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I'm trying to teach myself statistics online @ khan academy and although its a wonderful resource I have no one to turn to for help when I don't understand. I hope someone here might be able to give me some advice.

I thought I understood it all and thought would test myself so I've made some questions up to try and work out the theoretical probability, then would cross x the answer against a simulation i've written in VBA/Excel to really understand it all.

But.... I'm really stuck on working out the probability of picking card #1 with value 6, #2 with value 6 then #3 a Red card.

Working out P(6,6) is easy for me = 4/52 * 3/51 ≈ 0.4525%

My issue is that after having picked two 6's in a row, what is the probability of then picking out a red card - I just can't get my head around it. Because you could have already picked 2 red cards which would reduce the probability or you could have picked 1 or none.

I don't know how you work it out from here, and googling it doesn't help as i'm not wording my question right and not getting the results I want.

Can anyone help me / push me in the right direction.

ANYTHING would be great.

Thanks

Lewis

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    $\begingroup$ A shortcut is to say that symmetry makes the probability of the third card being red after two sixes is $\frac12$. But it would be more educative to calculate the different cases as @lulu suggests. Use the shortcut as a check $\endgroup$
    – Henry
    Commented Jun 9, 2019 at 18:59
  • $\begingroup$ Use LaTeX please. $\endgroup$
    – Michael Rozenberg
    Commented Jun 9, 2019 at 19:12

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You got it already. Since you are aware that first two picks being red or not affects the probability of third pick being red, you should consider it case by case:

Case 1: First two 6's are black. First, we can find the probability of having two black 6's in our two picks. This is nothing but $\frac{2}{52}\cdot\frac{1}{51}$. In this case, having a red card on the third pick is $\frac{26}{50}$ since we still have $26$ red cards.

Case 2: One of first two picks is red. In this case, probability of having one black 6 and one red 6 is nothing but $\frac{2}{52}\cdot\frac{2}{51}\cdot2$. Here, we multiplied by $2$ since we can have our first pick a red 6 and second pick a black 6; or first pick a black 6 and second pick a red 6. And these cases are just symmetric so instead of separating them, we just multiplied the result by $2$. Then having a red card on the third pick is $\frac{25}{50}$ because we are left with $25$ red cards after first two picks.

Case 3: I am leaving it to you to find the probability for this case and the answer. I can check it if you write it as a comment.

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  • $\begingroup$ Thank you so much for your explanation. Its really helped me and I think I have the answer. $\endgroup$
    – Lewis Morris
    Commented Jun 10, 2019 at 4:43
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    $\begingroup$ I've had to change it around a bit because the probability of picking a red card is higher as there could be between 24-26 red cards not 11-13 (two suits, not one). I believe that for case 3 it would be 2/52 * 1/51 * 24/50 ≈ 0.04% (slightly lower than case 1 as there are less red cards to choose from) . I hope this means the overall probability of P(6,6,Red) ≈ 0.2262% $\endgroup$
    – Lewis Morris
    Commented Jun 10, 2019 at 4:50
  • $\begingroup$ Oh, that is mistake on my part, I am really sorry for that. Yes you are absolutely right, in my answer too, it should be $\frac{26}{50}$ and $\frac{25}{50}$ in case 1 and case 2 respectively. Seems like you checked my answer here :) $\endgroup$
    – ArsenBerk
    Commented Jun 10, 2019 at 8:08
  • $\begingroup$ Thank you so much, you've shed some light on something I was having trouble understanding before. Amazing !! :) $\endgroup$
    – Lewis Morris
    Commented Jun 10, 2019 at 8:45
  • $\begingroup$ I thank you too for your correction and you are welcome :) $\endgroup$
    – ArsenBerk
    Commented Jun 10, 2019 at 9:08

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