[Guest]

So, Cody being a high-roller, is invited by Bellagio to participate in their complementary Jackpot game. The rules are simple-- Pick a card of black color (Spade or Clubs) out of a single deck of playing cards (with 1 Joker, so total 53 cards) and win a Million Dollars.

Now, Cody went ahead and picked his card and gave it to the host (Cody CANNOT see the card he picked until the results are announced). Now comes the interesting part -- From the remaining deck of cards (52 cards), the host removed the joker out of it, and she gave Cody a choice -- If he wants, pick up another card from the same deck (remaining 51) and swap it with the current pick.

Cody is confused -- Do he needs to stick with his choice, or go for the swap?? Which gives better chance to hit the Million jackpot? Can you help?

(This is somewhat similar to the 3-door game with a Car and 2 goats. However, the odds are different.)

[HoustonHokie]

If Cody picks the joker to begin with, does the host tell him when she goes to remove it from the deck?

[Guest]

Since there is only 1 Joker, there is no joker to remove, and the game must end there.

Apparently, Cody didnt pick up the joker on first attempt, since the host removed joker from the remaining deck afterwards. The funda, is host knows the cards, and Cody doesn't.

If Cody picks the joker to begin with, does the host tell him when she goes to remove it from the deck?

[Guest]

Since the joker was not the selected card (since the dealer removed it after the selection). Than his current choice offered a 50% change of being black. To swap a card from the remaining 51 would neither increase or decrease the odds.

Edited June 11, 2009 by shawnd1223

[Guest]

First pick he has a 26/53 (~49%) chance at grabbing a black.

If he chooses to swap, he now has a 26/51 (~51%)(at best) chance.

If he did draw black first time around, he still has a 25/51 (~49%) chance of drawing black if he decides to swap.

I would keep the first pick. Seems to be too simple of an answer to be right tho.

[Guest]

I don't think it makes a difference. Since the joker was removed, the odds are 26 out of 52 that he has already picked black card (or 50%). If he picks another card, the odds would average out to the same thing (26 out of 52) - especially since we don't know the card he already picked. if he had picked a black card, the odds are 25 out of 51 that he will pick a black card on the second try. If he already picked a red card, then the odds are 26 out of 51 that he will pick a black card on the second try. The average is 25.5 out of 51 (or 50%) same as the odds on the first card.

[Guest]

Since the host saw the card that he selected;

had it been red, she would've immediately announced that he had lost (and saved the casino 1 million dollars). Since it was black, she pulled the joker and tried to give the casino a 50% chance of keeping the million.

So keep the card. 100% chance it is a winner.

Edited June 11, 2009 by shawnd1223

[Guest]

Same situation as Let's Make a Deal Riddle.

Instead of opening a door w/a donkey, he's removing a joker.

Not sure what the best site is for explanation, but here's one.

http://www.increasebrainpower.com/lets-make-a-deal.html

Always switch.

(Alternatively 26/53 < 25/51 < 26/51, so the switching makes sense.)

[HoustonHokie]

First pick he has a 26/53 (~49%) chance at grabbing a black.

If he chooses to swap, he now has a 26/51 (~51%)(at best) chance.

If he did draw black first time around, he still has a 25/51 (~49%) chance of drawing black if he decides to swap.

I would keep the first pick. Seems to be too simple of an answer to be right tho.

Same situation as Let's Make a Deal Riddle.

Instead of opening a door w/a donkey, he's removing a joker.

Not sure what the best site is for explanation, but here's one.

http://www.increasebrainpower.com/lets-make-a-deal.html

Always switch.

(Alternatively 26/53 < 25/51 < 26/51, so the switching makes sense.)

The devil is in the details.

As several have pointed out, the chances of drawing a winning card on the first pick is 26/53, and the worst case chance on the second pick is 25/51. However, 25/51 (0.4901) is not greater than 26/53 (0.4905), so there is a small decrease in your chances of selecting a new black card if you already hold one. This is different than the typical Monty Hall/Let's Make a Deal kind of game, where your chances do improve.

That being said, I think the difference in those probabilities is neglible in comparison with the chance for improvement if you do not hold a winning card to begin with (0.4905 vs 0.5098). So, yes, switch. Unless you draw the joker to begin with...

[Guest]

The devil is in the details.

As several have pointed out, the chances of drawing a winning card on the first pick is 26/53, and the worst case chance on the second pick is 25/51. However, 25/51 (0.4901) is not greater than 26/53 (0.4905), so there is a small decrease in your chances of selecting a new black card if you already hold one. This is different than the typical Monty Hall/Let's Make a Deal kind of game, where your chances do improve.

That being said, I think the difference in those probabilities is neglible in comparison with the chance for improvement if you do not hold a winning card to begin with (0.4905 vs 0.5098). So, yes, switch. Unless you draw the joker to begin with...

I'll agree with that logic..

If you chose to switch, your odds of winning or losing become the same, since you are unaware of the first card drawn.

If it is red, then its 51% to win, 49% to lose.

If it is black, then its 49% to win, 51% to lose.

The 0.4905 vs 0.5098 that you mention could go both ways equally since the first card drawn is unknown.

That would average out to be about a 50/50 chance, and 50% is slightly better than the original first draw of 26/53 ~49%. But I would'nt temp it if I was playing. But technically, yes the odds will be in your favor if you switch.

[Guest]

This is not like the Monty Hall Problem, because in that situation the incorrect door (and in this case) card had to be one of the others. Because of the addition that if he picks the joker, he loses instantly, if he is offered the choice to pick another card once the joker is shown, he now has the additional information that he didn't pick the joker originally and therefore the odds on his picked card increase from (26/53) to (26/52)=50%. It makes no difference to switch.

If the OP intended this to be similar to the Monty Hall Problem with slightly different probabilities, he should have had the host pick out and show Cody a red card instead, in which case there would be a slight advantage to switching cards.

[Guest]

As of now he knows didn't get the joker, so he has a 50/50 chance that his card is red or black. The joker is now removed so it plays absolutely no role in this "riddle."

The remaining cards are now either 49/51% favor of red, or 49/51% favor of black, but he doesn't know. So it doesn't matter because he'll end up with a one card from a 52 card deck that's half black, half red. 50/50.

The end.

Edited June 11, 2009 by Shadax

[Glycereine]

As of now he knows didn't get the joker, so he has a 50/50 chance that his card is red or black. The joker is now removed so it plays absolutely no role in this "riddle."

The remaining cards are now either 49/51% favor of red, or 49/51% favor of black, but he doesn't know. So it doesn't matter because he'll end up with a one card from a 52 card deck that's half black, half red. 50/50.

The end.

Some people read to much into this. The above is correct. his chance to draw a black card was 26/53. BUT the chance he HOLDS a black card after the joker is removed is 50%. Which is the same as his chance to draw a black card if he switches cards (given that his current color is unknown).

Makes no difference to switch.

[Guest]

doesnt he kinda contradict himself

"Since there is only 1 Joker, there is no joker to remove, and the game must end there.

Apparently, Cody didnt pick up the joker on first attempt, since the host removed joker from the remaining deck afterwards. The funda, is host knows the cards, and Cody doesn't."

He says if it happens the games over but it is impossible to happen

but either way it is equal

if can pull the joker

chance to choose black then switch to black

26/53*25/51

chance to choose red then switch to black

26/53*26/51

so whole thing is

26/53*25/51+26/53*26/51=26/53

chance to pull on first try 26/53

if you cant pull the joker its like its never there because its not there to switch to either

black then switch black

26/52*25/51

red the switch black

26/52*26/51

add em

26/52*25/51+26/52*26/51=.5

which is the chance in the begining

voila

abra kadabra

woopdy doopy

[Guest]

It does not make a difference.

He has a 26/53 chance of winning on the first draw, which is a 49.06% chance of winning.

After the joker is removed, he will have a 26/51 (50.98%) chance of winning if he holds a red card and switches.

Also, he will have a 25/51 (49.02%) chance of winning if he holds a black card and switches.

The chance of him holding a red card is 26/53. The chance of him holding a black card is 26/53.

Therefore, the chance of him winning after a switch is:

(26/53)*(26/51)+(26/53)*(25/51)=0.4906, or 49.06%, the exact same as before.

(Original post was wrong due to rounding error from PEMDAS-deficient calculator)

This is kind of similar to Let's Make a Deal, which is explained as follows:

He had a 2/3 (67%) chance of picking goat, and a 1/3 (33%) of picking car and winning on first draw.

After a goat is shown:

If he picked car, a 0% chance of winning after a switch.

If he picked goat, a 100% chance of winning after a switch.

(2/3)*(100)+(1/3)(0)=0.67, or 67% of winning after a switch. He is more likely to win after a switch than your original draw.

This would be the exact same, but on a larger scale, if there were no Joker and instead the dealer showed him a red card (vinays beat me to it).

He had a 26/52 (50%) chance of picking red, and a 26/52 (50%) of picking black and winning on first draw.

After a red card is shown:

If he holds red, a 26/50 chance of winning after a switch.

If he holds black, a 25/50 chance of winning after a switch.

(26/52)*(26/50)+(26/52)(25/50)=0.51, or 51% of winning after a switch. He is more likely to win after a switch than your original draw.

Edited June 11, 2009 by smallbigtall

[Glycereine]

It does make a difference. My answer is do not switch.

He has a 26/53 chance of winning on the first draw, which is a 49.05660% chance of winning.

After the joker is removed, he will have a 26/51 (50.98039%) chance of winning if he holds a red card and switches.

Also, he will have a 25/51 (49.0196%) chance of winning if he holds a black card and switches.

The chance of him holding a red card is 26/53. The chance of him holding a black card is 26/53.

Therefore, the chance of him winning after a switch is:

(26/53)*(26/51)+(26/53)*(25/51)=0.4905659, or 49.05659%.

The post-switch winning percentage, compare that to the original 49.05660% winning percentage, is less by about 0.00001%.

Only issue with your math here is that the chance he is holding a red or a black card is not 26/53 but 26/52.

edit: for quick explanation. If you say he has a 26/53 chance of red and 26/53 chance of black then he has a 1/53 chance of neither, which is not possible since the joker is in the remaining cards. Since the joker is automatically not the one he picks (in this scenario) then you can completely ignore it.

Edited June 11, 2009 by Glycereine

[Guest]

Since you know you didn't pick the joker then:

Chance of picking black at the start: 26/52

Chance of picking another black after: 25/51

Chance of picking red at the start: 26/52

Chance of picking balck after: 26/51

So if you switch: (26/52)*(25/51) + (26/52)*(26/51) = 26/52 = 1/2

If you dont: 26/52 = 1/2

So it makes no difference.

[Guest]

Only issue with your math here is that the chance he is holding a red or a black card is not 26/53 but 26/52.

edit: for quick explanation. If you say he has a 26/53 chance of red and 26/53 chance of black then he has a 1/53 chance of neither, which is not possible since the joker is in the remaining cards. Since the joker is automatically not the one he picks (in this scenario) then you can completely ignore it.

Valid point. I did retract my statement (see above) due to calculator and rounding error.

Your argument 52 vs. 53 is correct, but does not make any difference in the calculation, as the denominators cancel anyway. Switching does not matter. Thanks for the clarification.

Edited June 11, 2009 by smallbigtall

[Guest]

Hi final,

I didnt mention anywhere that picking a joker is impossible to happen. Obviously, we all know that it is a possible scneario. What I meant was -- in the real time situation, it didnt happen during that iteration, and hence the host did allow a second chance to Cody with Joker removed.

Neverthless, your solution (along with many others) looks cool to me...

Peace...

doesnt he kinda contradict himself

"Since there is only 1 Joker, there is no joker to remove, and the game must end there.

Apparently, Cody didnt pick up the joker on first attempt, since the host removed joker from the remaining deck afterwards. The funda, is host knows the cards, and Cody doesn't."

He says if it happens the games over but it is impossible to happen

but either way it is equal

if can pull the joker

chance to choose black then switch to black

26/53*25/51

chance to choose red then switch to black

26/53*26/51

so whole thing is

26/53*25/51+26/53*26/51=26/53

chance to pull on first try 26/53

if you cant pull the joker its like its never there because its not there to switch to either

black then switch black

26/52*25/51

red the switch black

26/52*26/51

add em

26/52*25/51+26/52*26/51=.5

which is the chance in the begining

voila

abra kadabra

woopdy doopy

Edited June 11, 2009 by Terminator