1
$\begingroup$

I have seen this and this, but my setup is a little different so I'm not sure how to proceed.

Setup:

We are using a single standard deck of 52 cards.

Suppose I have been dealt two cards, and one of them is a 5. The other is neither a 4 nor a 9.

Suppose my opponent has also been dealt two cards (I cannot see them).

Suppose the flop (subsequent deal of three cards face up on the table for everyone to use) contains 6, 7, 8.

Suppose two more cards will be dealt face up on the table for everyone to use.

Question:

What is the probability that I will obtain a straight, or a straight flush, of 5 cards? That is, what is the probability that one or both of the next two cards is a 9 or a 4?

I know this is wrong because the probability is way too high, but here's how I've started:

1) There are 52-7 = 45 cards left in the deck.

2)

P(1st card is 4 | opponent has no 4s) = 4/45

P(1st card is 4 | opponent has one 4) = 3/45

P(1st card is 4 | opponent has two 4s) = 2/45

If the first card is a 4, then I don't care what the second card is - I have my straight.

3)

P(1st card is 9 | opponent has no 9s) = 4/45

P(1st card is 9 | opponent has one 9) = 3/45

P(1st card is 9 | opponent has two 9s) = 2/45

4)

So the probability that the 1st card is either a 4 or a 9 is: (2(4 + 3 + 2)/45) = 0.4

5) If the first card is not a 4 and not a 9, then I don't have my straight yet. I need to wait for the second card.

P(2nd card is 4 | opponent has no 4s AND 1st card was neither 4 nor 9) = 4/44

P(2nd card is 4 | opponent has one 4 AND 1st card was neither 4 nor 9) = 3/44

P(2nd card is 4 | opponent has two 4s AND 1st card was neither 4 nor 9) = 2/44

6)

P(2nd card is 9 | opponent has no 9s AND 1st card was neither 4 nor 9) = 4/44

P(2nd card is 9 | opponent has one 9 AND 1st card was neither 4 nor 9) = 3/44

P(2nd card is 9 | opponent has two 9s AND 1st card was neither 4 nor 9) = 2/44

7)

So the probability that the second card is either a 4 or a 9 will be:

$$2((4+3+2)/44) \approx 0.4091$$

8) Adding these together, the probability that I get a straight is:

0.4091 + 0.4 = 0.8091.

Obviously this is way too high. Why am I wrong and how do I fix it?

EDIT:

Okay, thanks for the hint. The comment implies that what my opponent has doesn't matter, but I'm not sure why.

I condition on what my opponent has because if they have, say, a four, then there are only 3 fours left in the deck, so the probability that I get a four is not going to be the same as if there were 4 fours in the deck. If there is no opponent, then there would be 52-5=47 cards in the deck, and 4 of them would definitely be fours, so I think that changes both the denominator and the numerator.

Let's pretend that I don't have an opponent.

Then there are 52-5=47 cards left in the deck.

The probability that the first card is either a 4 or a 9 would be 8/47

The probability that the second card is either a 4 or a 9, given that the first card was neither a 4 nor a 9, would be 8/46.

So then my probability of getting a straight would be $$8/47 + 8/46 \approx 0.3441$$, which I think is still too high. I'm clearly missing something, and I feel like it's very obviously staring me in the face, but I just don't see it.

Improve this question
$\endgroup$
6
  • $\begingroup$ Hint: Try to explain why you condition on what your opponent might have. In particular, can you point to how that differs from the situation where there is no opponent but the cards you have observed are exactly the same as described? $\endgroup$
    – whuber ♦
    Commented Jan 25, 2020 at 18:45
  • $\begingroup$ @whuber Thanks for the hint! I've taken a stab at it, but I still think I'm wrong. Could you please have a look? $\endgroup$
    – StatsSorceress
    Commented Jan 26, 2020 at 22:14
  • $\begingroup$ Think of it this way: the probabilities don't care how the deal is made, so first deal your hand, then deal the two common cards, then deal the flop (keep the cards upside down if you like), and finally deal all the other hands. Now it is physically obvious that the flop does not depend on what is in your opponents' hands; it doesn't even depend on whether you have any opponents at all. The lesson is that the probabilities only depend on the information you have. $\endgroup$
    – whuber ♦
    Commented Jan 26, 2020 at 22:57
  • $\begingroup$ Thank you, but I'm still missing something. Why don't the probabilities care how the deal is made? Doesn't it matter whether, say, a 4, is already on the table somewhere? $\endgroup$
    – StatsSorceress
    Commented Jan 26, 2020 at 23:06
  • $\begingroup$ The probabilities depend only on the information you have; they do not use any information about where the rest of the cards might be physically located. You can put them all in the deck or all in your opponent's hands; it doesn't matter. It can be helpful to consult a discussion of the axioms of probability: look for any reference to time or even physical manifestation of a population or process: you won't find any. $\endgroup$
    – whuber ♦
    Commented Jan 26, 2020 at 23:15

1 Answer 1

Reset to default
0
$\begingroup$

You've got it. $0.3441$ is the correct answer.

I'll explain why your reasoning was wrong in the first place. Let $M$ be the chance that a 4 is drawn, and let $V_0$, $V_1$, and $V_2$ be the chance that your opponent has zero, one, and two 4s, respectively.

The calculation you did was

$P(M) = P(M|V_0) + P(M|V_1) + P(M|V_2),$

but the correct formula is actually

$P(M) = P(M|V_0) P(V_0) + P(M|V_1) P(V_1) + P(M|V_2) P(V_2),$

or, equivalently,

$P(M) = P(M \land V_0) + P(M \land V_1) + P(M \land V_2).$

As for why you don't actually consider the number of 4s that your opponent has...

The number of 4s your opponent has matters in the sense that if you knew it, then that would affect the final answer. But it doesn't matter in the sense of you needing to consider it when doing your calculation.

As an analogy, consider the situation where you're playing with Alice, Bob and Carol, and everyone has been dealt their hand, but you haven't looked at either of your cards yet. What is the probability that your first card is red?

Well, there are a lot of things that "matter" in this case. The number of red cards Alice has matters; likewise, the number that Bob has, and the number that Carol has. For that matter, it also matters where the Queen of Spades is—if it's anywhere besides your first card, that increases the chance that your first card is red. And every other card matters, too.

But all of that is totally irrelevant, of course, since the probability that your first card is red is $1/2$.

Simply put... if there's a way to calculate the probability without considering your opponent's hand, then you don't need to consider your opponent's hand in order to calculate the probability!

In this case, there are 2 unseen cards that will be dealt and 45 unseen cards that will not be dealt. For each of those 45 cards, you don't care whether it's in the deck or your opponent's hand; it's all the same to you. So you don't need to consider your opponent's hand.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks Tanner, but I'm not sure that the formula you gave is what I calculated. Thta is, I don't think I calculated 0.3441 = 𝑃(𝑀)=𝑃(𝑀|𝑉0)𝑃(𝑉0)+𝑃(𝑀|𝑉1)𝑃(𝑉1)+𝑃(𝑀|𝑉2)𝑃(𝑉2) (for 4) plus the same again but in the case of 9. I'm also confused because you say that we don't consider the cards the opponent has, but that's whats used in the formula (V0, V1, V2). Could you please clarify? $\endgroup$
    – StatsSorceress
    Commented Jan 26, 2020 at 23:35
  • $\begingroup$ You have two options: one option (harder) is to use the formula $P(M) = P(M|V_0) P(V_0) + P(M|V_1) P(V_1) + P(M|V_2) P(V_2)$, and the other option (easier) is to just do 8/47 + 8/46. Both options are correct, and you used the second option and got the correct answer. You would have also gotten the correct answer if you had used the first option. $\endgroup$
    – Sophie Swett
    Commented Jan 26, 2020 at 23:50
  • $\begingroup$ Oh! Okay, so the denominator really should be 47, and not 45, because I consider the deck as "any unseen cards" (which would include my opponent's cards). $\endgroup$
    – StatsSorceress
    Commented Jan 27, 2020 at 0:11
  • $\begingroup$ @StatsSorceress Yep. Or, to be precise, the number of cards in the deck doesn't matter; what matters is the number of unseen cards. $\endgroup$
    – Sophie Swett
    Commented Jan 27, 2020 at 0:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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