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My gf and I bought into a friend’s two box pools. She got in on the first one and I got in the second so they were separate pools. We each had $2$ squares in each pool, and each pool had $100$ squares numbered $0-9$ along the top and side. These numbers were randomly assigned. We each wound up having the $1$ eagles-$3$ patriots square which pays if the final score has those as the last digits. We each won a jackpot in the pool we entered. We obviously thought this was pretty crazy, and so did our friends.

I understand all scores are not equally likely but not sure if that even plays a part here since the numbers for each box were randomly assigned. Are the odds based on the potential outcome? Is there an element where the fact that we had randomly been assigned the same point combo must be factored in?

What are the combined odds that both of us would win this?

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Assuming scores are randomly distributed, the probability than any particular square (score combination) wins is $1/100$.

The probability that two people have the same square is also $1/100$. To see that, imagine the squares are chosen one at a time. After the first person has her square, the chance that the second person gets the same square is $1/100$.

The probability that two people have the same square and that square wins is $(1/100)^2 = 1/10,000$. Pretty unlikely.

But you have to be careful when looking back at a gamble and saying that what happened was strange. You can always find some seeming coincidence after the fact. Maybe your ticket and your girlfriend's were reversed and one of you won? Maybe your ticket and hers bracketed the actual final scores by just $1$ digit. Since there are lots of those kinds of coincidences, some of them will happen, just by chance. After all, when someone wins the lottery that's a strange coincidence for that person - but not for the lottery itself.

Edit: Here are the calculations for your two pairs of squares, with the real probability $0.009$ for the square $1-3$.

Once you have your two squares chosen at random, your girlfriend's probability of no match is $$ 1 - \frac{ 98 \choose 2}{ 100 \choose 2} \approx 0.96 $$ so there's a $4\%$ chance that you share a square. That's four times the probability if you had just one square each, not double. That shared square happened to have a winning probability $0.009$ according to your research, so your coincidence happened with probability $$ 0.009 \times 0.04 = \frac{36}{10,000}. $$

My caveat on coincidences still applies.

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  • $\begingroup$ Interesting. I think you are right about the odds but is there not two disjoint events here? There's the random assignment of the squares in which we both ended up with the same square (which you calculated as 1/10,000). Then after the squares are set, they do not have an equal chance of paying out (some scores are statistically more or less likely than others in the game). That seems to be a separate probability. Is it not a part of this? $\endgroup$ – Hippocrates Feb 7 '18 at 15:33
  • $\begingroup$ The first sentence of my answer says I'm not dealing with the actual probabilities of the squares. You could find out what they are by looking at all scores over the years (all super bowls, or all NFL games) . That would be an interesting exercise. I'm sure the data are on the web. Let me know if you do it. $\endgroup$ – Ethan Bolker Feb 7 '18 at 15:39
  • $\begingroup$ I re-read your answer and you assume the scores are equally likely. Makes sense. I believe the probability for the 1-3 pair is .009. We each has two squares so the probability of having the same square is 2/100 or .02. So 1 / (.02 * .009 ) = ~ 1:5,554 odds? Does that seem right? Here is my source for the .009 figure: superbowlsquares.org/odds $\endgroup$ – Hippocrates Feb 7 '18 at 15:45
  • $\begingroup$ See my edit.... $\endgroup$ – Ethan Bolker Feb 7 '18 at 16:24
  • $\begingroup$ Thanks for your answer. This does seem correct to me. Point taken on the coincidences. If nothing else, this coincidence and subsequent exploration has taught me something about mathematics. $\endgroup$ – Hippocrates Feb 7 '18 at 19:09

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