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# Very tough question on Probability (and I mean it)

[FONT=Verdana, Arial, Helvetica, sans-serif]Puzzle Source: Mind and Visual Puzzles and Activities that enhance critical, logical, and creative thinking skills, and keep your brain active!

There are on the table 3 playing cards face-down. You have to guess which one is a King of Spades. You pointed the card C... OK, card B is turned over by our game Host and it isnโt a King of Spades. Now, if you change your mind and select card A instead of C, what are your winning probabilities?

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There are on the table 3 playing cards face-down. You have to guess which one is a King of Spades. You pointed the card C... OK, card B is turned over by our game Host and it isnโt a King of Spades. Now, if you change your mind and select card A instead of C, what are your winning probabilities?

**[ When posting your solution, do post detailed explanation ]**[/FONT]

Btw, let me know if this one explodes your skull. We'll try to provide you with free bandage.

Ahhh .. I wont give out the answer to this, but this question is like the game show that used to have Aman as the host. The one with the term

Hint: The host knows.

*Tai-Tai-Phis!!*Hint: The host knows.

Are we basing this on the probability that one of these three cards IS the king card or the probability of the king from the deck of cards is in these three?

If we know that The King of Spades is one of the three cards then:-

Biggie tell the answer.

The probability that one of the remaining two is The King of Spades is 0.5If we don't know then:-

The total number of ways we can draw 3 cards from a deck is 52C3

One card is disclosed. Assuming a deck of 51 cards and choosing 2 cards from it. The chances are 51C2

Chances that one card is The King of Spades is 1C1 as there is only 1 such card.

Since the other unknown card can be any from the remaining deck, the chances are 51C1

The total Probability that King of Spades is one of the remaining to cards is

( 1C1 * 50C1 ) / 51C2 = 50/1275

The Probability that the card selected is the King of Spades is

1C1 / ( 51C2 * 2) = 1/2550

I multiplied by 1/2 figuring that the probability of choosing 1 card from 2 will also be considered.

If there was only 1 card left then we would not have the 1/2 and the Probability would have been 1/1275

Biggie tell the answer.

So, Whats the answer?

Hi all,๐๐๐

posting after really a long time...

So, the problem is how will the probability of success changeif u change ur choice ur choice in the middle... Let's solve it step by step:

Let's assume card c is the king.

So, the conclusion is that, unless you choose the correct card in the first try itself, U win in all other cases. there are 3 cases and u win in 2, so

ANS: probability of success increases if you change ur choice after the host opens one of the cards. If u change ur choice, probability of winning is 2/3. If not, it's 1/3.

this is the famous scenario called the monty hall problem. for more on this,

Monty Hall problem

How's that??

posting after really a long time...

So, the problem is how will the probability of success changeif u change ur choice ur choice in the middle... Let's solve it step by step:

Let's assume card c is the king.

**possibility 1**: you select card A. The host opens card B(oblviously he wont open card c because he knows that is the king and he has to preserve the thrill in the game). Now if u change ur mind and opt for card c, u win.**possibility 2**: you select card B. The host opens card A(same reasoning as above). And then u choose card C, u win again.**possibility 3**: you select card C. The host opens either A or B. Then if you change ur mind and choose the other card, you lose.So, the conclusion is that, unless you choose the correct card in the first try itself, U win in all other cases. there are 3 cases and u win in 2, so

ANS: probability of success increases if you change ur choice after the host opens one of the cards. If u change ur choice, probability of winning is 2/3. If not, it's 1/3.

this is the famous scenario called the monty hall problem. for more on this,

Monty Hall problem

How's that??