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# Hurry up and Wait

Go to solution Solved by Molly Mae,

## Question

A discrete event (like rolling a fair die and wanting a 3 to appear) has a probability p of success (1/6 in this case.) The first roll is likely to fail, so let's keep rolling the die until we do get a 3, Then stop and write down the number of rolls that it took. Let's repeat the experiment a large number of times, each time recording the required number of rolls. So we have a bunch of 1s (the number of times 3 appeared on the first roll,) 2s (the number of times a 3 appeared on the second roll,) and so forth.

What number will most appear most often?

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• Solution

First, I have strep throat, so curse you for making me think about this.  I have no logical way to prove it on paper, but it is counter-intuitive.  My initial assumption was that it was either 3 or 4.  I wanted to see which would come out more often without using brain power.  I washed my hands and started rolling dice.  After 250 times, the number of times that comes up most often is 1.  I probably should have thought that, since anything that ends the trial removes the odds for future results, and the first time always has a chance, even if it is only 1/6.  Considering the other 5/6 of the time has to be distributed to all numbers above 1, it does now make sense in my head.

Thanks, Bonanova!

Edit: And now that I've posted this, I think proving it might be pretty trivial.

Edited by Molly Mae
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Perhaps

5/6 is prob of rolling one non-3

the first power of 5/6 that is lower than .5 is 4.

so I’ll guess 4 is the highest frequency.

I wish I weren’t double spacing, durn iPhone.

Edited by CaptainEd
Typo
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Spoiler

Either 4 or 5, but no proof yet!

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