Hi guys. I'm not a particularly skilled mathematician but, being a poker player, probabilty interests me. So, when I was stumped by the following puzzle, I thought I'd let you lot have a crack at it. Here's the question:

In a fair lottery of 100m people, with a single winner picked at random each day, what is the probability that any one other person will win it three times before you win it once?

I started trying to work it out but got to a point where I simply didn't know if I was even approaching the problem in the right way. I'm far more interested in understanding how to solve the problem than finding the actual answer. Can anyone help?

I'll give it a try (someone correct me if I'm wrong). The answer depends on the way that the problem is formulated. I'll give a solution in the case that one specific person wins 3 times before you. Let's say we have N people (N=100 million), and we perform m draws.

• First, we need the probability that you never win in m draws; that's P1 in the formulae below.
• Then, we need the probability that one specific person among the N-1 others, wins exactly 3 draws out of m; that's that's a binomium, given by P2.
• Combining these two, and adding all probabilities for all values of m, one arrives at the probability P3. I think that's the answer.
• Alternatively, one can ask the probability that a specific person wins at least 3 draws before you win. In that case, P2 has to be replaced by P4, and the final answer will be P5 instead of P3.

If you want the probability that any random person wins 3 times, then the problem becomes more complicated. I'll have to think about that a bit more.

What is the probability that a random person wins it three times on a row in the first three days? 1/n^3

What is the probability that you win it on day 4? 1/n

What is the combined probability? 1/n^4

Now the first winner can be any one of the (n-1) people other than you.

So the answer should be: (n-1)/n^4

Ain't it? I mean, this looks like the minimalist version of the problem to me. Insert any other winner in between and you decrease the probability...

Thanks guys.

Pulsar, I'm afraid I can't quite follow your formulae (not being a mathematician) but I'm pretty sure I understand the gist of each of them. I'm sure they're correct for the situations you describe. However, I believe that what the puzzle was geting at was the situation you said you'd need to devote more time to.

I think it's asking for the probability that any one of the other 99,999,999 players manages to hit the jackpot three times (not necessarily consecutively) before you manage to scoop it once.

Natselrox, I think your solution provides the probability that a specific individual wins the first three draws consecutively, followed by another specific person winning the fourth (I may be totally wrong). Again, I don't think that's quite what the puzzle is looking for.

It seems, from a layman's perspective, that it ought to be possible to answer this with an exact, numerical probability. Though there's probably a good reason why I'm wrong about that, too.

Whats the odds that a particular person wins it before you? - 0.50

Once that has happened we can disregard what has happened in the past, so what the probability that one person wins it before you again? - 0.50

Repeat again - 0.50

Therefore (assuming we pick the person who has to win it before you before they win the first time) surely its 0.5^3 = 0.125

Reading again, it seems to say that it's any one person winning it 3 times, so it's actually essentially 1*0.5*0.5 = 0.25

95Theses wrote:Whats the odds that a particular person wins it before you? - 0.50

Surely the most likely option is that both of you die without having won it? Similarly, Pulsar's calculation includes a sum to infinity so doesn't account for human mortality. You'd need more information or more assumptions to work out a probability.

logical bob wrote:Surely the most likely option is that both of you die without having won it? Similarly, Pulsar's calculation includes a sum to infinity so doesn't account for human mortality. You'd need more information or more assumptions to work out a probability.

Let's not complicate things with such practical issues
Actually, the general problem, when it doesn't matter who wins three times, is decided after a finite number of draws: you need at most 2N-1 draws. Indeed, in the worst-case scenario, all of your N-1 rivals will have won twice after 2N-2 draws; after one extra draw, one of them has won three times (or you've won).

So, a possible strategy would be this: the game is decided and stopped if either you've won (case A) or someone else has won three times (case B). If neither happens, the game is undecided (case C), and a new draw is needed.

Start with 1 draw, and calculate the odds for A1, B1 and C1 (that's 1/N, 0, and (N-1)/N respectively).
If C1, draw again, and calculate the conditional odds for A2, B2 and C2, given C1.
If C2, draw again...
Continue until you have reached 2N-1 draws.

The total probability that you win is then Atot = A1 + A2 + ... + A2N-1. The probability that someone else has won three times is Btot = B1 + B2 + ... + B2N-1. And Atot + Btot = 1.

This is pretty complicated, and it looks like a partition problem.

I think I finally solved it. If P(S) is the probability of success (you win once before someone wins three times), and P(F) the probability of failure (someone wins three times before you win once), and there are N people, the odds are



In particular, the odds are:

N P(S) P(F)
2 0.875 0.125
3 0.802 0.198
4 0.752 0.248
5 0.714 0.286
6 0.683 0.317
7 0.658 0.342
8 0.636 0.364
9 0.618 0.382
10 0.601 0.399
11 0.587 0.413
12 0.574 0.426
13 0.562 0.438
14 0.551 0.449
15 0.541 0.459
16 0.532 0.468
17 0.523 0.477
18 0.515 0.485
19 0.508 0.492
20 0.501 0.499

For more than 20 people, the odds are against you.

As you can see, the equations are complicated: they are based on the strategy in my previous post. Basically, you need to count all possible scenarios in every draw. It's pretty difficult, and it would take some effort to explain all the steps. But if anyone's interested, I'll try to post the details.