# How much should I pay in this poker game to make the expectation zero?

I am new to the idea of expectation, and I want to make sure that I understand it. I came across this question stating:

You are playing a version of poker where you are dealt five cards and you would win $100 if you are dealt a three of a kind. Otherwise you will lose and you have to pay some amount of money. How much money should you have to pay to make this game a fair game (i.e. have expected value \$0)?

So if I have understood expectation correctly, the expectation of this kind of poker game would be: $$\mathbb{E}[x] = -m\left(1-\frac{{13 \choose 1} \cdot {4 \choose 3} \cdot {12 \choose 2} \cdot {4 \choose1}^2}{52\choose 5}\right) +100\left(\frac{{13 \choose 1} \cdot {4 \choose 3} \cdot {12 \choose 2} \cdot {4 \choose1}^2}{52\choose 5}\right)$$ Where $$m$$ is the amount of money that we have to pay in this game, then I would set the expectation to zero and solve for $$m$$?

• I can't follow your calculation. Are you saying that there are $13\times 12\times 4^3$ ways to get three-of-a-kind? That's not right. – lulu May 9 at 18:50
• Yes, you are correct I made a mistake there. Will fix it now. Thank you. – Mashpa May 9 at 18:51
• If you have four of the art do you win or lose? – user May 9 at 20:47
• @user you would lose. – Mashpa May 9 at 20:56
• You have wrongly computed the probability to win. – user May 10 at 13:20

Yes, if $$p$$ is the probability of success you get $$0 = \mathbb{E}[X] = 100p-m(1-p) \iff m = \frac{100p}{1-p}$$ and it only remains to find $$p$$...