Use Equation (2.1) to compute, in a hand of bridge, the conditional probability that East has 3

Use Equation (2.1) to compute, in a hand of bridge, the...

Key Concepts

Conditional Probability

Conditional probability is the measure of the likelihood of an event occurring given that another event has occurred. In probability theory, it is defined as the probability of the intersection of two events divided by the probability of the known event, and it is crucial for updating probabilities based on new information.

Combinatorial Analysis

Combinatorial analysis involves counting the number of arrangements or selections in a given scenario, usually through combinations and permutations. This concept is fundamental in calculating probabilities where the number of favorable outcomes and total outcomes must be determined, particularly in situations involving selection without replacement.

Hypergeometric Distribution

The hypergeometric distribution models the probability of drawing a specific number of successes from a finite population without replacement. This concept is commonly applied in card games and other real-world problems where the probability changes with each draw from a limited set of items.

Conditional Distributions in Card Games

Conditional distributions in card games examine how the probability of a specific card grouping or hand configuration changes when additional information about the distribution of cards among players is known. This concept enables the calculation of probabilities in scenarios like card deals, where known outcomes restrict the possible configurations for the remaining hands.

Transcript

00:02 From the question, our experiment is a duck of 52 cars is there to four players.

00:09 So 13 cars each.

00:11 We call these four players north, east, south and west.

00:16 So let's event a, b, north and south together having eight spades.

00:27 So eight spades.

00:31 And event b, b.

00:34 Each which have three spades so we want to find a probability of b given that event a has already occurred from our question so the outcome space let's say s contains every division of do those 52 cuts into the four other growth of 13 so so if all event in s are considered equally likely, then probability of event a is a subset of the universal certificate s and a probability of a would then be given as the length of a by the length s.

01:38 So where this denote the number of elements in a, hence this will also denote the number of elements in s.

01:58 So the number of elements in s given as this will be given by 52, choose 13 and 39, choose 13 then 29 choose 13 which will be equal to 52 factorial over 13 factorial to the power 4 so it's like choosing 13 cars for north then for east then for south and the remaining cars goes to west the the formula for conditional probability for a, b events is given us b, probability of b given a is equal probability of a intersection b, all over probability of a.

03:03 So we will have to find the various probabilities.

03:09 So let's count the number of elements in a.

03:14 So we choose first the cards for north and south.

03:18 The eight space can be chosen in 13, choose a ways.

03:27 And the remaining 18 cards for the non -spates is 39, 18 ways.

03:37 Okay.

03:40 Then determine the north card for those 26 will be 26, choose 13 ways.

03:50 So whichever choice are done in choosing the cars for north and south, the remaining 26 cars can be distributed among east and west in any of the 26 choose 13 ways.

04:06 So using the formula, using the formula probability of a is given by a number of elements in a or by the number of elements in the universal asset...

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