# A very strange deck - probability and expected number of draws

Say we have a virtual deck of 70 cards of four suits and each player has access to his/her own unique independent deck (one players' actions do not affect another player's):

$$\begin{array}{r|rr} &\text{probability}& \text{quantity of card}\\ \hline \text{Heart}&0.5\%& 1\\ \text{Spade}&5\hphantom{.0}\%& 9\\ \text{Diamond}& 50\hphantom{.0}\% & 30 \\ \text{Clubs}& 44.5\%& 30 \end{array}$$

Here's where it gets interesting: the probability of the card does not change as cards are pulled out (without replacement!).

For example if I pull a club, the probability to draw another club remains 44.5%...same if I pull 2 clubs, still 44.5%. This holds until I pull all the clubs, at which point, there are no more clubs in the virtual deck and the probability drops to 0.

How would we calculate:

Expected number of draws it would take to draw a heart

% of players who will draw a heart on the 0-69th draw

Probability of an individual user drawing the heart on their last (70th) draw

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It is a very strange deck, if the $30$ diamonds are $50\%$ and the $30$ clubs are $44.5\%$. –  André Nicolas Jun 6 '13 at 21:02

Let $X$ = # of draws to draw a heart. Let's consider $P(X)$. Note that $X$ could be $1,2,...,70$. Since draws are independent (provided that cards don't run out), we have: $$\begin{array}{c|c} X & P(X) \\ \hline \\ 1 & 0.005 \\ 2 & (0.995)(0.005) \\ 3 & (0.995)^2(0.005) \\ 4 & (0.995)^3(0.005) \\ \vdots & \vdots \\ 69 & (0.995)^{68}(0.005) \\ 70 & ? \end{array}$$

Note that the % of players who will draw a heart on the ($1$ to $69$)th draw is: $$P(X<70)=\sum_{x=1}^{69} 0.005(0.995)^{x-1} = \dfrac{0.005(1-0.995^{69})}{1-0.995} = 1-0.995^{69}\approx 0.2924$$

Thus, the probability of an individual user drawing the heart on their last ($70$th) draw is: $$P(X=70)=1-P(X<70) \approx 1 - 0.2924 = 0.7076$$

Hence, the expected number of draws is: $$E(X) = \sum_{x=1}^{70} xP(x) = 70(0.7076) + \sum_{x=1}^{69} x[0.005(0.995)^{x-1}] \approx 59.2$$

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There is the complication that for example the probability of spade drops to $0$ once $9$ spades have been drawn, so the probability of heart is not constant. Your calculation should, however, give a good approximation. –  André Nicolas Jun 6 '13 at 21:52
Thanks! this is exactly what I was looking for! Could you remind me how to do the summation of x[.0005(.995)^(x-1)]? –  Derek Jun 6 '13 at 22:38
What if the probabilities changed, how would that affect the answer? For example: if heart was changed to 20%, and/or spade was also raised to 20%. Would the change in the probability of the heart be significant enough to throw off this approximation? –  Derek Jun 6 '13 at 23:01
I agree with André; this is only an approximation. Note that: $$\sum_{n=1}^k nr^{n-1} = \dfrac{k r^{k+1}-(k+1) r^k+1}{(r-1)^2}$$ (this is based off taking the derivative of the power series corresponding to an infinite geometric series). Thus: $$\sum_{x=1}^{69} x[0.005(0.995)^{x-1}] = 0.005 \left[ \dfrac{69 \cdot 0.995^{70}-70 \cdot 0.995^{69}+1}{(0.995-1)^2} \right]$$ –  Adriano Jun 6 '13 at 23:36
Changing the probability of getting a heart will significantly affect the answer; just replace every $0.005$ with $0.2$ and replace every $0.995$ with $0.8$. Changing the probability of getting a spade will not affect the answer at all (assuming that the probability of getting a heart is still $0.005$). –  Adriano Jun 6 '13 at 23:42