# very stupid confusion in probablity or combination?

i confuse very much on these when any question from probability is asked for example suppose you have three beggars let they be x,y,z and 2 biscuits and you have to distribute between them so total number of ways should be the numbers of solution of equation x+y+z=2 which is 6 and i know it i am certain of that but if we look at this way

consider first biscuit it has three options and the other one it also has three options so it should be 3*3 that is 9 but it isn't i know it isn't but i am still very confused what it means mathematically and practically

to explain what i am talking about consider these question i am not making these up they are in my textbooks

1 find the probability that a particular beggar gets exactly 1 biscuit

2 number of ways of distributing biscuits between beggars

now on the first one the solution uses 9 as total number of ways but in second its 6 which is according to my intuition and which i feel should be used in first too ... what is going on here can someone explain me please examples of both cases with proper logic would be much appreciated

• You're overcounting when you say that the first biscuit has 3 options and the second one has 3 options. If beggar 1 gets the first biscuit and beggar 2 gets the second biscuit, then it's the exact same as if beggar 2 gets the first biscuit and beggar 1 gets the second biscuit. It may help to write out the different scenarios to convince yourself that there are only 6. Commented Jan 20, 2016 at 3:21
• You are right that the first biscuit has 3 options and the second biscuit has 3 options. However, your mistake is in treating the biscuits as distinct. Commented Jan 20, 2016 at 3:21

The number of non-negative integer solutions of the system:

$\begin{cases}x_1+x_2+\dots+x_r=n\\ 0\leq x_1\\ \vdots\\ 0\leq x_r\end{cases}$

is known to be $\binom{n+r-1}{r-1}$ and can be proven using a stars and bars argument.

Assuming the biscuits are considered identical and may not be broken apart, the number of ways of distributing the biscuits is then $\binom{2+3-1}{3-1}=\binom{4}{2}=6$ as you correctly found.

However: assuming each biscuit is given to one of the beggars uniformly at random, note that these six possibilities are not equiprobable. It is twice as likely to arrive at the scenario $(1,1,0)$ than you are to arrive at the scenario $(2,0,0)$ (where the numbers correspond to the number of biscuits received by the respective beggar).

To answer the question of the probability, again assuming that the biscuits are distributed uniformly at random (this really should have been specified in the problem. Perhaps in reality the shortest man is more likely to receive the biscuits since he looks more pathetic), imagine for a moment that the biscuits are in fact different colors.

You have nine different ways of distributing the biscuits now, and each of these are equiprobable. The probability that the first beggar receives exactly one biscuit is then $\frac{1}{3}\cdot \frac{2}{3}+\frac{2}{3}\cdot\frac{1}{3}=\frac{4}{9}$ (he either receives the red biscuit but not the blue one or he doesn't receive the red biscuit but does receive the blue one).

Remember that the scenarios $(\color{red}{1},\color{blue}{1},0)$ and $(\color{blue}{1},\color{red}{1},0)$ are considered different in this case, whereas in the other problem where the biscuits are considered identical.

• there is one more similar question it says "if n biscuits are distributed among N different beggars what is the probability that a particular beggar gets exactly r biscuits" this is exact question not a word changed and the answer is {nCr*(N-1)^n-r} /N^n so the question should have mentioned that all the biscuits are different from each other. and i get the solution {if i don't overthink it} but still i want to read your comment explaining it and remember you are explaining 17 year old noob so walk me through as if i am a stupid potato Commented Jan 20, 2016 at 4:00
• sorry for inconvenience i don't know mathjax or latex or whatever the heck it is Commented Jan 20, 2016 at 4:03
• @user306284 Here is a link to learn MathJax. In the same way, we assume that each biscuit is given to a beggar with uniform probability, so it is okay if we temporarily assume that there is a "first biscuit" a "second biscuit" etc... At this point, you should use the binomial distribution which gives the immediate answer as above. Commented Jan 20, 2016 at 4:08

It depends on whether you are counting the biscuits as distinguishable or not.

There are six ways to distribute two indistinguishable biscuits among three distinct beggars. Counted as $\binom{2+3-1}{2}$ by a method known as "stars and bars". $$**!\;!,*!*!,*!\;!*,!**!,!*!*,\;!\;!**$$

However, the probabilities of each of these events may not be equal.

So when evaluating the probabilities we count ways of assigning beggars to each distinct biscuit.   That's counted as $3^2$ : three options for two unique items.   These nine outcomes are each equally likely; however, several of them may seem to be the same event if we can't distinguish between the biscuits.

$$\circ\bullet!\;!,\underline{\circ!\bullet!, \bullet!\circ!}, \underline{\circ!\;!\bullet,\bullet!\;!\circ},!\bullet\circ!,\underline{!\circ!\bullet,!\bullet!\circ},\;!\;!\circ\bullet$$

This can indeed be quite confusing.
The central message is that the ways of distributing using stars and bars is not equiprobable.

To compute probabilities, it is clear, for instance, that taking distribution to be random,
P(A gets both biscuits) $= \frac19$, i.e. it now follows ~bin($2, \frac13$), leading to the desired answer of $\frac49$

A further example

An example that may strike a chord is: randomly choose 3 children, and group them as $B-G$.

There will only be $\binom{3+2-1}{2-1}= 4$ such groups: $3B-0G,\; 2B-1G,\; 1B-2G,\; 0B-3G$

but P(3 boys) $= \frac18$

To calculate probability: (no. of favorable outcomes)/(total no. of outcomes).

Probability that beggar gets exactly 1 biscuit is calculated as follows:

Total no. of favorable outcomes are : 6 (3 * 2 first biscuit can be distributed in 3 ways while second biscuit can be distributed in 2 ways since none beggar can get more than one)

Total no. of outcomes : 9( 3 * 3 since first biscuit can be distributed in 3 ways and second biscuit can be distributed in 3 ways and each beggar can get more than one biscuit )

So probability : 6/9