The Mathematics Behind the Birthday Game

For those of you who've ever been to the Stampede, you should be familiar with the Birthday Game. That's the one where you place a buck or two in the square with the name of your favourite month, and hope the 14-sided die will show the same month. Some of you are still placing money in one square at a time, while some of the shrewder ones have decided to cover all the bases by buying up all twelve months plus New Years and Christmas. I'm here to discuss strategy for this particular game, so hopefully next year you'll get a better return on your investment.

First, we need to make an assumption regarding the die. In order to keep things simple, we'll assume that we have in our hands a fair die; that is, the outcome of a roll is truly random and each face has an equal chance of landing on top on any given roll. This means the probability of each outcome is assume to be P(NY) = P(Jan) = P(Feb) = ... = P(Dec) = P(X'mas) = 1/14.

Now a quick refresher course in probability.

1. P(2 faces) = 2 x P(1 face) = 2/14 = 1/7 (Probability of winning if we buy two months)
2. P(7 faces) = 7 x P(1 face) = 7/14 = 1/2 (Probability of winning if we buy seven months)
3. P'(1 face) = 1 - P(1 face) = 1 - 1/14 = 13/14 (Probability of losing if we buy one month)
4. P(1 face & 1 face) = 1/14 x 1/14 = 1/196 (Probability of winning twice in a row if we buy one month each time)
5. P(1 face 1 face) = 1/14 x 13/14 + 13/14 x 1/14 = 26/196 = 13/98 (Probability of winning once if we play twice, buying one month each time)

As we already know, buying all 14 squares results in P(14 faces) = 1. This is the safest route for somebody simply looking to score a stuffed animal for his girlfriend who's throwing a tantrum, but it's certainly not the wisest investment.

Let's begin by splitting up our bets into halves. That means (2) where we now have a 1/2 chance of winning in each round by buying seven squares. That also means we can now play two rounds with the same amount of money. Combining (2), (4), and (5), we see that:

P(7 faces & 7 faces) = 1/2 x 1/2 = 1/4 (Probability of winning twice in a row)
P(7 faces 7 faces) = 1/2 x 1/2 + 1/2 x 1/2 = 1/2 (Probability of winning once out of two rounds)

Voila! By introducing a 1/4 chance of not winning anything, we've also introduced a 1/4 chance of winning twice in a row, while keeping the chance of winning at least once at a respectable 3/4.

What if we took it one step further? How about three rounds of 5 squares? This requires us to buy a total of 15 squares, one more than in previous cases, but let's just see how our chances are:

P(5 faces) = 5/14
P(three wins) = (5/14)^3 = 125/2744 (~4.56%)
P(two wins out of three) = (5/14 x 5/14 x 9/14) x 3 = 675/2744 (~24.6%)
P(one win out of three) = (5/14 x 9/14 x 9/14) x 3 = 1215/2744 (~44.3%)
P(no wins) = (9/14)^3 = 729/2744 (~26.6%)

By introducing a further 1.6% chance of not winning, we see that we've bought ourselves a nearly 5% chance of winning three times in a row. Granted, this is a rather slim chance, and you have to be pretty lucky to win three stuffed animals for that tantrum-throwing girlfriend of yours, but your chances of winning at least once remains relatively unchanged at around 73.4%, fairly close to the 3/4 (75%) in the previous case. So, in my opinion, no reason to be parsimonious over two bucks, and who knows, maybe you'll be the one in twenty who gets to go home with three stuffed animals and one happy girlfriend.

Onto four rounds of 4 squares: (that's 16 squares total)

P(4 wins) = 256/38416 (~0.666%)
P(3 wins) = 2560/38416 (~6.66%)
P(2 wins) = 9600/38416 (~25%)
P(1 win) = 16000/38416 (~41.65%)
P(no wins) = 10000/38416 (~26%)

The improved odds over the last (3 rounds x 5 squares) scenario is partly due to the fact that we're now buying 16 squares total. Nevertheless, the improvement in the odds of three wins out of four makes this an attractive option, since the odds of winning at least once is also increased to just under 74%.

Of course, I've tried to keep the total number of squares as close to 14 as possible here for the sake of comparisons, but there's no stopping you rich folks from increasing the stakes now is there? So what if we decided to increase the stakes by 50% over two rounds? That is, let's buy 9 squares each round instead of 7. So here's two rounds of 9 squares:

P(9 faces) = 9/14
P(two wins) = (9/14)^2 = 81/196 (~41.3%)
P(one win) = (9/14 x 5/14) x 2 = 90/196 (~45.9%)
P(no wins) = (5/14)^2 = 25/196 (~12.8%)

By not scrimping on your stakes, we've cut the winless chances by nearly half, simply by upping your stakes by less than half (from 14 to 18). What if we went a little further and did 10 squares each time for a total of 20?

P(10 faces) = 10/14 = 5/7
P(two wins) = (5/7)^2 = 25/49 (~51%)
P(one win) = (5/7 x 2/7) x 2 = 20/49 (~40.8%)
P(no wins) = (2/7)^2 = 4/49 (~8.16%)

Eureka! We now actually have better than one in two chances of winning twice in a row. And still we've upped our expenses by less than 50%!

I hope this entry has helped you rethink your strategies for the next time you visit the games at the Stampede. Certainly, these carnies have ripped us off collectively long enough. It's time to play the odds and try to win one back for the masses. Or two. Or three.

P.s. Of course, as I stated at the beginning, I've assumed a fair die for simplicity. To improve your chances further, I would suggest you try to observe the patterns of the die. I've heard reports of certain faces showing up more often than others. Use this to your advantage when deciding which squares to buy; this will increase your chances even more!