0
$\begingroup$

In a game where card color determines each player's team, there are 5 shuffled cards -- 3 blue and 2 red -- and 4 players. The first player chooses one card from the pile and keeps it. This goes around until each player has chosen a card. At the end, the one remaining card gets discarded. In a single round of the game, what is the probability that one red card is chosen? What is the probability that both red cards are chosen? Please explain your methodology.

So far, I (no mathematician) have reasoned that this problem requires some sort of Bayesian decision tree. For the first person, the probably of red is 40%. The second person's card depends on what the first player drew. If the first player drew red, then the second player has a 25% chance of drawing red; otherwise, s/he has a 50% chance of drawing red. This continues on. However, I'm having trouble generalizing this into a function of N cards, N-1 players. Additionally, I'm not sure how to figure the probability of 2 reds. Probability of 1 red is simpler.

$\endgroup$
  • 1
    $\begingroup$ This question would be better if it included your thoughts so far $\endgroup$ – Henry Oct 16 '16 at 23:07
  • $\begingroup$ Good point -- just added some thoughts. $\endgroup$ – acs254 Oct 16 '16 at 23:24
  • $\begingroup$ You could look at the probabilities that the last card is red or blue $\endgroup$ – Henry Oct 16 '16 at 23:25
  • $\begingroup$ @Henry awesome hint. So the probably of 2 red is 68.75% and the probability of only 1 red being drawn is 31.25%. Correct? That was easily done by drawing a Bayesian tree. Now how could I generalize this to N cards and N-1 players where I would like to determine the probability of drawing Y red cards? $\endgroup$ – acs254 Oct 17 '16 at 0:11
  • $\begingroup$ Maybe using a Markov Chain could help. The chains could keep track of the amount of blue cards remaining. This completely determines the amount of remaining red cards if you know the original difference between blue and red cards. I'm also unsure of where you want to go with this result, but you could also try computing by conditioning on the first step and working up with the law of total probability. $\endgroup$ – Cehhiro Oct 17 '16 at 0:45
0
$\begingroup$

Total number of possible situations (including all permutations) is $5!$, because we have $5$ variants for the first card, $4-$second, $3-$third, $2-$forth and $1$ for the one discarded.

So, we'd like to calculate number of situations when $a)$ one red card picked; $b)$ both red cards piked

$$ a)\quad 4!*C^{2}_{1}C^{3}_{3} = 4!\frac{2!}{1!(2-1)!}\frac{3!}{3!(3-3)!}=4!*2 \\ b)\quad 4!*C^{2}_{2}C^{3}_{2} = 4!\frac{2!}{2!(2-2)!}\frac{3!}{2!(3-2)!}=4!*3 $$ Explanation:
consider case $a)-$there are $C^{2}_1$ combinations to choose $1$ red card from $2$ and $C^{3}_{3}$ combinations to choose $3$ blue cards out of $3$. Their multiple is number of combinations to choose $1$ red card out of $2$ and $3$ blue cards out of $3$ (without permutations). Number of permutations of $4$ elements is $4!$.

Thus, probability that one red card is chosen: $$ \frac{4!2}{5!}=\frac{2}{5}=0.4 $$ And, for both red cards chosen: $$ \frac{4!3}{5!}=\frac{3}{5}=0.6 $$ Solution can be generalized: let there be $N(=A+B)$ cards: $A$ red, $B$ blue. $N-1$ players. Last card discarded. Then, probability to pick exactly $C$ red cards: $$ \frac{(N-1)!C^{A}_{C}C^{B}_{N-1-C}}{N!}=\frac{C^{A}_{C}C^{B}_{N-1-C}}{N} $$ Constraints: $A>0,B>0,C\in[A,A-1]$

$\endgroup$
0
$\begingroup$

Kostiantyn is right but there might be a simpler argument in your specific case.

Just look at the color of the discarded card: it is blue if and only if both red cards have been picked and it is red if and only if exactly one red card has been picked. Since everything is made randomly, the last card has a probability 0.4 to be red and 0.6 to be blue !

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.