# A problem of points

An urn contains $c$ elements of 3 different types: There are $\alpha>0$ elements of type $A$, $\beta>0$ elements of type $B$ and $\gamma>0$ elements of type $G$, and $c=\alpha+\beta+\gamma$.

Alex bet $S$ dollars to get, in $n>0$ independent trials (i.e. with replacement), at least one element of kind $A$. Therefore, he can win the game with probability $P(Alex)=1-\left(\frac{c-\alpha}{c}\right)^n$.

Bart bet $S$ dollars not to get, in $n>0$ independent trials (i.e. with replacement), any element of kind $B$. Therefore, he can win the game with probability $P(Bart)=\left(\frac{c-\beta}{c}\right)^n$.

For the sake of simplicity, we assume that $\alpha,\beta,\gamma,n$ are such that $P(Bart)=P(Alex)$. Thus, since Alex and Bart have the same chance to get the $S+S$ dollars at stakes, if they both win or if they both lose, they get back their money, i.e. $S$ dollars.

But if the game is interrupted at the trial $k<n$, how should Alex and Bart correctly divide the stakes?

The solution should be given both in case the trials are performed from the same urn, and in case the trials are performed separately from two identical urns (one for Alex, one for Bart).

HINTS:

Alex gets this much money: $M_A = S \times \mathbf{1}[Alex\ wins] + S\times\mathbf{1}[Bart\ loses]$

Bart gets this much money: $M_B = S \times \mathbf{1}[Alex \ loses] + S \times \mathbf{1}[Bart \ wins]$

Here, the notation $\mathbf{1}[event]$ is the indicator random variable for the event, i.e., its value is $1$ if the event occurs, and $0$ if not.

Taking advantage of linearity of expectation: $E[M_A] = S \times P(Alex\ wins) + S \times P(Bart \ loses)$, and similarly for $E[M_B]$. The $2S$ total should be split according to $E[M_A]$ and $E[M_B]$.

The answer now splits into 4 cases, depending on whether Alex has won already, and, whether Bart has lost already.

After $k<n$ trials, Alex could have won already, i.e. $P(Alex \ wins) = 1$, or his outcome could still be undecided in which case he still has $n-k$ trials left, so $P(Alex \ wins) = 1-\left(\frac{c-\alpha}{c}\right)^{n-k}$.

Similarly, Bart could have lost already, or he still has to survive $n-k$ trials to not lose.

In each of the 4 cases it should now be easy to calculate the answers.

Re: the version of the question where they play from the same urn - this makes their winning/losing dependent, but expectations are still linear even for dependent variables, so the answer does not change at all.

• Thanks a lot, antkam. It is a very illuminating answer! I study it, then I come back to you, in case.
– user559615
May 10, 2018 at 19:46
• I have some troubles in understanding how to finalize the values of $E(M_A)$ and $E(M_B)$. Should I simply sum up the four possibilities for each player? I can do it for Alex, $$E[M_A]=(S\times 1 + S\times1)+(S\times 1 + S\times \left[1-\left(\frac{c-\beta}{c}\right)^{n-k}\right])+(S\times \left[1-\left(\frac{c-\alpha}{c}\right)^{n-k}\right]+S\times1)+(S\times\left[1-\left(\frac{c-\alpha}{c}\right)^{n-k}\right]+S\times\left[1-\left(\frac{c-\beta}{c}\right)^{n-k}\right])$$ but not for Bart...
– user559615
May 11, 2018 at 7:19
• No. If you do that, your $E[M_A] > 2S$ which makes no sense. The truth is when the game ends at round $k$, the game can be in one of 4 different states: (Alex won already, Bart lost already), (Alex won already, Bart undecided), (Alex undecided, Bart lost already), (Alex undecided, Bart undecided). Each case the math is slightly different. The way I read the OP question, it did not specify which case happened, so IMHO there are 4 different answers, one per case. (Unless the OP actually implicitly specified the (Alex undecided, Bart undecided) case...? But I didn't read it that way.) May 11, 2018 at 11:41
• I see your point. I ask you then two more questions: Is there a way to give a unique answer, which takes into account the four possible cases you mentioned? Also, can you just sketch me one of the cases explicitly? I am trying to develop the math but I end always with nonsenses as above...
– user559615
May 11, 2018 at 11:51
• Also, my reading of your approach is that $E[M_B]=S\times P(Bart \ wins)+S\times P(Alex\ loses)$. But Bart cannot win at the $k$ trial (the next extracted elements can always be a $B$) and Alex cannot lose (the next extracted element can always be an $A$). Is therefore $E[M_B]=0$?
– user559615
May 11, 2018 at 12:07

[1] Alex has already won, i.e. $P(Alex \ wins) = 1$, and Bart's outcome is undecided. He will lose only if, in the remaining $n-k$ trials, at least one element of kind $B$ is extracted. This can occur with probability $P(Bart \ loses)=1-\left(\frac{c-\beta}{c}\right)^{n-k}$. Therefore, $$E[M_A]=S\times 1+S\times \left[1-\left(\frac{c-\beta}{c}\right)^{n-k}\right],\,\,\,\, E[M_B]=S\times \left(\frac{c-\beta}{c}\right)^{n-k}+S\times 0.$$

[2] Alex has already won, i.e. $P(Alex \ wins) = 1$, and Bart has already lost, i.e. $P(Bart \ loses) = 1$. Therefore, $$E[M_A]=S\times 1+S\times 1,\,\,\,\, E[M_B]=S\times 0+S\times 0.$$

[3] Alex's outcome is undecided, i.e. $P(Alex \ wins)=1-\left(\frac{c-\alpha}{c}\right)^{n-k}$. Bart's outcome is also undecided, i.e. $P(Bart \ loses)=1-\left(\frac{c-\beta}{c}\right)^{n-k}$. Therefore, $$E[M_A]=S\times \left[1-\left(\frac{c-\alpha}{c}\right)^{n-k}\right]+S\times \left[1-\left(\frac{c-\beta}{c}\right)^{n-k}\right],$$ $$E[M_B]=S\times \left(\frac{c-\beta}{c}\right)^{n-k}+S\times \left[\left(\frac{c-\alpha}{c}\right)^{n-k}\right].$$

[4] Alex's outcome is undecided, i.e. $P(Alex \ wins)=1-\left(\frac{c-\alpha}{c}\right)^{n-k}$. At the same time, Bart has already lost, and $P(Bart \ loses) = 1$. Therefore, $$E[M_A]=S\times \left[1-\left(\frac{c-\alpha}{c}\right)^{n-k}\right]+S\times 1,\,\,\,\, E[M_B]=S\times 0+S\times \left(\frac{c-\alpha}{c}\right)^{n-k}.$$

• @antkam Did you mean something like this?
– user559615
May 11, 2018 at 13:59
• yes :) -- that is if i read (interpret) the OP question correctly. May 11, 2018 at 14:40
• Got it. Thanks a lot for your kindness!
– user559615
May 11, 2018 at 15:02