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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?
it's the branching probabilities that I don't know how to deal with. Choosing 2 is straightforward because either the game ends or you do exactly the same thing again,

but choosing 3 does either the same thing again, or puts you onto the 2 path, or ends the game. So with choosing three the chance that you'll end the game on the 2nd round is greater than you ending it on the first.

1/3 chance of winning round one, OR 1/3 chance you have a 1/3 chance of winning round two, OR 1/3 chance you have a 1/2 chance of winning round two.

How do I collapse that into a single probability over the 4 rounds?

Or does it somehow balance out in a spooky way? JoeSapphire: How do I collapse that into a single probability over the 4 rounds?
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... ZFR: 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.
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. ZFR: 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. JoeSapphire: 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. ZFR: Going to try and do the simulation today.
Yay! Excited ZFR: Going to try and do the simulation today. JoeSapphire: Yay! Excited
I said simulation. Not stimulation. JoeSapphire: Yay! Excited ZFR: I said simulation. Not stimulation.
oh. Disappointed. ZFR: 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. JoeSapphire: 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.
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. JoeSapphire: 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...
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.
Post edited March 14, 2019 by ZFR
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? JoeSapphire: 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. ZFR: 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.
I should have known better than to think an idea from ZFR was worth considering. HAVE I LEARNED NOTHING??

;p JoeSapphire: Now why the heck is that? What's the chances of getting a 1/69 chance in 4 attempts? ZFR: 5.7%

But consider this:
There is 16% chance of rolling 3 or less at least once in 4 attempts.
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? JoeSapphire: So as you roll the lower numbers it has a much greater impact on the problem than if you roll a higher number.
Yes, exactly! JoeSapphire: 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?
Too much work. And Perunage isn't even a word. It's an unword.
Post edited March 14, 2019 by ZFR ZFR: 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.