This is my question that I am trying to conquer: A, B, and C draws a card in that order from a well-shuffled pack of 52 cards. The first to draw a diamond wins ₽740. If 'A' starts, find their mean and the variance.

Now frankly, my probability-assessing skills are pretty weak, so I am unable to figure out a solution to this. I thought it'll require a binomial distribution, only to notice that the information isn't enough. What method should I employ and why?

I did try searching for questions of a similar kind but to no avail.

I understand that when A draws, he draws with a probability of $\frac{13}{52}$.

Do B and C also draw with the same probability? Or am I even starting out incorrectly?

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    $\begingroup$ If A doesn't draw a diamond is the card replaced in the deck, or does B draw from $51$ cards? $\endgroup$ – saulspatz Aug 25 at 18:43
  • $\begingroup$ The thing is, it isn't really mentioned in the questioned. Usually, we assume that the card is not replaced. @saulspatz $\endgroup$ – Bladarc Aug 25 at 18:44
  • $\begingroup$ A wins if the first card is a diamond, or if the first $3$ cards are not diamonds and the fourth is, or if the first six cards are not diamonds, but the seventh is, ... $\endgroup$ – saulspatz Aug 25 at 18:50

This is an interesting question, or at least, it has an interesting answer.

Let $a,b,c$, respectively, be the probabilities that A,B,C win. If a non-diamond is replaced in the deck, and the deck is reshuffled, we see that $b=\frac34a$, for if A doesn't draw a diamond, we effectively start all over with B as the first player. Similarly, $c=\frac34b={9a\over16}.$ Since $a+b+c=1,$ (the probability that the game goes on forever is zero,) we have $$a\left(1+\frac34+{9\over16}\right)=1$$ so that $$\begin{align}a &= {16\over37}=0.432432\dots\\ b&={12\over37}=0.324324\dots\\ c&={9\over37}=0.243243\dots\end{align}$$

On the other hand, if the card is not replaced, let $p_n$ be the probability that the first $n$ cards are not diamonds, but the next one is. Then $$p_n={{39\choose n}\over{52\choose b}}\cdot{13\over52-n}$$ We have $$\begin{align} a &= p_0+p_3+p_6+\cdots+p_{39}\\ b &= p_1+p_4+p_7+\cdots+p_{37}\\ c &= p_2+p_5+p_8+\cdots+p_{38} \end{align}$$

Numerical calculation gives$$ \begin{align} a&=0.43127307363406425\\ b&= 0.3251444699292056\\ c&= 0.24358245643673027 \end{align}$$ so we see that replacing the card doesn't change the probabilities much at all. (I also performed the calculation using exact arithmetic to be sure that the values are actually different -- they are.)

I leave it to you to compute the players' expectations and variances.


The critical thing is to compute the probability of $A$ winning, the rest is standard and I'll omit it.

We can proceed recursively. Let $P_A(m,n)$ denote the probability that $A$ wins given that we start with $m$ non-diamonds and $n$ diamonds. Of course we want $P_A(39,13)$.

I'll assume we are working without replacement (the computation with replacement is similar but simpler). It is easy to deduce $P_B, P_C$ from $P_A$ as $$P_B(m,n)=\frac m{n+m}\times P_A(m-1,n)\quad \& \quad P_C(m,n)=\frac m{n+m}\times \frac {m-1}{m+n-1}\times P_A(m-2,n)$$

recursively, consider the effect of the first draw. $A$ wins on the first draw with probability $\frac n{n+m}$. If $A$ misses on the first draw, $A$ then counts on $B,C$ also missing, at which point we're back at the start, only with $m$ replaced by $m-3$. The probability of all three missing in the first round is $\frac m{n+m}\times \frac {m-1}{n+m-1}\times \frac {m-2}{n+m-2}$. It follows that $$P_A(n,m)=\frac n{n+m}+\frac m{n+m}\times \frac {m-1}{n+m-1}\times \frac {m-2}{n+m-2}\times P_A(m-3,n)$$

For initial conditions, we have $$P_A(0,n)=1\quad P_A(1,n)=\frac n{n+1}\quad P_A(2,n)=\frac n{n+2}$$

Fixing $n=13$ we can use a spreadsheet or a program to compute $$P_A(39,13)\approx \boxed {.4313}$$

Hard to see a useful sanity check on this...of course for small $m$ we expect $P_A(m,13)$ to be very high, as $A$ has an excellent chance of winning on the first round. As $m$ gets extremely large, we expect $P_A(m,13)\to \frac 13$, since the first few draws are all expected to fail and have minimal impact on the distribution (so all three players start out very close to equal). But that really does require large $m$...trusting my (hastily written) code I see $P_A(300,13)\approx .3475$ which is at least getting there. And $P_A(1000,13)\approx .3376$ which, to me, strongly suggests that the code is working.

  • $\begingroup$ I got the same answer for $P(39,13)$ as you did, by a somewhat different approach, so it look correct. In my case, I computed the probabilities for all three players independently, and they sum to $1$. $\endgroup$ – saulspatz Aug 25 at 19:46
  • $\begingroup$ @saulspatz Ah, good. Thanks. Yes, I think your method is different enough to count as a check. And, yes, $P_A+P_B+P_C=1$, numerically for all the $m$ I looked at...I didn't try to confirm that algebraically. $\endgroup$ – lulu Aug 25 at 20:08

Without replacement, $A$ can win at the $1,4,7,...,34$ or $37$th draws. Hence: $$P(A)=P_A(1)+P_A(4)+\cdots+P_A(37)=\\ \frac{13}{52}+\frac{39}{52}\cdot\frac{38}{51}\cdot\frac{37}{50}\cdot\frac{13}{49}+\cdots+\frac{39\cdots4}{52\cdots17}\cdot\frac{13}{16}=\\ 0.25+13\sum_{k=1}^{12} \frac{{39\choose 39-3k}}{{52\choose 51-3k}\cdot (3k+1)}\approx 0.431273.$$


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