Consider the game:

  • 100 (n=100) people (inclusive you) are putting their names in a bowl for drawing a lot
  • There is 5 (p=5) prices, thus 95 wins nothing


  • You can just throw instead of 1 namecard, 2 (c=2) namecards and your chance will increas from 5/100 to 6/101 which slightly higher than one namecard (c=1)
  • Since there will be 5 draws. If your name is drawn twice, you will be discarded


How much namecards should you throw into the bowl to increase your chance but not get caught. Optimal value for "c".


Define $X_m$ as the number of draws (out of $n=5$) of your own name given you put $m$ times your own name in the bowl while there are $N=99$ other names in the bowl. Then $X_m$ is hypergeometric distributed with parameters $N+m$, $m$ and $n$: $$X_m \sim H(N+m,m,n),$$ with corresponding positive probability mass function $$P(X_m=k) = \frac{{{m}\choose{k}} {{N}\choose{n-k}}}{{N+m}\choose{n}}, \text{ for } 0\leq k \leq n.$$

The probability to win is: $$p(m)=P(X_m=1) = \frac{{m \choose 1}{99 \choose 4}}{{99+m \choose 5}}$$ The probability of not getting disqualified is $$q(m)=P(X_m \leq 1) = (P(X_m=0)+P(X_m=1)).$$

While $p(m)$ is maximized for $m=24$, $q(m)$ is strictly decreasing in $m$. The sum of $p(m)+q(m)$ is maximized for $m=12$. R-code:

q<-choose(m,1)*choose(99,4)/choose(99+m,5) + choose(m,0)*choose(99,5)/choose(99+m,5)


lines(q,type="l", col="red",ylab="",xlab="m")
legend("topright", legend=c("p(m)", "q(m)"),
        col=c("black", "red"), lty=c(1,1), cex=1)

legend("topright", legend=c("p(m)+q(m)"),
         col=c("black"), lty=1, cex=1)

which.max(p) #24
which.max(p+q) #12

Two plot showing p(m),q(m) and p(m)+q(m)

  • 1
    $\begingroup$ Thank you for your time and you have my respect dude. pitty cant vote it $\endgroup$ – Andy Feb 8 '19 at 21:37

You should throw $X$ cards until the benefit of cheating outweighs the potential loss of getting caught.

The probability of winning with 1 name card as you ask is 5/100. If you have two name cards the probability of winning is 6/101.

So you would have to work out the probability of getting caught (i.e. multiple probabilities):

The probability of drawing two cards out of six is a little bit complicated:

YYYYYY = (6/101)*(5/100)*(4/99)*(3/98)*(2/97)*(1/96)

You can follow the steps in a similar fashion and then work of the probability of being caught under each scenario.

You then would cheat up to the probability of winning is greater than the probability of being caught.

This would take you a long time by hand, so probably consider using a PC.

  • 1
    $\begingroup$ i had just the same strategy in mind but was lazy to calculate. @chRrr had done a nice calculation $\endgroup$ – Andy Feb 8 '19 at 21:38

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