+EV Lottery Side-Bet

[RocketManJames]

A friend of mine a while back told me that significantly more than 50% of the time the California "Fantasy 5" lottery numbers will contain 2 (or more) that are consecutive... basically there are 39 numbers and 5 drawn.

First, is this true? If so, how +EV is this side-bet to bet on the next draw containing consecutive numbers.

-RMJ

[pzhon]

[ QUOTE ]
A friend of mine a while back told me that significantly more than 50% of the time the California "Fantasy 5" lottery numbers will contain 2 (or more) that are consecutive... basically there are 39 numbers and 5 drawn.

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To choose 5 nonadjacent numbers out of 39 to mark with an X rather than an O, choose 5 positions out of 35, then replace the first 4 with XO and the last with X. Every 5 nonadjacent numbers can be obtained this way, and every choice of 5 out of 35 gives you a different nonadjacent 5 out of 39.

It looks like the numbers are not adjacent (35 choose 5)/(39 choose 5)=56.38% of the time.

[RocketManJames]

Thanks, I guess my friend was wrong. And, this is a -EV side-bet if you take the "consecutive" side of the bet.

-RMJ

[BruceZ]

[ QUOTE ]
[ QUOTE ]
A friend of mine a while back told me that significantly more than 50% of the time the California "Fantasy 5" lottery numbers will contain 2 (or more) that are consecutive... basically there are 39 numbers and 5 drawn.

[/ QUOTE ]

To choose 5 nonadjacent numbers out of 39 to mark with an X rather than an O, choose 5 positions out of 35, then replace the first 4 with XO and the last with X. Every 5 nonadjacent numbers can be obtained this way, and every choice of 5 out of 35 gives you a different nonadjacent 5 out of 39.

It looks like the numbers are not adjacent (35 choose 5)/(39 choose 5)=56.38% of the time.

[/ QUOTE ]

I agree with this answer. Nice solution, pzhon. Did you think up this method yourself, or have you seen it before?

This reminds me of the exercise of counting the number of distinguishable ways to place r indistinguishable balls in n cups, where any number of balls can go in each cup. Photons, nuclei and some atoms behave as though each of these arrangements is equally probable. Particles which behave this way are said to obey "Bose-Einstein statistics" and are called "Bosons".

<font color="white"> C(n+r-1,r) = C(n-1,r-1). Hint: Represent each cup with the letter C followed by a series of B's representing balls in that cup. </font>

[BruceZ]

I forgot to mention that I put the answer to the Bose-Einstein counting problem in white at the bottom of my last post.

[RocketManJames]

Hi Bruce... your Boson problem is the same as a "multi-choose" problem I ran into in school years ago. The way we learned to solve this was to solve an equivalent problem of permutations of |'s (Blockers) and o's (Items). I do not believe I'd be able to come up with this equivalent problem on my own.

Say you have 5 indistinguishable particles and 4 cups.

Some arrangements follow:

A B C D
ooo | o | | o
o | | oooo |
| oo | oo | o

WLOG, the number of permutations is then just as you have it, since you have (n + r - 1) total characters, and r of them being o's.

So I thought a bit about how you've proposed it be done with the hint. It is very similar to the above. Basically, there are (n + r) C's and B's. But, we know that the first spot must be a C. So, we have (n + r - 1) spots to choose r of the B's.

-RMJ

[bunky9590]

I'm not sure about that side bet, looks like a math problem to me.

Me and another entrepenurial kid ran a lottery in High school for 3 years. Talk about profitable!!!!

It was mega stacked in our favor but people didn't care!

You bought a number of your choice between 000- 999 for a quarter per number (No str8 and box crap) You hit, you win $100.00. price per number $.25. We were paying out 400:1 on a 1000:1 longshot. We even had teachers playing. Kids would put in like a dollar a day every day. We usually had between 100-300 numbers per day. Someone would hit on average about once every 2 weeks (but it ran for over a month one time)

We had a scientific calculator that would choose a random number. Every day at the beginning of 6th period we would punch it up and record the number. Run it down the list, announce the number and verify any winners. We were averaging around $50.00 per week after any payouts. (EACH!)

Senior year we got the hammer dropped down on us. Administraton found out 3/4 of the way through the school year and we were told to cease and desist or face expulsion.

I pulled the plug, much to the dismay of the highschool compulsive gambler crowd. Man, that was some nice bank for back in high school.

[pzhon]

[ QUOTE ]
[ QUOTE ]
To choose 5 nonadjacent numbers out of 39 to mark with an X rather than an O, choose 5 positions out of 35, then replace the first 4 with XO and the last with X.

[/ QUOTE ]
I agree with this answer. Nice solution, pzhon. Did you think up this method yourself, or have you seen it before?


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Thanks. I came up with it myself, but I'd be surprised if the argument were new. This problem is a dimension lower than the hard square problem, which has been studied by many people. The 1-dimensional analogue of the hard square constant is phi = (sqrt(5)+1)/2, since the total number of nonadjacent subsets of {1,...,n} is (n+1 choose 0) + (n choose 1) + (n-1 choose 2) + ... = the n+2nd Fibonacci number.

[fnurt]

If you knew then what you know now, you probably would have cut the principal in on the take and kept on raking it in. Maybe you even would have gotten some free advertising over the PA between periods.

[bunky9590]

The only reason i didn't get expelled is because I was honor society, Leaders club, top 5% of my class, Outstanding High school student of America, and never in trouble before. I took the entire fall for the deal and never sold the other guy out. They wanted him to take the fall. I wound up with some school priveleges revoked but no real consequences to speak of. Mother was disapointed by Dad shook his head and laughed. He wished He thought of it.

The Principal was unbuyable. The thought did cross my mind but i dismissed it just as quickly.