This past week, I helped an associate with a presentation in which a portion of was going to rely on what is called the birthday paradox.
Some of you may have seen this math teaser before. It essentially goes something like this: In a classroom of 23 students, what is the probability that at least two of them share a single birthday?
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I think the average person trying to take a reasonable guess at this problem will miss by a mile. This is already a significant clue to the puzzle. There are 365 days in a year (we’re ignoring leap years). If you have 23 kids and 365 days, it seems like maybe there is a 10 percent probability at most?
If you guessed more like 20 percent, you’d still have missed by that mile. There is roughly a 50 percent chance that at least two kids will share a birthday.
How can this be? Well, think of it this way. The first kid can have any day as his birthday. The second kid can have any of the other 364 days as his birthday. So, we have a 364/365 chance that two kids don’t share a birthday or a 1 in 365 chance that they do. If we throw a third kid into the mix, that one can have his birthday on any of the other 363 days of the year to keep from any overlaps.
So, now we have (364/365) times (363/365) as the probability of no overlaps, or one minus that amount for the probability of one duplicate birthdate. We still haven’t moved the needle much.
If we continue this to five kids, we get a 97.29 percent chance of no duplicates and a 2.71 percent change of at least one duplicate. When we get to 23 students, we get to the tipping point. There is a 49.27 percent chance of no duplicates and a 50.73 percent chance of a duplicate.
There is another way of looking at this problem that might explain better what is going on. When we have 23 students in a class, there’s 253 different pairings of two students. These are, in essence, the combinations we need to check to see if there are any duplicates.
Well, if you were to re-phrase the question as — what if you have 253 pairs of people, what is the probability that at least one pair would have the same birthday — a 50/50 proposition might not seem so farfetched.
What is the point of today’s math teaser? A lot of probabilities are very misunderstood. Several years ago, over a 10-day span, a major jackpot was hit twice at a casino’s properties. The odds of each occurring was high — 20 million to 1 or so.
They wanted to use this information and the ‘astronomical’ probability of this event occurring in their marketing material. After all, this 20 million to 1 jackpot hit twice in a 10-day span.
They asked me to figure out the exact probability. Fortunately, they knew exactly how many hands had been played across their multiple casinos in this time. From my recollection, it was about 160,000 hands.
So what was the probability that this 1 in 20,000,000 event happened twice in 160,000 hands? The mathematical answer was 1 in about 30,000. In those 160,000 hands, something with a probability of 0.000005 percent happened twice while something with a probability of 99.999995 percent happened 159,998 times.
Throw in the fact that the two events could have happened anywhere in those 160,000 hands and the probability is simply not all that impressive anymore that the event happened.
I’ll throw one last example out there. Have any of you ever seen the same number come up three times in a row on a roulette wheel? I’ve never seen it. The probability is 1 in 1444 on a double 00 wheel.
This is not all that high, but not all that common either. If there are 200 roulette wheels in Las Vegas, over a 12 hour ‘day,’ there are roughly 72,000 spins per day. This would mean that on any given day in Las Vegas, the same number may be coming up three times in a row 50 times per day!
Math is a funny thing.
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