25 Points to whoever can solve this

[GELATINOUS CUBE]

There are actually only 32 possible outcomes...

I think you should give this up..

true... you listed the outcomes.

actually I have 10 parlays.
1,2
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5

here are the parlays

and these are the possible outcomes that would lead to a profit;

1+2+3
1+2+4
1+2+5
1+3+4
1+3+5
1+4+5
2+3+4
2+3+5
2+4+5
3+4+5
1+2+3+4
1+2+3+5
1+2+4+5
1+3+4+5
2+3+4+5
1+2+3+4+5

remember there is a .6 chance of each leg winning
and a .4 chance of each leg losing.

so the (prob. of winning) to the power of (wins) times the (prob of losing) to the power of (loses) times possible outcomes = probability of that win/loss outcome.

+

(prob. of winning) to the power of (wins) times the (prob of losing) to the power of (loses) times possible outcomes = probability of that win/loss outcome

+

(prob. of winning) to the power of (wins) times the (prob of losing) to the power of (loses) times possible outcomes = probability of that win/loss outcome (to the power of N total win/loss) outcomes

=

probability of winning win loss outcomes.

[GELATINOUS CUBE]

1+2+3
1+2+4
1+2+5
1+3+4
1+3+5
1+4+5
2+3+4
2+3+5
2+4+5
3+4+5

these hit at .6 x .6 x .6 x. 4 x .4 = 3.5% x 10 =35%

plus

1+2+3+4
1+2+3+5
1+2+4+5
1+3+4+5
2+3+4+5

.6 x .6 x .6 x.6 x .4 = 5.2% x 5 poss. outcomes = 26%

plus

1+2+3+4+5
.6 x .6 x.6 x .6 x .6= 7.8%

for a total prob. of winning at 68.8%

if you hit the first 2 legs.

you just need one more,
so the only way you lose is if all 3 final bets lose
WWW
WWL
WLW
LWW
LWL
[B]LLL

THAT HAPPENS .4 X .4 X .4 = 6.4% OF THE TIME X 1 POSSIBLE OUT COME = 6.4% CHANCE OF LOSING.
WHICH GIVES YOU A 93.4% CHANCE OF WINNING. IF YO HIT YOUR FIRST TWO LEGS.
[/B]

[GELATINOUS CUBE]

AGAIN, I THINK I GOT THIS.

YOUR CHANCES OF TURNING A PROFIT ARE 68.8%

IF YOU HIT YOUR FIRST TWO LEGS, YOUR CHANCES IMPROVE TO 93.4%

find the percentage probability of each of the 32 outcomes and add up the winning combos.

then find the percentage probability of each of the 6 outcomes for question 2, and just subtract the 1 loser from 100%, you should get about the same exact answer.

Goodnight.

[E-ftw]

Here is a better solution.... straight bet all the plays and take the win...

[Metalhead]

68.26%

93.6%

[steveC3]

wow,makes it sound like theres no fun in gambling

[hhsilver]

1. you need at least 3 wins out of the 5 plays.

prob = 5C3 x .6^3 x .4^2 + 5C4 x .6^4 x .4 + 5C5 x .6^5 = .68256 or 68.256%

2. given 1st 2 win, you need at least 1 of remaining 3 plays to win

prob = 1 - all 3 lose = 1 - .4^3 = .936 or 93.6 %

metalhead already got these answers .... just showing it's not as complicated as some make it

[Rich Boy]

Good work cube,

how long did you spend on this question?

[GELATINOUS CUBE]

An hour.

Hey, 8th grade Algebra was a long time ago, and I don't bring my textbooks home with me (from work).

You kinda gave me a hint with the 32 outcomes.

I was thinking 5 factorial (5x4x3x2x1), that's where I got 120 at first.

But it came back to me, probability of outcome times total outcomes... added up.

Sorry it took so long.

[GELATINOUS CUBE]

thanks Hi Ho Silver.

But I try to explain the solution in English.

Most people don't remember 8th grade either, so your notations just will lose them.

Good job metalhead, that answer 68.2, is more accurate than 68.7

I maybe don't deserve 25 points, because I guessed wrong at first, but I'll take them.

[Fishhead]

wow,makes it sound like theres no fun in gambling

No crying in baseball either

[semibluff]

1) 68.256%.

2) 93.6%.

[DrunkHorseplayer]

There are only 10 2-team parlays involved; given the fact that you're getting 5-1 ($50 bet to make $250 profit), if the first two plays hit you're 100% guaranteed to make a profit. As far as the first question, I don't know how to calculate it but given the fact that each parlay is 36% to hit (60% x 60%), there is a 98.85% chance that you'll hit at least one out of ten.
Having said that, don't count on getting 5-1 on two teamers where each individual play is a 60% chance; it simply isn't going to happen.

[gauchojake]

I m just happy we still have posters repping the 805 albeit posthumously