May 17th, 2020 at 1:00:48 AM
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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?

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?

May 17th, 2020 at 2:03:37 AM
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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).

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).

May 17th, 2020 at 2:43:35 AM
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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.

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.

May 17th, 2020 at 5:59:31 AM
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Quote:sodawaterThanks 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

May 17th, 2020 at 1:33:11 PM
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Thanks, Chester, looks good.

May 17th, 2020 at 8:23:56 PM
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I think the way you're wording the problem is confusing. May I suggest this alternative.

There are two sets of three cards. One set has two red cards and one black. The other set has two black cards and one red. A set is chosen at random and a card from that set is chosen randomly and revealed. The revealed card is red. What is the probability it came from the deck with two red cards?

There are two sets of three cards. One set has two red cards and one black. The other set has two black cards and one red. A set is chosen at random and a card from that set is chosen randomly and revealed. The revealed card is red. What is the probability it came from the deck with two red cards?

It's not whether you win or lose; it's whether or not you had a good bet.