Basically, with n=2, the probability of the game lasting 5 rounds or more is 6.25%. With n=3, it's greater than that. As n increases this probability increases. At what n does the probability cross the 33% boundary?

I... don't know. But generally you're right.

This seems to be an exercise in recursion.

Assuming P(n,k) means the probability that for n the game will last at least k rounds.

STEP I: For n = 2, it is easy.

P(2,1) = 1/2 = 50%

For the general case P(2,k) there is:

a) 1/2 chance of getting a 1 and ending the game.
b) 1/2 chance of getting a 2 and coming to P(2,k-1) state.

Generalizing we get that
P(2,k) = 1/(2^k)


STEP II: For n = 3

For a general case P(3,k) there is
a) 1/3 chance of getting a 1 and ending the game.
b) 1/3 chance of getting a 2 and coming to state P(2,k-1) (SOLVED IN STEP I)
c) 1/3 chance of getting a 3 and coming to state P(3,k-1)

Generalizing we get that
P(3,k) = (1/3)*(1/(2^k)) + (1/3)*P(3,k-1)

This is pretty much what I got so far

Then you should do a general generalization for P(n,k).
(and finally to solve our problem substitute k = 4).

Maybe if I have time I can spend more time on this.


However I do have 2 other "ugly" things I can share:

1) A quick and dirty approximation
This is not a solution, but might help to give an approximation
I think the number we're looking for is about 17
Why? In general doing a "roll n" on average returns (n+1)/2. For example roll 2 returns on average 1.5, while throwing a dice on average returns 3.5.
So... the way I see it 17 "on average" returns a 9 which "on average" returns a 5 which "on average" returns a 3 which returns 1 33% of the time.

2) A simulation.
If I have time tomorrow I can quickly write a program that simulates this. If we run it for example a billion times for different values of n we can get a pretty good approximation (very close to the real mathematical answer). I wonder how close it will be to 17...

Yeah, this makes a lot of sense. I was thinking "but each time you do it there's a chance you'll get the 1 earlier so that has to be taken into account" but, like in the "Vitek ruins the party" problem, the chance that they'll get a one is balanced by the chance they'll get a higher number and make it more difficult. I think...

So yeah I reckon you're right, and I feel silly for how simple your solution is.

It's no solution by any means. J7st an approximation. Going to try and do the simulation today.

Yay! Excited

I said simulation. Not stimulation.

oh. Disappointed.

Well, just to show how wrong it was, here is a simulation of a million rolls.

For each n, I'm showing smaller/larger where smaller is the number of games that took 4 rounds or less, while larger is number of games that took 5 or more rounds.

n=2 937449/62551 Percentage: 0.937449 (theoretical result = 15/16 = 0.9375).
n=3 887449/112551 Percentage: 0.887449
n=4 845572/154428 Percentage: 0.845572
n=5 810433/189567 Percentage: 0.810433
n=6 779713/220287 Percentage: 0.779713
n=7 752507/247493 Percentage: 0.752507
n=8 727434/272566 Percentage: 0.727434
n=9 704812/295188 Percentage: 0.704812
n=10 685429/314571 Percentage: 0.685429
n=11 667229/332771 Percentage: 0.667229
n=12 650215/349785 Percentage: 0.650215
n=13 634763/365237 Percentage: 0.634763
n=14 621552/378448 Percentage: 0.621552
n=15 608001/391999 Percentage: 0.608001
n=16 595224/404776 Percentage: 0.595224
n=17 583227/416773 Percentage: 0.583227
n=18 571480/428520 Percentage: 0.57148
n=19 561338/438662 Percentage: 0.561338
n=20 551960/448040 Percentage: 0.55196
n=21 542085/457915 Percentage: 0.542085
n=22 533090/466910 Percentage: 0.53309
n=23 524290/475710 Percentage: 0.52429
n=24 516926/483074 Percentage: 0.516926
n=25 508300/491700 Percentage: 0.5083
n=26 501516/498484 Percentage: 0.501516
n=27 494586/505414 Percentage: 0.494586
n=28 487870/512130 Percentage: 0.48787
n=29 481219/518781 Percentage: 0.481219
n=30 473662/526338 Percentage: 0.473662
n=31 469201/530799 Percentage: 0.469201
n=32 462481/537519 Percentage: 0.462481
n=33 456636/543364 Percentage: 0.456636
n=34 452328/547672 Percentage: 0.452328
n=35 447009/552991 Percentage: 0.447009
n=36 441666/558334 Percentage: 0.441666
n=37 436331/563669 Percentage: 0.436331
n=38 431871/568129 Percentage: 0.431871
n=39 426974/573026 Percentage: 0.426974
n=40 423005/576995 Percentage: 0.423005
n=41 418365/581635 Percentage: 0.418365
n=42 413676/586324 Percentage: 0.413676
n=43 410717/589283 Percentage: 0.410717
n=44 406217/593783 Percentage: 0.406217
n=45 403128/596872 Percentage: 0.403128
n=46 398642/601358 Percentage: 0.398642
n=47 394557/605443 Percentage: 0.394557
n=48 391553/608447 Percentage: 0.391553
n=49 388079/611921 Percentage: 0.388079
n=50 384641/615359 Percentage: 0.384641
n=51 381825/618175 Percentage: 0.381825
n=52 378256/621744 Percentage: 0.378256
n=53 374682/625318 Percentage: 0.374682
n=54 371962/628038 Percentage: 0.371962
n=55 368556/631444 Percentage: 0.368556
n=56 366024/633976 Percentage: 0.366024
n=57 363788/636212 Percentage: 0.363788
n=58 360211/639789 Percentage: 0.360211
n=59 358143/641857 Percentage: 0.358143
n=60 355311/644689 Percentage: 0.355311
n=61 352405/647595 Percentage: 0.352405
n=62 350069/649931 Percentage: 0.350069
n=63 346524/653476 Percentage: 0.346524
n=64 344817/655183 Percentage: 0.344817
n=65 342302/657698 Percentage: 0.342302
n=66 338894/661106 Percentage: 0.338894
n=67 337806/662194 Percentage: 0.337806
n=68 334423/665577 Percentage: 0.334423
n=69 333534/666466 Percentage: 0.333534
n=70 331525/668475 Percentage: 0.331525
n=71 328403/671597 Percentage: 0.328403
n=72 326086/673914 Percentage: 0.326086
n=73 324302/675698 Percentage: 0.324302
n=74 322612/677388 Percentage: 0.322612
n=75 321499/678501 Percentage: 0.321499
n=76 317983/682017 Percentage: 0.317983
n=77 316758/683242 Percentage: 0.316758
n=78 314248/685752 Percentage: 0.314248
n=79 312127/687873 Percentage: 0.312127
n=80 310778/689222 Percentage: 0.310778
n=81 309492/690508 Percentage: 0.309492
n=82 306543/693457 Percentage: 0.306543
n=83 305287/694713 Percentage: 0.305287
n=84 303303/696697 Percentage: 0.303303
n=85 301689/698311 Percentage: 0.301689
n=86 300091/699909 Percentage: 0.300091
n=87 298270/701730 Percentage: 0.29827
n=88 295769/704231 Percentage: 0.295769
n=89 295357/704643 Percentage: 0.295357
n=90 293826/706174 Percentage: 0.293826
n=91 292769/707231 Percentage: 0.292769
n=92 290302/709698 Percentage: 0.290302
n=93 288692/711308 Percentage: 0.288692
n=94 288343/711657 Percentage: 0.288343
n=95 286579/713421 Percentage: 0.286579
n=96 284439/715561 Percentage: 0.284439
n=97 283310/716690 Percentage: 0.28331
n=98 281429/718571 Percentage: 0.281429
n=99 280048/719952 Percentage: 0.280048
n=100 279409/720591 Percentage: 0.279409

The simulation answer is 69. Though it could be somewhere in the vicinity given the simulation sample size.
Yeah, thinking it over it's not really balanced. My quick and dirty approximation is completely wrong.

Look at n = 10 for example.

There is a 10% chance of first roll being 1. 10% of second roll being 1. 10% of third roll being 1. And 10% of fourth roll being one. The combined % that at least one of those rolls being one is 34%.

And this is assuming we make four %roll 10 rolls. i.e. all rolls are made in the 1-10 range. i.e. without the range getting smaller due to a smaller number being rolled. i.e. with no "balancing" whatsoever.

lololol. Good to see the universe has a sense of humour.

Now why the heck is that? What's the chances of getting a 1/69 chance in 4 attempts?

5.7%

But consider this:
There is 16% chance of rolling 3 or less at least once in 4 attempts.

I should have known better than to think an idea from ZFR was worth considering. HAVE I LEARNED NOTHING??

;p


ah yeah! The balancing thing I mentioned before is very different to the Vitek-ruins-a-perfectly-good-party-out-of-jealousy problem - in that one rolling 1 or 10 both reduce the problem to 100%s, as it were. I think we've got curves going on in this one? So as you roll the lower numbers it has a much greater impact on the problem than if you roll a higher number.
Does that make sense?

So if you and Beatrice do 69 the average is 34, and if you compare how the outcomes of 68, 34, and 2 affect the probability of reaching 1 with the other results, I think you'd find that 68 and 34 have less difference than 34 and 2.
Because of the curve...


Also it's making me itch that you've expressed the probabilities as a portion of 1, but labelled them percentages. Can you change it to Perunages?

Yes, exactly!
Too much work. And Perunage isn't even a word. It's an unword.

it's un anword


We solved two problems in one day. The world is a better place because of what we did.