1
$\begingroup$

I'm playing a card game where you begin with 5 cards and if you mulligan, you place them at the bottom and draw 5 new cards. If I were to use a standard 52 poker deck what I'd be asking is:

What is the probability to draw at least 2 diamond cards if I mulligan on a failure?

Keep in mind on mulligan we have known information. If we hit 0 or 1 diamonds, they would go to the bottom giving use 2 possible mulligan scenarios. I'm looking for a formula so I can apply it to a card game to determine the amount of a card type to include to reach a 90% success rate of this scenario.

$\endgroup$
  • 1
    $\begingroup$ As with many problems of this type, it’s likely easier to compute the probability of drawing at most one diamond. $\endgroup$ – amd Apr 11 '17 at 0:20
  • 1
    $\begingroup$ What's "a failure"? If you bury $n$ diamonds, then you can compute the probability of getting at least $2$ in the next $5$. But is that what you are asking? $\endgroup$ – lulu Apr 11 '17 at 0:30
  • 1
    $\begingroup$ So you're sure you mulligan exactly one time? That means you know for a fact you got either one or zero diamonds, and it becomes a very straight forward problem adding the two mutually exclusive probabilities-(zero diamonds of five cards of 52)*(two diamonds of five cards of 47 cards 13 diamonds included)+(one diamond of five cards of 52)*(two diamonds of five cards of 47 cards 12 diamonds included) $\endgroup$ – Arby Apr 11 '17 at 0:42
  • $\begingroup$ A failure is simply not obtaining 2 diamonds in the first attempt. The situation I'm trying to determine needs 2 successes. I'm attempting to apply this formula to another game where I have 50 cards and need to know how many of a card type must I run to reach 90%+ success of drawing 2 between both the first and second attempt of an opening hand. Determining one card is easy, as you just check the distribution of seeing 10 drawn cards. And yes you can only mulligan once. $\endgroup$ – David Cox Apr 11 '17 at 16:31
1
$\begingroup$

Let $X$ be the # of diamonds among the initial 5 cards.

Let $Y$ be the # of diamonds among the next 5 cards.

It seems you're looking for

$\mathbb{P}(X+Y\geq 2\mid X\leq 2) =\mathbb{P}(Y\geq 2-X\mid X\leq 2) = \\ \mathbb{P}(X=0)\cdot\mathbb{P}(Y\geq 2\mid X=0) + \mathbb{P}(X=1)\cdot\mathbb{P}(Y\geq 1\mid X=1) = \\ \frac{\binom{39}{5}}{\binom{52}{5}}\cdot (1- \frac{\binom{34}{5}}{\binom{47}{5}} - \frac{\binom{34}{4}\cdot \binom{13}{1}}{\binom{47}{5}}) + \frac{\binom{39}{4}\cdot \binom{13}{1}}{\binom{52}{5}}\cdot (1- \frac{\binom{35}{5}}{\binom{47}{5}})$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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