relationship:

Odds = (# of Favorable Outcomes) : (# of Unfavorable Outcomes)
Probability = F / (F + U)

So the probability of winning on any single turn is:

P(5000) = 1 / (1 + 42.86) = 0.0228 = 2.28%
P(2000) = 1 / (1 + 28.57) = 0.0338 = 3.38%
P(1000) = 1 / (1 + 14.29) = 0.0654 = 6.54%
P(500) = 1 / (1 + 1.18) = 0.4587 = 45.87%

In other words, you can expect to win SOMETHING only about 58% of the time.

The expected point value for each turn would then be SUM(P(q) * q) for each possible outcome q.

EV(5000) = 0.0228 * 5000 = 114
EV(2000) = 0.0338 * 2000 = 67.6
EV(1000) = 0.0654 * 1000 = 65.4
EV(500) = 0.4587 * 500 = 229.4
EV(0) = 0.4193 * 0 = 0
EV(Turn) = 476.4 points per turn

which would SEEM to indicate a total Expected Value of 44,782 points over the maximum 94 turns.

HOWEVER, that overstates what’s possible. You can only win a maximum of 5 total prizes > 500, so even if you maxed out by hitting five 5K prizes (astronomically unlikely), your MAXIMUM possible value would be

MAX(5x5K) = (5 * 5000) + (500 * 89) = 69,500 points

with an approximate expected value of

EV(5x5K) = (5 * 5000) + (0.4587 * 500 * 89) = 45,412

which is just barely above the naive EV.

And if you were “unlucky” enough to hit five 1K prizes, your maximum possible value would be

MAX(5 * 1K) = (5 * 1000) + (500 * 89) = 49,500 points

with an expected value of approximately 25,412 points, or 270 points per turn.

JEM continued with:

FWIW, I modeled 94 turns with the probabilities above and the cap of 5 high-value prizes. Over a couple of million iterations, I got these results:

MIN: 17,500
MAX: 54,000
AVG: 31,657
STD DEV: 4,158

If I’m correct, that would predict that 95% of people who play 94 times will receive between 23,000 and 40,000 points, with less than 0.2% exceeding 44,000.

Good stuff from JEM!  To summarize:

  • For those who do not win the group A prizes, the chance of winning something is about 58%
  • If you “play” 94 times, you are very likely to win a total of between 23,000 and 40,000 points.

Later, ‘Dem Flyers followed up with a similar analysis (see: Before you go off writing 94 postcards…).  In this post, they came up with slightly different numbers, but the same overall theme:

  • Across all entries which do not win the group A prizes, the average (mean) result will be 476.4 IHG points.
  • If you “play” 94 times, you are very likely to win between 30,000 and 50,000 points.

DemFlyersIHGPromoDistribution

Why did ‘Dem Flyers come up with a higher range than JEM?  JEM answered in this comment:

I can’t tell for sure since I can barely read Java code, but from your data set, it appears that you’re significantly overestimating the expected value. The Ts&Cs state that you can only win a total of 5 (any combination) of the “Group B” prizes (1000, 2000, or 5000 points). After you win 5, you’ll only have the 46% chance of winning 500 points. Thus the maximum possible return is 5*5000 + 89*500 = 69,500. Your data set shows small, but non-zero probabilities all the way up to 445,000.

I modeled the contest using that constraint, and my expected value for 94 entries dropped to about 31,600 (336 points per entry, well below your stated 476) with a StdDev of 4100, so 95% of the total returns should fall within the range 23K-40K points and less than 0.2% of results should go higher than 44K.

Aha!  I had missed that, but sure enough, as JEM wrote, there is a 5 prize per person limit to Group B prizes:

Limit: One (1) prize from Group A and five (5) prizes from Group B prize per person.

Chance of winning summary

I’ll leave it to others to check JEM’s math.  It looks reasonable to me.  So:

    • The chance of winning something with each entry is about 58%
    • If you “play” 94 times, you are very likely to win a total of between 23,000 and 40,000 points.

Is it worth it?

To enter the contest without any IHG stays, you must hand write your entry on a 3 X 5 piece of paper and mail it in.  If each entry takes 2 minutes of your time and 49 cents for a stamp, then your total cost for 94 entries will be:

  • 188 minutes (let’s call it 3.5 hours, if we include breaks)
  • $46.06

In exchange for that time and expense, you’ll likely get between 23,000 and 40,000 points.  With the current Fair Trading Price of IHG points at .56 cents each, that means that you’ll likely get between $128 and $224 “worth” of points.

I’m not sure it’s worth the effort, but I’ll probably do it anyway.  I haven’t yet decided, though, whether I’ll submit 94 entries for my wife as well… or my son…  or my dog…

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