Would you bet on this lottery puzzle?

Question

(a) There is a 49/6 lottery that is offering even odds (ie 1:1 or 2 for 1) for betting that at least 2 of the 6 randomly chosen #s from 49 will be consecutive.
Is it advisable for you to bet ? 
(b) If the system is changed so that 49 & 1 are also deemed to be consecutive, would your decision change ?
(c) What is your expected value for a $1000 bet on (a) and (b)

Brian · Accepted Answer

(a) I think it is easier to first determine the probability that no 2 of the 6 numbers
are consecutive and then subtract that from 1 to get the desired probability.

I've looked at this as a 'stars and bars' type of calculation. First, we write out
the numbers 1 to 49 and then circle 6 randomly chosen numbers. The variables
in the equation represent the number of unchosen numbers between each successive
pair of chosen numbers as well as 1 variable representing the number of unchosen 
numbers preceding the lowest chosen number and 1 variable representing the number
of unchosen numbers after the largest chosen number. (That was a lousy
explanation but it is getting rather late here.) We then need to look separately
at the cases where (i) 1 is chosen but not 49, (ii) 49 is chosen but not 1,
(iii) both 1 and 49 are chosen and (iv) neither 1 nor 49 are chosen.

As an example, in case (iv) we then need to solve the equation

a + b + c + d + e + f + g = 43, where all the variables must be at least 1.

The number of solutions in this case ((36 + 7 - 1) C (7 - 1)) = (42 C 6).

I end up with a probability of

1 - [((42 C 6) + (42 C 4) + 2*(42 C 5)) / (49 C 6)] = 0.4952 < 0.5, so I would

not take the bet.

(b) This would eliminate the (42 C 4) term and result in a probability of

0.5032 > 0.5, so I would take this bet.

(c) I would expect to lose about $9.60 in scenario (a) and win $6.40 in (b)
after wagering $1000, (at least that's how I think it works; I'm not really a
betting man. :) )

I'm tired so I may have screwed up somewhere. I'll check through my
calculations again tomorrow if you think I've made an error.

Hillary · Answer

1

String · Answer

EDIT: Great fun! Brian and I came up with solutions simultaneously :o)

Clever question, I might leave the expected value part (part (c)) to others. Let b(n,k)=n!/[k!·(n-k)!] be the binomial coefficient corresponding to choose k non-ordered elements from a set of n elements:

Part (a)
-----------
You can construct all ways to choose 6 non-consecutive numbers from 1 through 49 by choosing 6 numbers a<b<c<d<e<f from 1 through 44 and map these to (a,b+1,c+2,d+3,e+4,f+5). Hence there will be b(44,6) = 44!/(6!·38!) such non-consecutive outcomes. Thus you win with a probability of

1-b(44,6)/b(49,6) = 22483/45402 ≈ 49.52%

Part (b)
-----------
49 and 1 are considered consecutive. If we choose non-consecutively among 1 through 48 we're good. This makes b(43,6) outcomes. If 49 is chosen, the remaining 5 must be chosen non-consecutively among 2 through 47. This makes b(42,5) additional outcomes. Thus the probability of winning now becomes:

1-[b(43,6)+b(52,5)]/b(49,6) = 71803/142692 ≈ 50.32%

Whether you want to bet in either case, I can't tell. I wouldn't, but if I should, I would prefer case (b) which has very subtle tendency towards winning :)

Anonym · Answer

I play the $5 poker ;)

Would you bet on this lottery puzzle?

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