Weekend Probability Puzzle 1?

Nothing that is revealed tells you anything that defines a preferred ordering for the remaining cards. So just randomly choosing cards is fine.

There are only 3 cards that matter: The two repeated colors, and black.

Of these three:

Pr[black is picked first] = 1/3

Pr[black is picked second] = (2/3)(1/2)

Pr[black is picked third] = (2/3)(1/2) = 1/3

So Pr[win] = Pr[black is picked third] = 1/3

***

@Julius: Your compuation for the probability of loosing on turn 3 is incorrect. You give:

Pr[loose on turn 3] = 8/9*6/8*1/7

It should really be:

Pr[loose on turn 3]

=Pr[loose on turn 3 | pick black on turn 1](1/9)

+ Pr[loose on turn 3 | pick repeated color on turn 1](2/9)

+ Pr[loose on turn 3 | pick neither black nor repeated color on turn 1](6/9)

actually you COULD win on the second turn if you picked the pair of same colored cards on the first two turns. also, the probability of losing the game on turn 9 has to be zero since you must have picked both cards by then.

anyway, the probability of losing on each turn is

1/9= 1/9

8/9*1/8= 1/9

8/9*6/8*1/7= 2/21

8/9*6/8*5/7*1/6= 5/63

8/9*6/8*5/7*4/6*1/5= 4/63

8/9*6/8*5/7*4/6*3/5*1/4= 1/21

8/9*6/8*5/7*4/6*3/5*2/4*1/3= 2/63

8/9*6/8*5/7*4/6*3/5*2/4*1/3*1/2= 1/63

8/9*6/8*5/7*4/6*3/5*2/4*1/3*0/2= 0

equals

1/9+1/9+2/21+5/63+4/63+1/21+2/63+1/63= 5/9

so the probability of winning is 4/9

You could also calculate the probability of winning on each turn:

win:

0/9= 0

8/9*1/8= 1/9

8/9*6/8*1/7= 2/21

8/9*6/8*5/7*1/6= 5/63

8/9*6/8*5/7*4/6*1/5= 4/63

8/9*6/8*5/7*4/6*3/5*1/4= 1/21

8/9*6/8*5/7*4/6*3/5*2/4*1/3= 2/63

8/9*6/8*5/7*4/6*3/5*2/4*1/3*1/2= 1/63

8/9*6/8*5/7*4/6*3/5*2/4*1/3*1/2*0/1= 0

equals

0+1/9+2/21+5/63+4/63+1/21+2/63+1/63+0= 4/9

tadaa!

there is no strategy that can improve on this .

Well first of all the probability that you will lose on the first try is 1/9 and 8/9 to not lose... for the second turn the chances of losing is 1/8 so the chance to lose at the second turn is 8/9 * 1/8

You have a chance of losing the game up until the 8th Turn

So the chances of losing is 1/9 + (8/9 * 1/8) + (8/9 * 7/8 * 1/7)..... And so on but if you will notice each term is just 1/9 so the sum is 8/9.... The chances of losing is 8/9 while the chances of winning is 1/9.

As for the strategy, im not really sure. You can pick any card you want but theres no way of knowing which card will give me a pair or the black one. And there's no strategic choice of bettering your chances for each action you take...

One thing is for certain, for each turn you take without losing, you get lesser and lesser chances of losing... The longer you play the greater tour chances of winning