sodawater Joined: May 14, 2012
• Posts: 3318
May 17th, 2020 at 1:00:48 AM permalink
I do not know the answer to this one, but I am pretty sure it's 50 percent. However, is this one of those unintuitive, Monty-Hall-type problems where new information changes the odds?

A robot deals out three poker cards face down, with two possibilities of equal probability: 2 red cards and 1 black, or 2 black cards and 1 red.

The robot is programmed to reveal the first two cards one at a time, and must show both a red card and a black card in the first two cards (to maintain suspense), but the order is random.

Does the conditional probability of having the set with 2 red cards change from 50 percent if we see the first card is red?
ksdjdj Joined: Oct 20, 2013
• Posts: 864
Thanks for this post from: May 17th, 2020 at 2:03:37 AM permalink
IMO, yes and no (it depends how you program the robot).

If the robot is programmed like this:

1. If it picks the first card at random and it ends up being red, before it reveals the 2nd card (black card) then I think it is a 2/3 chance of having 2 reds and 1 black.

2. If it picks the first two cards from a "batch of two cards that must have a red and a black card", and reveals the red card 1st, then I think it is a 50% chance having 2 reds and 1 black.

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Just got back from my first night out from dinner since the Corona virus, so a bit tipsy (whether I am correct or not, I hope this post made sense).
sodawater Joined: May 14, 2012
• Posts: 3318
May 17th, 2020 at 2:43:35 AM permalink
Thanks for your answer, ksdjdj. Now I think the way I was originally thinking about the problem means the answer is actually 2/3.

So for clarification:

* The robot knows the color of all three cards.

* The robot first randomly selects any card to turn over.

* The robot then turns over the card that is the different color from the first one.

So now the third card has a 2/3 chance of matching the first card color, correct?

It is crazy how I can understand the Monty Hall problem for decades and still get this one wrong initially.
ChesterDog Joined: Jul 26, 2010
• Posts: 851
Thanks for this post from:  May 17th, 2020 at 5:59:31 AM permalink
Quote: sodawater

Thanks for your answer, ksdjdj. Now I think the way I was originally thinking about the problem means the answer is actually 2/3.

So for clarification:

* The robot knows the color of all three cards.

* The robot first randomly selects any card to turn over.

* The robot then turns over the card that is the different color from the first one.

So now the third card has a 2/3 chance of matching the first card color, correct?

It is crazy how I can understand the Monty Hall problem for decades and still get this one wrong initially.

Yes.

And here is the equation:

P(red picked first) * P(RRB | red picked first) = P(RRB) * P(red picked first | RRB)

Three of the above probabilities are easy:
• P(red picked first) = 1/2 since RRB and BBR are equally likely, there are equal numbers of reds and blacks.
• P(RRB) = 1/2 because RRB and BBR are equally likely.
• P(red picked first | RRB) = 2/3 because the robot will pick a card at random, and there are twice as many reds in RRB.

Substitute:
P(red picked first) * P(RRB | red picked first) = P(RRB) * P(red picked first | RRB)
(1/2) * P(RRB | red picked first) = (1/2) * (2/3)

Solve:
P(RRB | red picked first) = 2/3
sodawater Joined: May 14, 2012
• Posts: 3318
May 17th, 2020 at 1:33:11 PM permalink
Thanks, Chester, looks good.
Wizard 