Ask The Wizard #356

The average time needed for an occurrence to take place can be calculated as the total of the durations in which it hasn't occurred yet. This principle applies to both discrete and continuous data.

The answer is 712830140335392780521 / 6621889966337599800 = 107.6475362712258

If the card is drawn at unpredictable intervals rather than strictly once per time unit, the outcome remains unchanged as long as the average interval adheres to an exponential distribution with a mean of 1.

The expected time until a specific card is drawn averages 52 units. In accordance with the characteristics of the exponential distribution, the likelihood that this card has not been drawn after t units of time can be expressed as exp(-t/52).

At any given t time frame, the chance that a particular card has been drawn at least once can be represented as 1-exp(-t/52).

The probability that all 13 designated cards have been drawn at least once after t time units is mathematically calculated as (1-exp(-t/52))^13.

After t time units, the likelihood that at least one of the 13 specific cards has not been drawn is given by 1-(1-exp(-t/52))^13.

The probability that every suit is missing at least one card after t time units would be represented as (1-(1-exp(-t/52))^13)^4.

Putting this equation into an integral calculator , being careful to set to set the bounds of integration from 0 to infinity, yields 712830140335392780521 / 6621889966337599800 = 107.6475362712258

The table below illustrates the survival probabilities for each player based on their turn order. The bottom right cell indicates that the expected number of players to survive is approximately 0.23884892.

My approach to solving this problem involved employing a Markov chain, which may be complex and time-consuming to detail.

The answer is 1.9565784%, or 1 in 51.1096316.

Among the eight cards dealt to the four opponents, the probability that all four are aces is calculated as combin(46,4)/combin(50,8) which results in 0.000303951.

Subsequently, the probability that none of the four aces are in the same hand is established as 1-2^4*4!*4!/8! = 0.228571429. Therefore, the probability that at least one pair of aces exists among them is calculated as 1 - 0.228571429, yielding 0.771428571.

The likelihood that all four aces are in play and at least one hand holds two of them equals 0.000303951 * 0.771428571, which results in 0.000234477.

In the scenario involving eight cards for the four opponents, the probability of three of them being aces is given by combin(4,3) * combin(46,5)/combin(50,8), yielding 0.010212766.

From that point, the calculation for two aces existing in the same hand results in 4*3*COMBIN(3,2)*5*COMBIN(4,2)/(COMBIN(8,2)*COMBIN(6,2)*COMBIN(4,2)) = 0.428571429.

The probability that three of the aces are drawn and two belong to the same hand is 0.010212766 * 0.428571429, which equals 0.0043769.

For the scenario with eight cards among the four opponents, the probability that two of the cards are aces, expressed as combin(4,2) * combin(46,6)/combin(50,8), is 0.104680851.

The probability that both aces are contained within the same hand is calculated as 1/7, giving 0.142857143.

The chance that two of the aces are accounted for and both belong to a single hand amounts to 0.104680851 * 0.142857143, resulting in 0.014954407.

Summing the probabilities of scenarios where at least one opponent holds two aces yields a final total of 0.000234477 + 0.0043769 + 0.014954407, culminating in an overall probability of 0.019565784.