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Ask the Wizard #356

A card is draw with replacement from a deck of 52 cards. What is the expected number of draws needed until all 13 cards of any one suit has been drawn? Please use calculus for your solution.

"Anonymous" .

The expected time for an event to happen equals the sum over all time that it hasn't happened yet. This is true for both discreet and continuous variables.

The answer is 712830140335392780521 / 6621889966337599800 =~ 107.6475362712258

Instead of a card being drawn exactly once per unit of time, the answer will be the same if a card is drawn with a random period of time between draws if that average time follows an exponential distribution with mean of 1.

The time between any given card being drawn will have a mean of 52. Given the properties of the exponential distribution, the probability the card will not have been drawn after t units of time is exp(-t/52).

After t units of time, the probability any specific card will have been drawn at least once is 1-exp(-t/52).

After t units of time, the probability 13 specific cards will have been drawn at least once is (1-exp(-t/52))^13.

After t units of time, at least one of 13 specific cards will cards will NOT have been drawn is 1-(1-exp(-t/52))^13.

After t units of time, the probability every suit will have at least one card missing is is (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

This question was asked and discussed in my forum at Wizard of Vegas.

In Ask the Wizard colulmn 355 a question was asked about the glass bridge problem in the Squid Game. The question assumed players remembered where the safe steps were. My question is what would be the answer if the players did not remember.

"Anonymous" .

Let me reword what you are asking without referencing the previously problem first.

16 players face a game with a glass bridge. The bridge is divided into 18 pairs of glass panels. In each pair, one panel of glass is tempered and can support the weight of a player. The other piece of the pair is ordinary glass and will break under the weight of a player. If a player steps on a piece of ordinary glass, he will break it and fall to his death.

The players must advance, one at a time, in a predesignated order. Players do not remember where the safe steps are, except if it is obvious because one panel in a pair is broken.

Assuming random guessing at each pair of glass steps, what is the expected number of players to cross safely?

Please click the button below for my answer.

The following table shows the probability of each player, in order they had to play, surviving. The lower right cell shows the expected number of survivors is 0.23884892.

Memoryless Squid Game Bridge Puzzle

1 0.00000381
2 0.00000763
3 0.00001526
4 0.00003051
5 0.00006094
6 0.00011911
7 0.00023545
8 0.00046159
9 0.00089886
10 0.00175139
11 0.00345091
12 0.00693198
13 0.01418276
14 0.02923634
15 0.05993762
16 0.12152477
Total 0.23884892

My solution used a Markov chain, which would be hard and time consuming to explain.

This question is asked and discussed in my column a Wizard of Vegas.

If I am holding pocket kings in Texas Hold 'em and have four opponents, what is the probability at least one of my opponents is holding pocket aces?


The answer is 1.9565784%, or 1 in 51.1096316.

In the eight cards for the four opponents, the probability that four of them are aces is combin(46,4)/combin(50,8) = 0.000303951.

From there, the probability that the four aces are all in different hands is 1-2^4*4!*4!/8! = 0.228571429. Thus, the probability of the alternative, that there is at least one pair of aces, is 1 - 0.228571429 = 0.771428571.

The probability that all four aces are out and at least one hand has two of them is 0.000303951 * 0.771428571 = 0.000234477.

In the eight cards for the four opponents, the probability that three of them are aces is combin(4,3) * combin(46,5)/combin(50,8) = 0.010212766.

From there, the probability two of them are in the same hand is 4*3*COMBIN(3,2)*5*COMBIN(4,2)/(COMBIN(8,2)*COMBIN(6,2)*COMBIN(4,2)) = 0.428571429.

The probability three of the aces are out and two of them are in the same hand is 0.010212766 * 0.428571429 = 0.0043769.

In the eight cards for the four opponents, the probability that two of them are aces is combin(4,2) * combin(46,6)/combin(50,8) = 0.104680851.

The probability they are both in the same hand is 1/7 = 0.142857143.

The probability two of the aces are out and are in the same hand is 0.104680851 * 0.142857143 = 0.014954407.

Taking the sum of the ways at least one opponent can get two aces, we get an answer of 0.000234477 + 0.0043769 + 0.014954407 = 0.019565784.

I saw a promotion at an online sports book where a money line bet in the NFL would automatically be graded a winner if the chosen team was up by 17 or more points. What is the value of this?

"Anonymous" .

This promotion will turn what would otherwise be a losing bet into a winner if the chosen team is up by 17 or more points and then loses. A good example of such a situation is a bet on the Atlanta Falcons in Super Bowl 51. At one point, in the third quarter, the Falcons were up 28 to 3, a lead of 25 points. However, they went onto lose 28 to 34.

To answer this question, I looked at 4,131 games played in every NFL season from 2000 to 2015. The following table shows the greatest deficit the winning team had during the course of the game. The probability column filters out the five games that ended in a tie.

Greatest Deficit Overcome

Deficit Games Probability
Tie 5 0.000000
0 1804 0.437227
1 100 0.024237
2 29 0.007029
3 560 0.135725
4 235 0.056956
5 23 0.005574
6 131 0.031750
7 622 0.150751
8 39 0.009452
9 34 0.008240
10 195 0.047261
11 84 0.020359
12 14 0.003393
13 49 0.011876
14 104 0.025206
15 10 0.002424
16 6 0.001454
17 36 0.008725
18 14 0.003393
19 2 0.000485
20 4 0.000969
21 22 0.005332
22 0 0.000000
23 2 0.000485
24 5 0.001212
25 1 0.000242
26 0 0.000000
27 0 0.000000
28 1 0.000242
Total 4131 1.000000

The "Tie" row represents the five games in the 16 seasons that ended in a tie, so let's not count those. The "0" row represents the 43.7% of games where the winning team was never behind.

The table shows 87 games saw a team losing by 17 or more and then went onto win. Over the 4126 games resolved (i.e. not counting the five ties), this probability is 2.11%.

Given these situations would flip a loss to a win, we double this probability to get a value of 4.22%. The house edge on money line bets is about the same as bets against the spread at 4.76%. Subtracting 4.22%, we get a very low house edge of 0.54% under this promotion.