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Ask the Wizard #333

There are an infinite number of light bulbs, all turned off. The time between light bulbs being turned on has an exponential distribution* with a mean of one day. Once a light bulb is turned on, its life expectancy also follows an exponential distribution with a mean of one day.

What is the mean time until the first light bulb burns out?

*: Random events that follow the exponential distribution have a memory-less property in that the past does not matter. In other words, a single event is never over-due and the probability of it happening is always the same.

Ace2

The answer is e - 1 = apx. 1.7182818...

 

On average, it will take one day for the first light bulb to be turned on.

From there, it will take half a day, on average, until the next significant event, either a new bulb being turned on or the first bulb burning out. We add 1/2 a day to the waiting time until that event. So, we're now at 1 + (1/2) = 1.5 days.

There is a 1/2 chance the second event was a second bulb being turned on. In that case, there is a 1/3 day waiting time until the next significant event (either one of the first two bulbs burning out or a new bulb being turned on). So, add the product of 1/2 (the probability of getting this far) and 1/3, which equals 1/6, to the waiting time. So, we're not at 1.5 + 1/6 = 5/3 = 1.66667 days.

There is a (1/2)*(1/3) = 1/6 chance the third signficant event was a third bulb being turned on. In that case, there is a 1/4 day waiting time until the next significant event (either one of the first three bulbs burning out or a new bulb being turned on). So, add the product of 1/6 (the probability of getting this far) and 1/4, which equals 1/24, to the waiting time. So, we're not at 5/3 + 1/24 = 41/24 = 1.7083 days.

Following this pattern, the answer is (1/1!) + (1/2!) + (1/3!) + (1/4!) + (1/5!) + ...

It should be common knowledge that e = (1/0!) + (1/1!) + (1/2!) + (1/3!) + (1/4!) + (1/5!) + ...

The only difference is our answer lacks the 1/0! factor. Thus, the answer is e - 1/0! = e - 1 = apx. 1.7182818...

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

On average, how many games of Hearts* must a single person play to see all 52 cards in his own hand?

*: Hearts is played with a single 52-card deck. Each hand consists of 13 cards.

"Anonymous" .

The average number of hands needed to see all 52 cards is approximately 16.41217418.

 

I used a Markov chain in Excel to solve this problem. The following table shows the probability of seeing all 52 cards in 4 to 100 hands. The left column shows the number of hands. The middle column shows the probability the player sees the 52nd card in this many hands exactly. The right column shows the probability the player sees all 52 cards in this many hands of less. For example, the probability it takes exactly 20 hands is 4.64% and the probability it takes 20 hands or less is 84.63%.

Hearts Question

Hands Probability
Exact
Number
Probability
by this
Number
4 0.0000000000 0.0000000000
5 0.0000000002 0.0000000002
6 0.0000007599 0.0000007601
7 0.0000746722 0.0000754323
8 0.0012814367 0.0013568690
9 0.0078648712 0.0092217402
10 0.0250926475 0.0343143878
11 0.0519205664 0.0862349541
12 0.0800617820 0.1662967361
13 0.1007166199 0.2670133561
14 0.1098088628 0.3768222189
15 0.1081357062 0.4849579251
16 0.0989810156 0.5839389408
17 0.0859323992 0.6698713400
18 0.0717845305 0.7416558705
19 0.0582992717 0.7999551422
20 0.0463771514 0.8463322937
21 0.0363346393 0.8826669329
22 0.0281478762 0.9108148092
23 0.0216247308 0.9324395399
24 0.0165110023 0.9489505422
25 0.0125489118 0.9614994539
26 0.0095051901 0.9710046441
27 0.0071815343 0.9781861784
28 0.0054157295 0.9836019079
29 0.0040783935 0.9876803013
30 0.0030680973 0.9907483986
31 0.0023062828 0.9930546814
32 0.0017326282 0.9947873096
33 0.0013011028 0.9960884124
34 0.0009767397 0.9970651521
35 0.0007330651 0.9977982171
36 0.0005500841 0.9983483012
37 0.0004127226 0.9987610238
38 0.0003096311 0.9990706549
39 0.0002322731 0.9993029280
40 0.0001742327 0.9994771607
41 0.0001306901 0.9996078508
42 0.0000980263 0.9997058771
43 0.0000735246 0.9997794017
44 0.0000551461 0.9998345478
45 0.0000413611 0.9998759089
46 0.0000310217 0.9999069306
47 0.0000232667 0.9999301974
48 0.0000174503 0.9999476477
49 0.0000130879 0.9999607356
50 0.0000098160 0.9999705516
51 0.0000073620 0.9999779136
52 0.0000055216 0.9999834352
53 0.0000041412 0.9999875764
54 0.0000031059 0.9999906823
55 0.0000023294 0.9999930117
56 0.0000017471 0.9999947588
57 0.0000013103 0.9999960691
58 0.0000009827 0.9999970518
59 0.0000007370 0.9999977889
60 0.0000005528 0.9999983416
61 0.0000004146 0.9999987562
62 0.0000003109 0.9999990672
63 0.0000002332 0.9999993004
64 0.0000001749 0.9999994753
65 0.0000001312 0.9999996065
66 0.0000000984 0.9999997048
67 0.0000000738 0.9999997786
68 0.0000000553 0.9999998340
69 0.0000000415 0.9999998755
70 0.0000000311 0.9999999066
71 0.0000000233 0.9999999300
72 0.0000000175 0.9999999475
73 0.0000000131 0.9999999606
74 0.0000000098 0.9999999705
75 0.0000000074 0.9999999778
76 0.0000000055 0.9999999834
77 0.0000000042 0.9999999875
78 0.0000000031 0.9999999907
79 0.0000000023 0.9999999930
80 0.0000000018 0.9999999947
81 0.0000000013 0.9999999961
82 0.0000000010 0.9999999970
83 0.0000000007 0.9999999978
84 0.0000000006 0.9999999983
85 0.0000000004 0.9999999988
86 0.0000000003 0.9999999991
87 0.0000000002 0.9999999993
88 0.0000000002 0.9999999995
89 0.0000000001 0.9999999996
90 0.0000000001 0.9999999997
91 0.0000000001 0.9999999998
92 0.0000000001 0.9999999998
93 0.0000000000 0.9999999999
94 0.0000000000 0.9999999999
95 0.0000000000 0.9999999999
96 0.0000000000 0.9999999999
97 0.0000000000 1.0000000000
98 0.0000000000 1.0000000000
99 0.0000000000 1.0000000000
100 0.0000000000 1.0000000000

There is an old electronic blackjack game at the casino in Cal Nev Ari with the following rules:

  • Wins, except blackjack, pays 3 for 2 (or 1 to 2)
  • Blackjacks pay 6 for 1 (or 5 to 1)
  • Single deck
  • Dealer stands on soft 17
  • Double on any initial two cards down offered
  • Splitting allowed
  • Do double after split
  • No re-splitting
  • No surrender

Rosebud

Interesting. I assume if the player doubles and wins he still only is paid 1 to 2 on the total amount bet.

First, here is the basic strategy for these rules:

  • Hard hands: Never double. Otherwise, play like conventional basic strategy, except stand on 12 vs. 3 and 16 vs. 10.
  • Soft hands: Never double. Hit soft 17 or less and soft 18 vs. 9. Otherwise, stand.
  • Pairs: Only split 8's against a 6 to 8. Always hit two aces. Otherwise, follow strategy for hard totals.

Under these rules and strategy, I get a house edge of 7.88%.

If the player had to achieve a point twice, before rolling a seven, to win the pass line bet in craps, how much would that increase the house edge?

Gary

That dreadful rule would increase the house edge from 1.41% to 33.26%.