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I'm working on a puzzle which describes a casino game where you draw 13 cards. You win if your first card is an ace, or your second is a two, or... all the way up to the 13th draw being a king. In order to analyse the problem, I started by simplifying to a two-card version. You draw two cards and win if the first is an ace or the second is a deuce.

The odds of drawing an ace on the first card is clearly 4/52 or 1/13.

My understanding, backed up by this question If you draw two cards, what is the probability that the second card is a queen?, suggests that the odds of drawing a 2 on the second card is also 1/13. What are the odds of drawing either? To my knowledge, you can't directly calculate that but instead calculate the inverse. The odds of not drawing a specific card should clearly be 12/13 and the odds of not drawing an ace followed by not drawing a two should be 12/13 * 12/13 = 144/169. The odds of drawing one or the other should therefore be 25/169.

But there's another approach to analysing the problem. There are 52 * 51 = 2652 different combinations for the first two cards. If A is an ace, T is a two and x is any other card, there are 5 combinations which give you a win:

AA Ax AT xT TT

Given that there are four aces, four twos and 44 other cards, the number of winning hands is:

4*3 + 4*44 + 4*4 + 44*4 + 4*3 = 392.

So the odds of winning should be 392 / 2652.

25/169 and 392/2652 are very close but they are NOT the same number. I believe the second number is accurate, but I can't see where the logic in the first method fails. I suspect that it has to do with the possibility of drawing both being double counted but I can't see how that should matter. It seems like you should be able to treat each draw as an independent event. Additionally, the second method doesn't scale well - it would be extremely tedious to directly calculate the number of winning hands for the full 13 card game.

edit: Clarified the rules for winning the game.

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  • $\begingroup$ The odds of firstly drawing an ace and secondly a two is $\frac4{52}\frac4{51}$. Can you understand why? $\endgroup$ – drhab Oct 30 '16 at 15:22
  • $\begingroup$ Arthur, I'm not counting the number of ways to draw aces and twos, I'm counting the number of ways to win. Neither xA nor Tx win the game. $\endgroup$ – Dan J. Oct 30 '16 at 15:44
  • $\begingroup$ Heads up - odds and probability are two different things. $\endgroup$ – Sean Roberson Oct 30 '16 at 15:56
  • $\begingroup$ drhab, I disagree. If you draw a two on the first draw, then the odds of drawing a two on the second draw are only 3/51, not 4/51. You have to take both possibilities into account and that gives you odds of 1/13. See the linked question in my submission. $\endgroup$ – Dan J. Oct 30 '16 at 16:04
  • $\begingroup$ The probability of firstly drawing an ace (not a two) and secondly a two is $\frac4{52}\frac4{51}$. If firstly an ace has been drawn then $4$ of the remaining $51$ cards are two's. $\endgroup$ – drhab Oct 30 '16 at 16:26
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Let $A$ denote the event that the first draw will be an ace and let $B$ be the event that the second draw will be a two.

Then:$$\Pr(A\cup B)=$$$$\Pr(A)+\Pr(B)-\Pr(A\cap B)=$$$$\Pr(A)+\Pr(B)-\Pr(A)\Pr(B\mid A)=$$$$\frac4{52}+\frac4{52}-\frac{4}{52}\frac{4}{51}$$

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The problem with your first method is that the events are not independent. Yes, the probability that you don't draw an ace from a full deck is $12/13$, and the probability that you don't draw a two from a full deck is $12/13$, but if you draw one card and then draw a second, the outcome of the second draw is dependent on the outcome of the first draw. Hence you cannot multiply the probabilities.

Your second method of calculating $P(\text{ace on first draw or two on second draw})$ is correct because it takes into account the outcome of the first draw and the second draw.

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  • $\begingroup$ Then is the question I linked to about drawing a queen on the second card wrong? In particular, look at A.J.'s accepted answer showing that the odds remain 1/13 even if you take the first draw into account. Is that incorrect or does it not apply here? And if doesn't apply, why not? Also, my description of the game was ambiguous. It should be OR, not AND. I'll edit. $\endgroup$ – Dan J. Oct 30 '16 at 15:50
  • $\begingroup$ I'm sorry, I misunderstood the condition for winning! As for the accepted answer in the link, the probability that the second card is a queen given that you know nothing about the first card is indeed $1/13$. Similarly, the probability that the second card is not a two given that you know nothing about the first card is $12/13$. But you cannot use this to calculate $P(\text{first card not ace and second card not two})$ because in that event you do have information about the first card. $\endgroup$ – kccu Oct 30 '16 at 20:23
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On questions like these you have to stop yourself from taking shortcuts and keep your eye on the big picture. Therefore always use both cards. First the number of probabilities of getting an ace on the first throw is 4 times the other 51 cards. Second the odds on getting a 2 on the second card if the first one isn't a ace or a 2 is 4 times 44 plus if the first card is a 2 then 4 times 3. Third add up all the 4 times and you get 4 times 98 which equals 392.

  • 4 * 51 +
  • 4 * 44 +
  • 4 * 3 =
  • 4 * 98 = 392

Or for simplicity you could just add the odds of an ace for the first card (4 times 51) and a deuce for the second card (51 times 4) then minus one of the double wins (4 times 4) and you get 4 times (51 + 51 - 4) equals 4 times 98 equals 392.

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