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I saw this interview question that I'm trying to do and am clearly missing something. It goes as follows:

Deck of $26$ cards consisting of two suits (say $13\times$Hearts and $13\times$Clubs). The aim is to get $5$ cards of a given suit. I am allowed to play in either of these two ways:

  • draw $1$ card, choose my suit.
  • draw $2$ cards, choose my suit.

The game then proceeds by turning over successive cards till either me or the opponent get $5$ of a kind.

Which of these is better? What about if this were a full deck of 52 cards?

In situation 1, I will have 4 more to get vs my opponent needing 5 more. In situation 2, either we're even in needing 4 more (with probability $\frac{13}{25}$) or I'm ahead, needing only 3 more cards (with probability $\frac{12}{25}$).

I believe the probability of winning in both cases is the same, because I can expand the first strategy in the second: having chosen the suit, either the next is of the same suit (in which case I would select it in strategy 2), or they are different, which doesn't favour one over the other.

So I assume the question is asking about the expected number of draws. But then this is where I'm having trouble as it seems unnecessarily complicated, so I must be missing something. I considered conditioning on the first two cards ie '1st and 2nd cards are the same suit' (and I'm then looking for the expected number of steps till I reach $3$ of a kind, without my opponent reaching $5$ of a kind) and '1st and 2nd cards are the same suit' (and I'm then looking for the expected number of steps till I reach $4$ of a kind, without my opponent reaching $4$ of a kind).

But computing these by hand looks like a pain, with lots of binomials (I tried to count by picturing the deck as an ordered list, where I'm looking at the first $n$).

So I feel like I'm making a mess of something that shouldn't be difficult. Would love some help.

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  • $\begingroup$ I'm not sure how to solve this, but I'll just point out that you're not even in situation 2. If you draw 2 cards of differing suits, then you've taken away one of your opponents cards, so he no longer has 13/25 probability. $\endgroup$
    – dperrin
    Feb 8 at 18:12
  • $\begingroup$ @dperrin Sorry I meant the probability of arriving at "we now each have 4 to get" after two draws is $13/25$, and the probability of arriving at "I now need 3 more, you need 5 more" after two draws is $12/25$. Is that not correct? $\endgroup$
    – moonerism
    Feb 8 at 18:20
  • $\begingroup$ I don't quite understand. If you choose Clubs, does your opponent also have to get $5$ Clubs, or does he have to get $5$ Hearts? $\endgroup$
    – saulspatz
    Feb 8 at 18:37
  • $\begingroup$ @saulspatz 5 Hearts $\endgroup$
    – moonerism
    Feb 8 at 19:43
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It doesn't matter which approach you choose, assuming that if only one suit has been seen, you will name that suit.

Let $P_1$ be the probability that you win if two cards have been drawn and they are of different suits, and let $P_2$ be the probability that you win, if two cards have been drawn and they are both of your suit.

Without loss of generality, the first card turned up is a Club. If you name Clubs, then with probability $\frac{12}{25}$ the next card will be a Club, and with probability $\frac{13}{25}$ it will be a Heart, so by the law of total probability, your probability of winning is $$.48P_2+.52P_1$$

Now suppose instead, you elect to see a second card. With probability $.48$ it will be a Club, and you will name Clubs, giving you a probability of winning of $P_2$. With probability $.52$, it will be a Heart, and you will name one of the suits, and have probability $P_1$ of winning. By the law of total probability, your overall probability of winning is the same as before.

The simplest approach is simply to name the suit of the first card turned up.

As to the question about a full $52$-card deck, the probabilities are exactly the same, assuming that when you name a suit before any card of another suit has been turned up, your opponent's suit will be the first new suit drawn. The cards in the other two suits (Diamonds and Spades in the original example,) have no effect on the outcome, so we may simply ignore them.

There's a substantial advantage to going first in this game. I calculate the probability of a first player win at approximately $0.613272$.

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  • $\begingroup$ Thanks, yes what you're saying is what ambiguously worded as 'expand strategy 1 in to strategy 2'. I found the same probability you did too ($268/437$). Is there are clever way of calculating the 'expected number of draws till someone wins' or is the only method to sum explicitly $\sum k\,\mathbb{P}(\text{win after exactly } k \text{ draws})$? It's not a big sum but by hand the denominators of the fractions get big $\endgroup$
    – moonerism
    Feb 9 at 0:43
  • $\begingroup$ @moonerism There may be a clever way, but if so, it's too clever for me. $\endgroup$
    – saulspatz
    Feb 9 at 4:00

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