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## Question

hey guys I've missed you all

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Two aliens, Aphid and Bronk, have a single 10-sided dice. If you haven't seen one, they look like this:  They have ten sides, labeled 0-9. They take turns throwing the die, Aphid-Bronk-Aphid-Bronk-etc. The first person to get a 0 wins.

(1) What is the probability that Aphid will win?

(2) Is it possible to arrange the dice order (ie, ABABAB... or ABBAABBAA..., etc) so that Aphid and Bronk have equal chances of winning? If possible, what is such an arrangement?

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• 0 probablity is simply favorable outcomes overtop of possible outcomes so the probablity of one of them rolling a zero would be 1/10 right?

if each rolls a 10 sided die and one of them wants to roll a zero you have one favorable outcome and two ten sided dice so you hve twenty possible outcomes. so is it 1/20?

could you explain the second question? the wording is a bit confusing

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• 0 hey guys I've missed you all

~~~

Two aliens, Aphid and Bronk, have a single 10-sided dice. If you haven't seen one, they look like this:  They have ten sides, labeled 0-9. They take turns throwing the die, Aphid-Bronk-Aphid-Bronk-etc. The first person to get a 0 wins.

(1) What is the probability that Aphid will win?

(2) Is it possible to arrange the dice order (ie, ABABAB... or ABBAABBAA..., etc) so that Aphid and Bronk have equal chances of winning? If possible, what is such an arrangement?

this is very slim chance, we are talking gambling and try to get seven dice in lv under their house rules. chances are they will run same nbr before they can arrive final zero. its like more better chance of get snake eyes than get nbr they intend, cuz if we can find a way to beat those nbr, you can bet we will be filthy rich.

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• 0

After each alien has one turn, Aphid's chances are .10 to Bronk's .09.

If that sequence is continued to infinity, [AB ...] p[Aphid] = (.1/.19) = .5263...

So Bronk suffers from letting Aphid go first.

We try to rectify by reversing the order of the next two throws and repeat to infinity: [AB BA ...]. Now p[Aphid] = .5028...

Looks good. Reverse the order of the next four throws and repeat that sequence: [ABBA BAAB ...]. Now p[Aphid] = .5006...

Reverse that group of eight throws and continue: [ABBABAAB BAABABBA ...]. Now p[Aphid] = .5002...

Reverse that and continue: [ABBABAABBAABABBA BAABABBAABBABAAB ...]. Now p[Aphid] = .5001....

The process converges to an even chance between the aliens.

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• 0 Bovanova beat me to it ( i actually got this one )

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• 0 )
Bovanova beat me to it ( i actually got this one

Isn't there a better answer for part 2 tho?

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• 0 Isn't there a better answer for part 2 tho?

I got the same thing... Curse my algebra 1.

But if i was starting..

change A to B and B to A so it\'s BAABABBA......

EDIT; typo

Edited by Riddle Master Zack

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• 0
I got the same thing... Curse my algebra 1.

But if i was starting..

change A to B and B to A so it\'s BAABABBA......

Switching A and B doesn't change the disparity tho, just the winner.

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• 0 it in groups of 5? such as ABABA BABAB....

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• 0

Giving each player an extra turn might converge also, but more slowly:

A BB p[Aphid] = .3690...

A BB AAA BBBB p[Aphid] = .4568...

A BB AAA BBBB AAAAA BBBBBB p[Aphid] = .4944...

A BB AAA BBBB AAAAA BBBBBB AAAAAAA BBBBBBBB p[Aphid] = .5089...

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• 0
it in groups of 5? such as ABABA BABAB....

That's simply the AB sequence. ##### Share on other sites
• 0

Note that only after some repeated sequence of flips can there be equity.

The cumulative win probability for each player changes on each individual flip.

There is a 20-flip sequence that when repeated gets to within 1 thousandth of a percent of equity.

ABBA BAAB BAAB ABBA BAAB ...

By adding 12 flips, we get a 32-flip sequence that when repeated gets to with 7 millionths of a percent of equal odds:

ABBA BAAB BAAB ABBA BAAB ABBA BAAB BAAB ...

The columns of the table below give, to 5 decimal places,

Flip #, who flips, the win prob for that flip, and cumulative probs for A and B:

.1 A p(A) = .10000 1.00000 .00000

.2 B p(B) = .09000 .52632 .47368

.3 B p(B) = .08100 .36900 .63100

.4 A p(A) = .07290 .50276 .49724

.5 B p(B) = .06561 .42221 .57779

.6 A p(A) = .05905 .49503 .50497

.7 A p(A) = .05314 .54647 .45353

.8 B p(B) = .04783 .50057 .49943

.9 B p(B) = .04305 .46540 .53460

10 A p(A) = .03874 .49720 .50280

11 A p(A) = .03487 .52275 .47725

12 B p(B) = .03138 .49989 .50011

13 A p(A) = .02824 .51882 .48118

14 B p(B) = .02542 .50172 .49828

15 B p(B) = .02288 .48727 .51273

16 A p(A) = .02059 .50023 .49977

17 B p(B) = .01853 .48910 .51090

18 A p(A) = .01668 .49913 .50087

19 A p(A) = .01501 .50782 .49218

20 B p(B) = .01351 .50001 .49999 [fairly close after these 20 flips]

21 A p(A) = .01216 .50684 .49316

22 B p(B) = .01094 .50069 .49931

23 B p(B) = .00985 .49528 .50472

24 A p(A) = .00886 .50014 .49986

25 B p(B) = .00798 .49584 .50416

26 A p(A) = .00718 .49971 .50029

27 A p(A) = .00646 .50314 .49686

28 B p(B) = .00581 .50005 .49995

29 B p(B) = .00523 .49731 .50269

30 A p(A) = .00471 .49978 .50022

31 A p(A) = .00424 .50198 .49802

32 B p(B) = .00382 .50000 .50000 [extremely close after these 32 flips]

To 10 decimal places, after 32 flips:

p(A) = 0.5000000352

p(B) = 0.4999999648

It's interesting to look at cumulative P[Aphid], and how close it is to 0.50000 ...

after each of the eight ABBA or BAAB groups:

ABBA .50276 + .00276

BAAB .50057 + .00057

BAAB .49989 - .00011

ABBA .50023 + .00023

BAAB .50001 + .00001

ABBA .50014 + .00014

BAAB .50005 + .00005

BAAB .50000 + .000000035

Finally, note this does not work for arbitrary geometries.

For the regular 6-sided die we have:

ABBA .50820 + .00820

BAAB .50286 + .00286

BAAB .50136 + .00136

ABBA .50178 + .00178

BAAB .50150 + .00150

ABBA .50159 + .00159

BAAB .50152 + .00152

BAAB .50149 + .00149

Which is not converging.

But for a 48-sided die, is there such a thing?

We have great equality after 16 rolls:

ABBA .500111 + .000111

BAAB .500005 + .000005

BAAB .499969 - .000031

ABBA .5000004 + .0000004 Hey UR, welcome back.

Tell us what you have for part 2?

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• 0 This area of math always confuzzles me.

Isn't there an inherent error in assigning any value to a random outcome based on order or sequence, other that the implied base probability.

I.E. lets refer to it as the Rosencrantz and Guildenstern coin flipping conundrum.

Isn't every roll of the die in and of itself a 10% (or flip of the coin being 50%) chance to generate a result of 0 and as such each roller has a effective 50% chance to win as they both have a 10% chance to roll regardless of when they roll?

I understand that going first, having initiative, is an advantage as it is in any game, but for part two, I am not sure how the sequence can impact the base odds of rolling a 0 on a ten sided die.

And bear in mind, I can only operate at a 5th grade math level. So an anectdotal proof is more effective that a long series of formulas that will make my eyes water. Mind you, I do except the rule of big numbers, in that more outcomes in your set, make the actual outcomes approach the predicted outcomes, but how does sequence change the base probablity of an event?

Thanks!

Edited by theharangue

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• 0

I think you can see - at 5th grade level - how this works.

A has a .1 chance of winning on his first roll.

Likewise, B has a .1 chance of winning on his first roll.

The difference between A's .1 chance and B's .1 chance is

that there is a .1 chance that B will not get his turn, whereas

A is certain to get his first turn. So B's chance of winning

on the second toss is .9 x .1 = .09. The factor of .9 is the chance

B will get his turn.

We've determined that A's chance of winning the game, not

just on the first toss, but after as many rolls as it takes, is .5263157895...

The interesting point is that when it is the turn of either

player, that player has exactly that probability of winning

the game from that point forward!

That's right. Each time a player gets the chance to roll the die,

he has a .1 chance of winning on that toss and a .5263157895...

chance of winning over the all the future tosses until finally a 0 is rolled.

What changes, as we said, is the probability of getting the

chance to roll the die. Only A's first roll is certain. The chance

to get a toss, after that, decreases as the game goes on, by

multiplying by .9 each time.

Does that help?

You just add those individual probabilities up for each player.

They get smaller and smaller - see the first column after the

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• 0 Okay, the base probability doesn't get modified, but the chance to roll at the base probablity does, based on order. Thanks! That does make sense.

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• 0

here I am! hehe

Yeah #1 can be solved by turning an infinite convergence into a geometric series

chances for victory at each turn:

A (.1)

B (.9)(.1)

A (.9)(.9)(.1)

B (.9)(.9)(.9)(.1)

A (.9)(.9)(.9)(.9)(.1)

etc

clearly, A's chance of winning is the sum of all of his chances of winning:

(.1) + (.9)(.9)(.1) + ...

which is clearly (.1) multiplied by the sigma of [n=0 to infinity] of (.9)^2n

this isn't a geometric series yet because we have the 2 in there:

(.9)^2n = ((.9)^2)^n = (.81)^n

therefore, our answer is (.1) multiplied by the sigma of [n=0 to infinity] of (.81)^n

This is a geometric series...

(.1) / (1 - .81) = .1/.19 = .5263157895... (or 10/19ths)

That is Aphid's overall chance of victory. If he rolls and fails, then that's Bronk's overall chance of victory, then A's, then B's, ad infinitum

I threw this one in there because honestly I don't know the answer However I think the answer is NO, that is, it is not possible. It may be possible as a limit of some method (as bonanova attempted above), and it is infinitesimally possible that the game will go on forever, but NOT if the game ends. If the game ends, it ended after some finite amount of turns, in which case the chances were not completely .5 yet

So there's a big amount of probability ambiguity here... can probability be expressed as a function of the number of turns? In that case, is the overall probability the average of that function?

So my answer is "no". This was more of a thought experiment for me, I figured the answer was more of a philosophical thing than an experimental thing

But I guess the question is.... even if the game ended after a finite amount of turns, does that mean that the probability of whoever won was still 0.5 because if they hadn't have won, the method would have continued to be followed?  ##### Share on other sites
• 0

As an interesting sidenote, the formula for any value p (in this case, p = 0.1) is:

p / ( 1 - (1-p)2 )

which simplifies to:

1 / (2 - p)

thus, for p=.1, it is 1/1.9

Edited by unreality

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• 0

A thought on helping even out the chances for for this...basically have each person roll the die one time...the person who gets a higher number gets to roll first in the sequence. That way it evens the chances of who gets to have the slight advantage. So, in a sense it evens out the chance of someone having an advantage to winning...so in the long run (multiple games played) they have an even chance of winning...

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• 0 AAAAABBBBBBBBBBBBBBBBBBBB...

In other words, A rolls 5 times and then B rolls until he wins. Half of the time A should win, and half of the time B should win. A's chances to win are 1/10+1/10+1/10+1/10+1/10=50% B has a 50% chance of even getting his turn and then a guaranteed victory if he gets a turn so his total chances are 50%. Does this make sense? Edited by flinchum

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• 0 AAAAABBBBBBBBBBBBBBBBBBBB...

In other words, A rolls 5 times and then B rolls until he wins. Half of the time A should win, and half of the time B should win. A's chances to win are 1/10+1/10+1/10+1/10+1/10=50% B has a 50% chance of even getting his turn and then a guaranteed victory if he gets a turn so his total chances are 50%. Does this make sense? I think A's chances are 50% only if he can be assured of rolling five unique numbers on his five rolls. So, to modify my previous game: A rolls until he gets 5 unique numbers (a zero ends the game with A winning). After A rolls his 5 numbers, B rolls until he gets a zero (and wins). This should equal a 50-50 chance for each player to win. Not a very fun game though.

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• 0
I think A's chances are 50% only if he can be assured of rolling five unique numbers on his five rolls. So, to modify my previous game: A rolls until he gets 5 unique numbers (a zero ends the game with A winning). After A rolls his 5 numbers, B rolls until he gets a zero (and wins). This should equal a 50-50 chance for each player to win. Not a very fun game though.

That seems to work. B doesn't need to roll, tho.

Nice. ##### Share on other sites
• 0 hey guys I've missed you all

(2) Is it possible to arrange the dice order (ie, ABABAB... or ABBAABBAA..., etc) so that Aphid and Bronk have equal chances of winning? If possible, what is such an arrangement?

Perhaps a simpler solution for (2) would be to only declare the winner if the next roll is not a zero?

Edited by G Threat

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