# Most Probable Sum [duplicate]

Possible Duplicate:
Probability of dice sum just greater than 100

A fair dice is rolled and the outcome of the face is summed up each time. We stop rolling when the sum becomes greater than 100. Which of the following is most probable sum?

1. 103
2. 102
3. 100
4. 101
5. All have equal probability

How best do I approach these types of problems?

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## marked as duplicate by Byron Schmuland, MJD, Ross Millikan, TMM, Pedro Tamaroff♦Jun 8 '12 at 19:09

I'd have said "A fair die", and reserved the word "dice" for the plural. (But some people have strange feelings about this one.) – Michael Hardy Jun 8 '12 at 17:42

The sum $100$ is not possible if we stop when the sum first becomes greater than $100$. Maybe you meant "greater than or equal to $100$." But we solve the problem as it stands.

Look at where we are just before we "go over." Maybe we are at $100$. Then $101$, $102$, $103$ are equally likely.

Maybe we are at $99$. Again, $101$, $102$, $103$ are equally likely. Maybe we are at $98$; same thing. Maybe we are at $97$; same thing.

Maybe we are at $96$. Now $103$ next is impossible, but $101$, $102$ are equally likely.

Maybe we are at $95$. Then only $101$ is possible among the three.

It is certainly possible that the sum just before we go over is $95$ or $96$. So $101$ is the most likely of $101$, $102$, $103$. And $102$ is next.

If the problem meant to say we stop when our sum is $\ge 100$, the same reasoning shows $100$ is the most likely "first over" number of our four choices.

Added Let $p_{100}$, $p_{99}$, $p_{98}$,, up to $p_{95}$ be the probabilities that we are respectively at $100$, $99$, and so on down to $95$ just before we go over. These $p_k$ are not equal, and would be fairly messy to compute. But we don't need to know them. The probability that we end up at $103$ is $\frac{1}{6}\left(p_{100}+p_{99}+p_{98}+p_{97}\right)$. The probability that we end up at $102$ is $\frac{1}{6}\left(p_{100}+p_{99}+p_{98}+p_{97}+p_{96}\right)$, clearly bigger. And the probability we end up at $101$ is even bigger.

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To complete this answer, you really should show that the result holds regardless of the probabilities of being at 100, 99, ..., which of course are not equal. (Trivial to do so of course, but still it should be done...) – jbowman Jun 8 '12 at 17:46
@jbowman: Yes, I wondered whether that would be obvious. – André Nicolas Jun 8 '12 at 17:48

Here's a simulation approach to the problem. The R function

f = function(n,k){s=0;
while(s <= n){s=s+sample(1:k,1)}
return(s)};


takes inputs n (number to be achieved) and k (number of sides of the die) and simulates this process, returning the final number achieved. In our case n = 100, k = 6. We can simulate this experiment one million times, and here's what we get:

table(replicate(10^6,f(100,6)))

101    102    103    104    105    106
285012 238015 189910 143634  95362  48067


So, for example, about 28.5% of the time we get 101. The empirical probabilities of getting 101, 102, ..., 106 are very close to being $6/21, 5/21, 4/21, 3/21, 2/21, 1/21$. Why should this be? Well, the average result from rolling a die is $7/2$, so if we just keep rolling a die and taking a running sum we should hit about $2/7$ of all integers. So the probability that our result is 106 is the probability of hitting 100 (about $2/7$) times the probability of rolling a 6 after that ($1/6$), or $1/21$. To get a result of 105, we have to either hit 99 and then roll a 6, or hit 100 and then roll a 5; these both have probability about $(2/7)(1/6) = 1/21$ and are mutually exclusive. And so on.

(The numbers of the form $i/21$ are not exact answers, because there's some approximation going on here.)

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if someone told me "hey, lets roll a die until it reaches 101 or more. If it is 101 or 102 I win, if it is 103-106 you win" I would instantly agree and more likely to lose – ajax333221 Jun 8 '12 at 23:49
As would I, if I hadn't paused to do this calculation. – Michael Lugo Jun 9 '12 at 0:50