Question

We start with 6 Black playing cards and we add cards to the point where the probability of drawing 2 Red cards in a row without replacement is `1/2`. How many cards do we add?

Answer 1

`15` red cards.

Explanation:

Let the number of red cards to be added to `6` black cards be `x`, where `x` is natural number.

Now probability of getting first red card is `x/(x+6)`

and conditional probability of getting second red card (assuming we have already got one red card) is `(x-1)/(x+5)`, as we are considering without replacement.

Hence probability of drawing `2` red cards, without replacement, is
`x/(x+6)xx(x-1)/(x+5)` and as this probability is `1/2`, we have

`(x(x-1))/((x+6)(x+5))=1/2`

or `2x(x-1)=(x+6)(x+5)`

or `2x^2-2x=x^2+11x+30`

or `x^2-13x-30=0`

or `(x-15)(x+2)=0`

i.e. `x=15` or `x=-2`

But as number of red cards can only be a natural number

`x=15`

Answer 2

If we only add Red cards, 15. If we don't limit the colour of cards added, there is no number that can be added that will achieve that probability.

Explanation:

Let me first point out that if we're talking about solely red cards being added, the answer is 15 and the details are here:

https://socratic.org/questions/how-many-red-cards-must-be-added-to-6-black-cards-so-that-the-probability-of-dra?source=search

That said, this question does not limit itself to red cards being added, and so we can have both black and red added. If we don't limit ourselves to a single pack of cards and can instead draw from an infinite supply of mixed cards , we can approach the question this way - first I'll lay out the starting argument the same way the question handling adding solely red cards does:

We have (B)lack cards and (R)ed cards. We start with `B=6` and want to add R so that the odds of drawing 2 R, without replacement, is `1/2`. How many R must we add?

The odds of drawing an R on a single draw is:

`R/(R+B)`

and so for example if we have `R=B=6`, we'd have the odds of drawing an R to be:

`6/(6+6)=6/12=1/2`

So now let's add that second draw into the mix. We've already drawn an R and so we have one R less, so the ratio for the second draw is:

`(R-1)/((R-1)+B)`

Which means that the two draws taken together are:

`(R/(R+B))((R-1)/((R-1)+B))`

Now in the analysis with working with only adding R, we know that `B=6`. We don't know that here. We know that `B >=6` and that, in general, the number of R we add will be equal to the number of B added, so we can express that as:

`B=6+R`

We still want the odds of the two draws to be `1/2`, so we'll get:

`(R/(R+6+R))((R-1)/((R-1)+6+R))=1/2`

`(R/(2R+6))((R-1)/(2R+5))=1/2`

`(R(R-1))/((2R+6)(2R+5))=1/2`

`(R^2-R)/(4R^2+22R+30)=1/2`

`2(R^2-R)=4R^2+22R+30`

`2R^2-2R=4R^2+22R+30`

`0=2R^2+24R+30`

And now I'll use the Quadratic Formula:

`x = (-b \pm sqrt(b^2-4ac)) / (2a)`

`x = (-24 \pm sqrt(24^2-4(2)(30))) / (2(2))`

`x = (-24 \pm sqrt(576-240)) / 4`

`x = (-24 \pm sqrt(336)) / 4`

`x~= (-24 \pm 18.33) / 4`

And all we end up with is negative results. This means there are no number of mixed cards that will give us the probability of drawing 2 R in a row with the odds of `1/2`.