[Guest]

Zack is entering this Million Dollar jackpot game with an opponent Jim. Here are the rules --

There are THREE doors, with one of them having a Million Dollar check. Other 2 doors contain nothing but pranks. The rule is to choose the right door and win the Million. Since there are 2 contestants, there is a toss of coin to determine who gets the first chance.

Contestant 1 -- have to pick 1 out of 3 doors.

Contestant 2 -- have to pick 1 out of the remaining 2 doors.

Results will be revealed only after both the contestants picked their doors. So, contestant 2 won't know what was in the door picked by contestant 1 until he picks his door.

Now, Zack won the toss, and he can decide whether to go first or be the second contestant.

So, here is the problem -- Zack is confused, whether he should go first or second. Which one gives him more chance to hit the jackpot?

[Guest]

Been awhile since i've done probability, but I think . . .

Zack should go first as this gives him a 33.3333...% chance to win whereas if he goes second, I believe he only has a 25% chance of winning.

[Guest]

Perhaps i'm a bit confused, the RESULTS won't be shown until after both people have picked or will the second chooser not even know what door the first chooser picked???

Shouldn't it be a 33% chance for both, considering that either way you choose you would have to get 1 out of the 3 possible doors?

Edited July 13, 2009 by 14U2NV

[Guest]

It doesn't matter by going first you have 33% chance or 66% chance of being wrong. This means second person has a 50% chance on a 66% chance which is 33%.

[Guest]

It's equal. The first to go has a 1 in 3 chance of winning. The second to go has a 1 in 2 chance of winning IF the first one is unsuccessful. So overall probability is 1/2 x 2/3 = 1/3.

Donjar

Edited July 13, 2009 by donjar

[Guest]

he should go second cause if he goes first he has a 33.3% chance of getting it right but a 66.6% chance of getting it wrong so odds are you're going to get it wrong on the first guess

[Guest]

This sounds similar to a question brought up on the Las Vagas film 21.

By going first the contestant has a 33% chance, so for the second contestant 1 door has been removed. Therefore he now has a 50% chance of winning the prize. Henceforth it will be better to go second.

[Guest]

If choosing first, there is a 1/3 chance that you would pick the correct door. Simple

If choosing second then there are only 2 doors to pick from… So there is a 50 % chance to pick the correct door. But first we must think about whether the first pick chose the door. Since the probability is a 33% chance for the first person to choose the correct door, we can assume that there is a 2/3 chance that the door with the check has not been picked. So, you have a 50% chance to pick the correct door with a 66% chance to have the check. Therefore, the probability of choosing the million-dollar check as Jim is also 33%. You have equal chances to grab it either way.

[Guest]

chanc3 its close to the question but what u have to remember about being the second contestant is that the first one does have a chance of getting it right. in the Las Vagas one in the movie 21, and it involves the showing of 1 door and what is behind it.

[Guest]

1st player has always 1/3 chances.

The second has 1/2 given the first one failed. So the second player odds to win are 2/3 (First player misses) by 1/2 (choosing the right remaining door).

2/3 * 1 / 2 = 1/3

So either way the got the same chances

[Guest]

equal chances.

[Guest]

The issue of probability and statistics.let me think

[Guest]

3 doors = 33.3% chance. Going first means you can pick the right door first, or the wrong door. It cancels out. The second contestant gets a 50/50 chance at 66% remaining choices, this equals 33.3% chance.

Edit: Upon further thought, I can see it would really only matter if one contestant really had his heart set on a specific door, but the first contestant chooses it. If it's a winner, than the contestant in that case "should have" gone first. But that door could also have been a WRONG one, which again balances out the chances.

/thread

Edited July 13, 2009 by Shadax

[Guest]

He should go first. Going first gives him a 1 in 3 chance. If he goes second, then he has 0% chance, if the other guy picked the million dollar door. However, if the other guy picked a wrong door, his chances improve to 50%. Overall, if going second, his chances would be two in 6 (1/3 x 2/3), which is the same odds as going first. I'd pick going first, though, since odds are the same either way and you don't want the other guy to get lucky and pick the right door first.

Edited July 13, 2009 by LogicMan

[Guest]

It seems as though a consesus for this problem but what if b=#of doors>>3 lets say there are 100 doors and 99 contestants. Is there an optimum time to pick? The first to go would probably not get it while the prize would most likely be selected before the last person guesses.

[Guest]

He should go first. Going first gives him a 1 in 3 chance. If he goes second, then he has 0% chance, if the other guy picked the million dollar door. However, if the other guy picked a wrong door, his chances improve to 50%. Overall, if going second, his chances would be two in 6 (1/3 x 2/3), which is the same odds as going first. I'd pick going first, though, since odds are the same either way and you don't want the other guy to get lucky and pick the right door first.

Upon personal preference, I think anyone would simply just chose to go first. It has nothing to do with statistics whatsoever but piece of mind knowing that if the first contestant gets it, that "could have" been the losers pick. It's all about perspective, but there is absolutely no probable difference.

[Guest]

Makes no difference. Pick first for a 1 in 3 chance. Pick second for a 1 in 2 chance of picking a from doors that between them have a 2 in 3 chance of being the winner. 1/2 * 2/3 = 1 in 3. So much for the obvious, now I'll go read what will undoubtably be the non-obvious correct answer.

Edited July 13, 2009 by Brighterthan1000rocks

[Guest]

It seems as though a consesus for this problem but what if b=#of doors>>3 lets say there are 100 doors and 99 contestants. Is there an optimum time to pick? The first to go would probably not get it while the prize would most likely be selected before the last person guesses.

It does not make a difference. If the winning door is revealed only once all the candidates have picked one, then the order is irrelevant. Imagine you have M doors and N contestants (assume M >= N). Then the first one to pick has 1/M chances to pick the right one. The second one has 1/(M-1) chances to pick the right one given the previous contestant didn't already pick it, i.e.: candidate number 2 has 1/(M-1) * (M-1)/M = 1/M chances to win. The third one has 1/(M-2) chances to pick the right door given the previous two candidates didn't pick it, that is: 1/(M-2) * (M-1)/M * (M-2)/(M-1) = 1/M And so on... (you could prove by induction that for all of the candidates, the probability to find the winning door is 1/M). Of course, the result is completely different is the outcome is revealed after each candidate picked a door...

Edited July 13, 2009 by math-o

[Guest]

Presumably both have a 1/3 chance, since contestant 2 doesn't know which door contestant 1 has chosen and the selected door is not removed as a possibility of selection.Also explained by the (1/2)*(2/3)= 2/6 or 1/3

These rules of gameplay are quite strange, no?

[Guest]

Good going guys! Many have got the answers right. So, if I were Zack, I would have gone first, eventhough the chances are equal -- to negate any chances of my opponent hitting the jackpot before me. What say?

Peace!

Edited July 13, 2009 by Terminator

[Guest]

it is irrelevent

Edited July 14, 2009 by PXSR13

[bonanova]

Player 1 has chance of 1/3.

Self explanatory.

Player 2 has chance of 1/3 x 0 + 2/3 x 1/2 = 1/3.

That is, there is a 1/3 chance [Player 1 chose correctly] that his chances are zero.

There is also a 2/3 chance [Player 1 chose wrong] that his chances are 1/2.

Net - doesn't matter - both Players have a 1/3 chance of getting the money.

If there were a 3rd player who got the unchosen door, his chances would be 1 - [1/3 + 1/3] = 1/3 as well.

[Guest]

Player 1 has chance of 1/3.

Self explanatory.

Player 2 has chance of 1/3 x 0 + 2/3 x 1/2 = 1/3.

That is, there is a 1/3 chance [Player 1 chose correctly] that his chances are zero.

There is also a 2/3 chance [Player 1 chose wrong] that his chances are 1/2.

Net - doesn't matter - both Players have a 1/3 chance of getting the money.

If there were a 3rd player who got the unchosen door, his chances would be 1 - [1/3 + 1/3] = 1/3 as well.

I agree, Bonanova!