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There is an urn with balls of one color. I draw one ball from this urn and put it back in. Somebody comes in and draws $m_0$ balls without replacement, when he is done, he then puts the balls he drew back into the urn, then somebody else comes in and draws $m_1$ times. Drawing a ball that nobody else drew gives me a value of x, if I draw the same ball as somebody else then we all have same chance of getting that object based on chance but only one of us will get it.

Let $C_i$ be the event that I draw the same ball as player i
Let a be the number of balls.

My expected value from this game is then: \begin{align} E = P(C_0 \cap C_1) \frac{1}{3}x + P(C_0 \cap \overline{C_1})\frac{1}{2}x+P(\overline{C_0} \cap C_1)\frac{1}{2}x + P(\overline{C_0} \cap \overline{C_1})x \\ =x (\frac{m_0}{a}\frac{m_1}{a} \frac{1}{3} + \frac{m_0}{a}(1-\frac{m_1}{a})\frac{1}{2} + (1-\frac{m_0}{a})\frac{m_1}{a}\frac{1}{2} + (1-\frac{m_0}{a})(1-\frac{m_1}{a}) ) \\ =x (\frac{m_0 m_1}{a^2} \frac{1}{3} + \frac{m_0-m_0 m_1}{a^2} \frac{1}{2} + \frac{m_1-m_0 m_1}{a^2} \frac{1}{2} + 1 -\frac{m_0}{a}-\frac{m_1}{a} + \frac{m_0m_1}{a} ) \\ =x (\frac{m_0 m_1}{a^2} \frac{1}{3} + \frac{m_0+m_1 }{a^2} +1 - \frac{m_1+m_0 }{a} ) \\ =x (\frac{m_0 m_1}{a^2} \frac{1}{3} + \frac{m_0+m_1 }{a^2}(1-a) +1 ) \end{align}

Question: How do I generalize the expected value for n players with each player having a specific number of draws? Trivially, if there are n players, the number of possibilities is $2^{n}$.

If there is also a better way to frame the problem than an urn problem, I am also interested.

Edit: Maybe a more useful way of phrasing probabilities is: Let $t_0$ be the event that no other person draws the same ball and $t_k$ the event that k other players draw the ball.
So the expected value is now:

$$E = \sum_{j=0}^n P(t_j)\frac{x}{j+1}$$

edit2: For the first 2 probabilities:

$$P(t_0) = \prod_{i=0}^n(1-\frac{m_i}{a}) $$

$$P(t_1) = \sum_{j=0}^{n} \left[\prod_{i\neq j}^n(1-\frac{m_i}{a})\frac{m_j}{a} \right]=\sum^n_{j=0} \prod^{j-1}_{i=0} \left(1-\frac{m_i}{a} \right)\prod^n_{k=j+1} \left(1-\frac{m_k}{a} \right)\frac{m_{j}}{a} $$

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Ok so I think I have the general formulation! Would appreciate if somebody can see if this is correct or find a better way to express this.

$$P(t_2) = \sum^{n-1}_{j_1 = 0} \left[\sum^n_{j_2= j_1+1}\prod^n_{i \neq j_1 \neq j_2 }\left(1-\frac{m_i}{a}\right)\frac{m_{j_1}}{a} \times \frac{m_{j_1+1}}{a} \right] $$

$$P(t_k) = \sum^{n-k+1}_{j = 0}\sum^{n-k+2}_{j +1}...\sum^{n-1}_{j +k-2}\sum^{n}_{j +k-1}\left[\prod^n_{i \neq j \neq j+1...\neq j +k-1 }\left(1-\frac{m_i}{a}\right) \right] \prod^{j+k-1}_{q = j}\left(\frac{m_q}{a} \right) $$

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  • $\begingroup$ I think the number of other players in this formulation is n+1 here instead of n, need to check... $\endgroup$ – Dio May 19 '17 at 15:02

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