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Question: You draw n cards at random from a standard deck of 52 playing cards. If the n cards contain either a J, Q or K then you win. If not then you put all n cards back in the deck, reshuffle and draw n − 1 new cards. You repeat until you win or until n = 0, when you lose. What are the probabilities of winning and losing if you start with (a) n = 1, (b) n = 2, (c) n = 8, or (d) n = 41?

I've approached this question and believe I've come up with a solution for part (a) and part (b), but (c) and (d) have really stumped me.

For part (a) I concluded that as you only select 1 card, the probability that you select a J, Q or K (4*3 = 12, so 12 total J/Q/K cards) is 12/52... and thus the probability that you lose is 1 - 12/52 = 40/52.

For part (b) I looked at the probability for losing first. On the first selection, you have to not select a J, Q or K card out of either of the 2 cards. So card 1 would be 40/52 probability, and card 2 would be 39/51 probability. So 40/52 * 39/51 = 10/17 so there is 10/17 chance of losing... and picking up 1 (n-1 cards) card next round. For total losing probability you would have to multiply probability of losing from first round, i.e. 10/17 by the probability of losing in the second round, i.e. 40/52... giving 100/222 as the probability for losing (for the n=2 case). So there is a 1 - 100/222 = 121/222 probability of winning (for the n=2 case).

For 8 cards and for 41 cards, I'm a bit stumped... and I'm not even sure if I got (a) and (b) totally correct. Any help would be greatly appreciated!

Thanks!!

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  • $\begingroup$ You have a typo. You mean the probability of losing is ${100\over221}$ in the $n=2$ case, not ${100\over222}$ $\endgroup$
    – saulspatz
    Commented Sep 4, 2018 at 16:41

2 Answers 2

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It is easier to find the probability that you lose : in order to lose the game you must not pick a J,Q or K in any of the draws which you make. Clearly, you either win or lose the game so calculating this probability we can subtract from $1$ to get the winning probability.


The way of thinking is this : when we pick $n$ cards, our sample space is $\binom{52}{n}$, the number of ways of picking $n$ cards out of $52$. We lose, while picking $n$ cards, if and only if our hand of cards does not contain a $J,Q,K$ : this means, it must be a subset of the other $40$ cards which aren't $J,Q,K$. Hence, the number of ways of achieving a selection of this kind is $\binom{40}n$ the number of ways of choosing all $n$ cards from that collection.

Thus, at each round with $n$ cards ,the probability of loss is $\frac{\binom{40}n}{\binom{52}n}$.


Now, the game at stage $n$ is independent at stage $n-1$, since we are shuffling and randomly picking again. Hence:

For $n = 1$, the answer is just $\frac{\binom{40}1}{ \binom{52}1}$.

For $n = 2$, we must lose when $n = 2$ and when $n = 1$. Since the probabilities are independent, they are multiplied so the answer is $\frac{\binom{40}2}{\binom{52}2} \times \frac{\binom{40}1}{\binom{52}1}$.

For $n = 3$, we must lose when $n = 3$ and when $n = 2$ and $n = 1$, so the answer is : $\frac{\binom{40}3}{\binom{52}3} \times \frac{\binom{40}2}{\binom{52}2} \times \frac{\binom{40}1}{\binom{52}1}$.

For $n = 8$, the answer is?

Now, by subtracting from $1$ you may obtain the winning probabilities.

Also, note that by using the formula $\binom nk = \frac{n!}{k!(n-k)!}$, we may provide great simplification to each of the expressions above.

For example, the probability for $n = 1$ is just $\frac{40}{52}$. Similarly, for $n = 2$ you get $\frac{40^2 \times 39}{52^2 \times 51}$. Next, you'd get $\frac{40^3 \times 39^2 \times 38}{52^3 \times 51^2 \times 50}$.


For $n = 41$, can you even lose? That is, can you show that every pick of $41$ cards must contain one of $J,Q,K$?

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  • $\begingroup$ Ah, that makes a lot of sense, thank you!! Is there any way to simplify the final fractional expressions or is it best to leave it as that? I can't really think of anyway in particular... but what I did notice in fact, as you commented, was that for n = 41 as there are 12 cards including J/Q/K then one of those 12 has to be selected as only 11 (52 - 41 = 11) remain! $\endgroup$
    – MKY6
    Commented Sep 4, 2018 at 20:19
  • $\begingroup$ No I do not think there can be any simplification of the expression I have written. However it is recursive in nature. Also, your second answer is correct! $\endgroup$ Commented Sep 5, 2018 at 2:50
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You did fine for a and b. For c it is the same approach but more work. You have to compute the chance of losing when you draw $8$ cards, which you do just like you did for $2$ cards except there are $8$ factors. A spreadsheet can make the computation easier because now that you have computed the chance of losing a two card draw the chance of losing a three card draw is just $\frac {38}{50}$ times it, then to get a four card draw you multiply by $\frac {37}{49}$ and so on.

The last part sounds like it would be a huge amount of work, but if you think about it there is a trick.

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  • $\begingroup$ So it's almost as though you're performing a geometric factorial? So for an n card draw, the probability of losing will be the probability of losing for the n-1 card draw multiplied by (40-(n-1))/(52-(n-1)).I'm not quite sure how to write this as a single expression series though? $\endgroup$
    – MKY6
    Commented Sep 4, 2018 at 16:54
  • $\begingroup$ That is correct. If you are doing it in a spreadsheet you can make a column for $n$ and a column for the chance of losing an $n$ card draw. Then just reference the $n$ column in your formula. it would be =up*(41-left)/(53-left) $\endgroup$ Commented Sep 4, 2018 at 16:57

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