Rules of the Game 1 2 Urns and one is chosen at random by a

Rules of the Game:

1.) 2 Urns and one is chosen at random by a \"dealer\"

2.) There are 20 total chips, 10 black chips (signifying a loss) and 10 white chips (signifying a win)

3.) The player (you) may allocate the chips however you wish in either urn, as long as each urn has at least one chip and all 20 are used in the experiment.

Describe the sample space and which way to allocate the chips such that you give yourself the highest probability of winning? Answer intuitively. Please check my answer below:

b1 = black chips in urn 1, b2 = black chips in urn 2, w1 = white chips in urn 1, w2 = white chips in urn 2

I defined the Sample Space with equations:

b1 + b2 = 10

w1 + w2 = 10

b1, b2, w1, w2 are all elements of set {0, 1, 2, 3, ... , 10}

If b1 = 0, w1 cannot also = 0 and vice-versa

If b2 = 0, w2 cannot also = 0 and vice-versa

I then intuitively (using a prob. tree) stated that the best allocation scheme for the player is (10, 0, 9, 1) or conversely (0, 10, 1, 9) with the first two digits being the white chips (winning chips). Because the two urns are chosen randomly, the player starts with a 50:50 chance to win as one of the urns has only 1 winning chip within it. If the dealer were to choose the other urn, the player still has a 9/19 chance of winning. Multiplying across I get that the player has a 14/19 chance of winning with a 5/19 chance of loss.

Does this seem like a decent answer?

Thanks

Solution

Yes