Q: Let&#39;s say the probability of making a shot at half court in basketball is 1%. How many shots would it take, on average, to get three in a row? What is the general formula for any probability and any number in a row?

Let&#39;s let: <ul> <li>a=expected more shots assuming initial state or last shot was a miss.</li> <li>b=expected more shots assuming last shot was made.</li> <li>c=expected more shots assuming last two shots were made.</li> </ul> We can set up the following equations on going from state to state: a = 1 + 0.01b + 0.99a b = 1 + 0.01c + 0.99a c = 1 + (1-p)a We now have three equations and three unknowns so we can solve it. I prefer matrix algebra. <img alt="" src="/media/content_images/572/determ-a.jpg" style="width: 560px; height: 207px;" /> <img alt="" src="/media/content_images/573/determ-b.jpg" style="width: 531px; height: 212px;" /> Without teaching a whole less on that, the solution can be expressed as the determ(A)/determ(B). The terms in the matrices are taken from the three equations above. The answer to this ratio of determinants is 101010. To answer the second question, the answer for any probability p and number n of consecutive successes is: (1/p)^n + (1/p)^(n-1) + (1/p)^(n-2) + ... + (1/p)^2 + (1/p)^1 In the case of this problem, the general formula shows the answer as 100^3 + 100^2 + 100^1 = 1000000 + 10000 + 100 = 1010100 This question is asked and discussed in my forum at <a href="https://wizardofvegas.com/forum/gambling/sports/39207-basketball-probability-question/#post928497" target="_blank">Wizard of Vegas</a>.

Question 1

Let's say the probability of making a shot at half court in basketball is 1%. How many shots would it take, on average, to get three in a row?

What is the general formula for any probability and any number in a row?

Accepted Answer

Let's let:

a=expected more shots assuming initial state or last shot was a miss.
b=expected more shots assuming last shot was made.
c=expected more shots assuming last two shots were made.

We can set up the following equations on going from state to state:

a = 1 + 0.01b + 0.99a
b = 1 + 0.01c + 0.99a
c = 1 + (1-p)a

We now have three equations and three unknowns so we can solve it. I prefer matrix algebra.

Without teaching a whole less on that, the solution can be expressed as the determ(A)/determ(B). The terms in the matrices are taken from the three equations above.

The answer to this ratio of determinants is 101010.

To answer the second question, the answer for any probability p and number n of consecutive successes is:

(1/p)^n + (1/p)^(n-1) + (1/p)^(n-2) + ... + (1/p)^2 + (1/p)^1

In the case of this problem, the general formula shows the answer as 100^3 + 100^2 + 100^1 = 1000000 + 10000 + 100 = 1010100

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

Question 2

The 13 cards of any given suit are removed from a deck.&nbsp; One card is dealt to two logicians each, Alex and Bob.&nbsp; 2&#39;s are low and aces are high.&nbsp; Each logician may look at his own card.&nbsp; Then, Alex may offer Bob to switch.&nbsp; If the offer is made, Bob may accept or reject it.&nbsp; What should be the optimal strategy of both players?&nbsp;

Accepted Answer

Alex should offer only with a 2. Bob should accept only with a 2.

To answer this question myself I played around with various strategies, as follows.

If Alex switches with a 4 or less, Bob should accept with a 2 and be indifferent at 3. The probability of Bob winning is 56.7%.

If Alex switches with a 3 or less, Bob should accept with a 2 only. The probability of Bob winning is 53.3%.

If Alex switches with a 2 only, Bob should always reject the offer. The probability of Bob winning is 50.0%.

The pattern is that Bob should be more picky about switching than Alex. If Alex switches with a 3 or higher, Bob can have an advantage with a lower criteria for switching. The only way Alex can defend against being defeated this way is to switch with a 2 only. Knowing this, Bob would never switch if an offer is made. Thus, if two logicians played, Alex should offer to switch with a 2 only. Bob should always reject said offer.

However, in the unlikely event Bob had a 2 and an offer to switch was made, of course Bob should take it, thinking that Alex either misread the card or isn't a true logician.

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

Question 3

How many spins in roulette does it take to see a number repeat in roulette, on average?&nbsp;

Accepted Answer

You didn't say the type of wheel, but here is the answer all three ways:

Single Zero = 8.306669466
Double Zero = 8.408797212
Triple Zero = 8.509594851

The following table shows the probability of a first repeat at each spin for all three wheels.

Probability of Repeat Number

Spin	Single Zero	Double Zero	Triple Zero
1	0.0000000000	0.0000000000	0.0000000000
2	0.0270270270	0.0263157895	0.0256410256
3	0.0525931337	0.0512465374	0.0499671269
4	0.0746253924	0.0728240268	0.0711070652
5	0.0914329132	0.0894330154	0.0875163879
6	0.1019353424	0.1000237672	0.0981754352
7	0.1057923554	0.1042352943	0.1027066091
8	0.1034096446	0.1024066049	0.1013898577
9	0.0958236089	0.0954768346	0.0950762036
10	0.0844931146	0.0847985044	0.0850200666
11	0.0710452616	0.0719051646	0.0726667236
12	0.0570282235	0.0582810281	0.0594376534
13	0.0437169674	0.0451747682	0.0465525677
14	0.0320000324	0.0334848063	0.0349144258
15	0.0223534530	0.0237240530	0.0250667672
16	0.0148879175	0.0160538705	0.0172161863
17	0.0094424270	0.0103646041	0.0113008813
18	0.0056941663	0.0063755953	0.0070811612
19	0.0032589823	0.0037306115	0.0042294718
20	0.0017665054	0.0020725619	0.0024039306
21	0.0009046116	0.0010908221	0.0012976683
22	0.0004364140	0.0005425405	0.0006638073
23	0.0001977062	0.0002542733	0.0003209618
24	0.0000837944	0.0001119289	0.0001462658
25	0.0000330845	0.0000461035	0.0000626155
26	0.0000121086	0.0000176932	0.0000250863
27	0.0000040842	0.0000062951	0.0000093656
28	0.0000012609	0.0000020644	0.0000032419
29	0.0000003534	0.0000006197	0.0000010345
30	0.0000000890	0.0000001689	0.0000003022
31	0.0000000199	0.0000000414	0.0000000802
32	0.0000000039	0.0000000090	0.0000000191
33	0.0000000007	0.0000000017	0.0000000040
34	0.0000000001	0.0000000003	0.0000000007
35	0.0000000000	0.0000000000	0.0000000001
36	0.0000000000	0.0000000000	0.0000000000
37	0.0000000000	0.0000000000	0.0000000000
38	0.0000000000	0.0000000000	0.0000000000
39	0.0000000000	0.0000000000	0.0000000000

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