Some dude in South Africa bought a ticket with the winning numbers just a little bit too late.

Deaf hardware store cleaner Stanley Philander had the numbers that won the record $12 million rollover (91 million rand) lottery in South Africa on Friday.

Problem was, he bought it after the numbers were selected, which means, that if those numbers just happen to come up again in next weeks drawing, Stanley is golden. Not quite as golden as if he had won this week, however. {…}

Let’s face it, not only is poor Stanley in the midst of a huge letdown at the moment, but that ticket of his is useless. The chances of the same numbers being drawn in back to back lotteries are astronomical.

The chances are, of course, just the same as any other set of six numbers!

On the other hand, if they did pick the exact same numbers back to back, Stanley probably still wouldn’t see any of that money until a lengthy investigation had concluded.


Category: Newsroom

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6 Responses to Decaf: What Are The Odds?

  1. web says:

    Actually, no.

    Logically, the likelihood of any six set of numbers, individually, is the same. The likelihood of drawing a specific set of numbers in a given week, is a known quantity. You are correct that his “odds” for the numbers picked have not changed.

    However, the coincidence of drawing the same set (or any two predefined sets you might choose) in a specific sequence, however, is the odds of one set, squared.

    This is in the same way that the odds of a specific ideal toin coss going heads or tails are 50-50 (excluding the “bizarrely improbably on edge landing”), but the odds of getting ‘Heads’ six times in a row (or any specific six-toss sequence) are 1 in 64 (1 in 2^6). And it’s the “coincidence” odds, that odds-squared phenomenon, that are why if he should win the gov’t would go into a lengthy investigation to determine if the balls had somehow been tampered with.

  2. trumwill says:

    But having already gotten the numbers once, you no longer square it. If you’re flipping a coin and get heads five times, the likelihood that you’re going to get heads on the next flip is 50%… which is the same likelihood that you’re going to get tails.

    In other words, the fact that heads landed on heads the five previous times has no beating on the sixth turn.

    The fact that the numbers came up on Monday make it no less likely that they will come up on Tuesday.

    As such, if it seems that the odds of those numbers coming up again are ridiculously low, it’s because the odds of any specific set of numbers coming up is ridiculously low.

  3. web says:

    In other words, the fact that heads landed on heads the five previous times has no bea[r]ing on the sixth turn.

    True, but the question of “What are the odds of X happening”, referring to the “two in a row”, is still the square.

    As such, if it seems that the odds of those numbers coming up again are ridiculously low, it’s because the odds of any specific set of numbers coming up is ridiculously low.

    Humans are amazingly good at getting this confused, it’s true. Seeing 5 coinflips in a row land heads, for instance, we “expect” to see a tails next because we have an expectation that for a large enough sample of flips, there should be an “even” distribution or pretty close to it.

    Put another way, a statistician who saw a coinflip generator consistently come up 60% or greater heads, would either wonder if the coins themselves are badly weighted, or if something’s wrong with his machine.

  4. trumwill says:

    True, but the question of “What are the odds of X happening”, referring to the “two in a row”, is still the square.

    No argument on that point. The case above, though, were talking about the “astronomical odds” of the number from last week getting selected again. Those odds are not squared because the first selection is fixed in the positive, so we’re only looking at one selection. Now, if the question is the likelihood of the same number getting chosen twice in the abstract, then it is squared.

  5. trumwill says:

    To further elaborate, we should distinguish between the odds of any number occurring twice in a row and a specified number occurring twice in a row.

    If we’re talking about any number, then you don’t square it because you can always take the first selection and then see if it occurs on the second selection. You would, however, square it once to see if you could get any number three times in a row.

    If we’re talking about a specific number, then you do square it for each selection.

    If you take a 6-sided die as an example, the chances of getting any number twice in a row is 1/6, the same as rolling any specific number (say “4”). The chances of rolling a “4” twice in a row is 1/36. The chances of rolling any number three times in a row is also 1/36. The chances of rolling “4” three times in a row is 1/216, which is the same as rolling any number four times in a row.

    I guess the best way to look at it is that if the number is unspecified, the first roll doesn’t really count because it’s just specifying the number for subsequent rolls.

    The same is true of a coin. The chances of flipping heads twice in a row is 1/4. The likelihood of flipping to the same side twice in a row is 1/2. Extend it to three flips and you’re looking at 1/8 for heads or 1/4 for flipping to the same side. If the side is unspecified, the first flip merely specifies.

    So in the event that the lottery balls produce the same numbers twice, the odds are the same as someone randomly buying one ticket. That’s not to say that it’s not incredibly unlikely, but rather to point out the unlikelihood of one ticket winning. You would still want to investigate it for the same reason that you would want to investigate it if the lottery ball vendor’s brother-in-law won. It’s perhaps no less likely than any other specific number being selection, but it’s still a very conspicuous event.

  6. Barry says:

    Having never taken anything close to a statistics class and getting easily lost in the math in the comments above, it seems to me the easy way to look at it is:

    If the odds of guessing a six-digit number are 1 in x, then the odds of guessing a twelve-digit number are even less… 1 in y, where y is much greater than x.

    This is because you’re not just drawing a random-six digit number, stopping time and restarting reality to draw another random six-digit number – you’re actually drawing a twelve-digit number over the span of two weeks. Whether the first six digits match the 2nd six digits is irrelevant… all twelve have to be drawn, matching the players winning card both times.

    I think I just confused and possibly hurt myself. I’m going to stop and let the professionals handle this…

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