[Guest]

Suppose we start with two identical urns. In one urn, we place 9 balls: 7 black and 2 white. In the other urn, we place 10 balls: 2 red, 2 white, 2 blue, 2 green, and 2 yellow.

Define a "trial" as the random selection of an urn followed by randomly drawing (one at a time, without replacement) all the marbles from that urn. A "terminal trial" is one where the LAST two marbles removed from the urn are the same color.

A game consists performing "trials" until a "terminal trial" is observed. You win the game if the FIRST two marbles drawn on the "terminal trial" are the same color. What is your chance of winning?

As an illustration, suppose you decide to play this game. You start by randomly selecting an urn. You then proceed to randomly draw out marbles from this urn one at a time and without replacement. Suppose the first two draws are the same color. You have not won yet. You need to keep drawing the remaining marbles, and if the last two drawn are the same color, the results become official and you win. If the last two drawn are not the same, you must replace all the marbles into the urn, randomize the urns, and start over. On the flip side, suppose the first two draws are not the same. You have not lost yet. Your only chance of winning now rests upon the last two marbles drawn being different so that you can start over. If the last two marbles drawn are the same color, the results become official and you lose.

[Guest]

Is it 5/21?

[Pickett]

Ok, it's been a while since I've done conditional probabilities...

First of all, my answer (for now) is you have a 11/63 chance of WINNING (approximately 17% chance). Here a few thoughts which led me to this:

IF you pick urn 1:

you have a 1/3 chance of winning, 5/18 chance of losing, and the 7/18 chance of starting over.

IF you pick urn 2

you have a 1/63 chance of winning, 8/63 chance of losing, and 8/9 chance of starting over.

Since you play until a terminal state, the starting over parts really have no impact, because you then are starting your odds all over again from scratch.

You also have a 1/2 chance of picking each urn...SO...

your chances of winning by urn 1 are (1/2 * 1/3) =1/6

your chances of winning by urn 2 are (1/2 * 1/63) = 1/126...

so, 1/6 + 1/126 = 11/63 which is your total chance of winning.

Like I said, been a while since I did probabilities, but that's my answer for now.

[Pickett]

Ok, changed my mind...the starting over definitely play a part in it...in fact, my new answer is:

infinite sum of (chance of winning * chance of starting over raised to the nth power)

sum from n=0 to infinity of (11/63 * (73/126)ⁿ)

And I don't know what that converges to...but I'm going to guess somewhere between 1/3 and 1/2...maybe 40%?

[CaptainEd]

Ok, changed my mind...the starting over definitely play a part in it...in fact, my new answer is:

infinite sum of (chance of winning * chance of starting over raised to the nth power)

sum from n=0 to infinity of (11/63 * (73/126)ⁿ)

And I don't know what that converges to...but I'm going to guess somewhere between 1/3 and 1/2...maybe 40%?

There's an alternative to the infinite sum, which has been useful in some of the other brain teasers.

Using your calculated probabilities (which I haven't checked--it looks hard to me), the expected value

Pwin = Purn1*Pwin_urn1 + Purn1*redo_urn1*Pwin + Purn2*Pwin_urn2 + Purn2*redo_urn2*Pwin

The key here is that, once you get back to restart, you're facing the same probability (Pwin). So, with this one equation, you can solve for Pwin--no infinite sum needed.

Separating out the terms with Pwin:

Pwin = Pwin * (Purn1*Predo_urn1 +

Purn2*Predo_urn2) + Purn1*Pwin_urn1 + Purn2*Pwin_urn2

Pwin * (1 -

Purn1*Pred_urn1 - Purn2*Pred_urn2) = Purn1*pwin_urn1 +

Purn2*pwin_urn2

Pwin = (Purn1*Pwin_urn1 + Purn2*Pwin_urn2) / (1 -

Purn1*Predo_urn1 - Purn2*Predo_urn2)

Using your numbers

Pwin =

(1/2 * 1/3 + 1/2 * 1/63) / (1 - 1/2 * 7/18 - 1/2 * 8/9)

= (1/6+1/126) / (1 -

7/36 - 4/9)

= (132/756)/(13/36) = 0.4835

Edited June 11, 2009 by CaptainEd

[CaptainEd]

There's an alternative to the infinite sum, which has been useful in some of the other brain teasers.

Using your calculated probabilities (which I haven't checked--it looks hard to me), the expected value

Pwin = Purn1*Pwin_urn1 + Purn1*redo_urn1*Pwin + Purn2*Pwin_urn2 + Purn2*redo_urn2*Pwin

The key here is that, once you get back to restart, you're facing the same probability (Pwin). So, with this one equation, you can solve for Pwin--no infinite sum needed.

Separating out the terms with Pwin:

Pwin = Pwin * (Purn1*Predo_urn1 + Purn2*Predo_urn2) + Purn1*Pwin_urn1 + Purn2*Pwin_urn2

Pwin * (1 - Purn1*Pred_urn1 - Purn2*Pred_urn2) = Purn1*pwin_urn1 + Purn2*pwin_urn2

Pwin = (Purn1*Pwin_urn1 + Purn2*Pwin_urn2) / (1 - Purn1*Predo_urn1 - Purn2*Predo_urn2)

Using your numbers

Pwin = (1/2 * 1/3 + 1/2 * 1/63) / (1 - 1/2 * 7/18 - 1/2 * 8/9)

= (1/6+1/126) / (1 - 7/36 - 4/9)

= (132/756)/(13/36) = 0.4835

(edited to removed unwanted line feeds)

[Guest]

You guys are correct, but it is more simply (perhaps) using conditional probability:

We seek P(first two draws the same color | last two draws the same color) = P(first two draws the same color AND the last two draws the same color) / P(last two draws are the same color). The probability in the numerator is 11/63 and the probability in the denominator is 13/36. The resulting ratio of these probabilities is 44/91 or about 0.4835.

[Guest]

Ok, it's been a while since I've done conditional probabilities...

First of all, my answer (for now) is you have a 11/63 chance of WINNING (approximately 17% chance). Here a few thoughts which led me to this:

IF you pick urn 1:

you have a 1/3 chance of winning, 5/18 chance of losing, and the 7/18 chance of starting over.

IF you pick urn 2

you have a 1/63 chance of winning, 8/63 chance of losing, and 8/9 chance of starting over.

Since you play until a terminal state, the starting over parts really have no impact, because you then are starting your odds all over again from scratch.

You also have a 1/2 chance of picking each urn...SO...

your chances of winning by urn 1 are (1/2 * 1/3) =1/6

your chances of winning by urn 2 are (1/2 * 1/63) = 1/126...

so, 1/6 + 1/126 = 11/63 which is your total chance of winning.

Like I said, been a while since I did probabilities, but that's my answer for now.

That looks like the odds of winning an individual trial, not the game. The outcome of the game is only determined by the terminal trial, so wouldn't non terminal trials be irrelevant?

If non-terminal trials are indeed irrelevant, then it simply boils down to the odds of the first two balls in a given trial being the same.