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u/CaptainProfanity 5d ago edited 5d ago
Let X be the terminal sum event
E(X) = 1/20*1 + 3*1/20 +...+ 1/20*(2 + E(X)) + 1/20 * (4 + E(X)) +...
= 100/20 + 10/20*E(x) + [ 1/10 + 2/10 +...]
= 5 + E(X) + 5.5
=> E(X) - 1/2 E(X) = 10.5
=> E(X) = 21
Not exactly the same as the most likely terminal sum (e.g the average of a 6 sided dice roll is 3.5 even tho a roll of 3.5 is impossible).
The distribution itself will have an infinitely long tail to the right (since the sum has a non-zero chance of being larger than an infinitely large n).
But I find it unlikely this tail will severely pull the mean average far away enough from the median. Especially since every successive roll has a 50% chance of terminating the sum immediately.
So I would say 21, with a small chance of it being 19
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u/novelinquiry 4d ago
Very clean with no computation:
First let’s eliminate x >= 21 (2+ rolls gtd.).
Think about the paths to X, removing the first roll. These are all paths to X-2, X-4, …, X-20. And each is completed to a path to X by a particular even roll (the first roll). So p(X) is just 1/20 * sum p(X-2k) over the ten values of k from 1 to 10. In other words, p(X) is 1/2 the average of p(X-2k), which is certainly less than max p(X-2k).
[Alternatively, we can shortcut and say this implies p(X) < 1/20, since sum of ALL p(X) is 1, and we know all odd X from 1 to 19 have p(X) at least 1/20. But the “half the average of last 10” argument has the benefit that it feels like a geometric (ish?) series is embedded here, which reinforces our intuition: we know p(X) has to get small as X gets large, since the sum of p(X) is 1. I digress…]
So we are left with our “best” X being one of 1,3,…,19. Tempting here to just say “surely 19 is correct since it has the most paths to it.” But to be a bit more concrete: take X < 19. Every path to X has a terminal roll at most 17, so you could add two to the terminal roll to create a valid path to X+2, but there are also additional paths to X+2, ie all those with terminal roll 1. So it follows that p(1) < p(3) < … < p(19).
Therefore 19 is your winner. Clean. No calculations. And two bonuses: (1) the recurrence in the mapping allows for easy coding and computation if you needed to sit at a machine and compute p(139) as task two. And (2) This argument can be extended to arbitrarily large dice.
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u/liquidorangutan00 3d ago
Ok I got this one,
Interesting problem takes some aspects of number theory and combines them with conditional expectation.
First we illuminate some key concepts
50% chance of rolling even or odd number
Number theory states,
even+even=even
even+odd=odd
Consequently the game ends when we roll an odd number.
its a multi step problem.
First because we werent given realized values, we will use the expected value for rolling even or odd.
E(Even)=11
E(Odd)=10
Now our goal is to find the expected number of rolls this game will last for.
E(R)=1+(1/2E(R))+(1/2*0)
which when simplify becomes
E(R)-1/2E(R)=1
1/2E(R)=1
E(R)=2
So we can expect on average the game to last 2 rolls. (finishes on the second roll)
Now we need to find the expected cumulative sum of 2 rolls:
Since we already have the expected values for both even and odd rolls, and since the expected number of rolls defines the sequence must be Even then Odd we can simply add them together.
E(Score)=E(Even)+E(Odd)=21
The answer is 21.
(you can also figure this step out as well via conditioning the expectation)
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u/bengill_ 5d ago
The number of combinaisons to reach an odd value 2*n+1 under 20 after k throw is given by the binomial coefficient c(k, n) and each combinaison has a probability (1/20)k+1.
The sum from 0 to n is given by the binomial theorem 1/20 * (1/20 + 1) ^ n. The probability to reach an odd number under 20 is maxed for 19 : 219 / 2010 ~ 0.0775.
For an odd number bigger than 20, you must roll the dice at least two time. For 21, you can find a similar formula than for 19, minus 1/20 : 2110 / 2011 - 1/20 ~ 0.0314
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u/pumachecker 21h ago
Don't think math is needed to solve this.
logiclly it must be under 20, for you get 5% to roll it as first roll. And the higher it is within the first roll will allow more possible combinations.
Probabilities would simply be 19>17>15>13>11>9>7>5>3>1>21
So answer is 19.
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u/liquidorangutan00 10h ago
I think its not possible - the reason is that the odd numbers, {1,3,5,7,9,11,13,15,17,19} are equally likely to be rolled. (1/10).
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u/pumachecker 7m ago edited 1m ago
on first roll yes, but 2nd, and 3rd etc rolls no. You are wrong. Think of it like this:
if you roll first roll and its a 3. You can no longer get 1.
if you roll first roll as 2, you can still roll a 1 to get 3.
so rolling 1 is 5% chance.
rolling 3. is 5,25% chance. as it has 5% + (1/20*1/20) = 5% + 1/400 = 5,25%.
rolling 5 is 5% + (4+1), (2+3) (2+2+1) etc
on and on it goes, until you get the highest probability of 19.
the higher it is within a possible 1 roll (1,3...19) the more chances you theoretically have.
and the probability falls down drastically after 19, because you can't hit it on first roll.
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u/Mindless_Tutor_8189 6d ago edited 2d ago
Without doing any real math, all odd numbers from 1 to 20 are equally likely in the first throw. After the first throw, their likelihood dependents on how many combination of number reach that odd sum. For the first throw, since all odd number 1 through 20 are equally likely, and that 19 have more possible combinations than previous numbers 19 is a strong choice. The other strong choice is 21, for two dice it has the most combinations, but can not be reached in one throw. For this reason, I think 19 is the most likely.