Any 7s Tip on Wizard of Odds

This is directed to the Wizard, but anyone is welcome to show me where my logic is incorrect.

On your craps page, you have the following tip:

"Tip: If you must bet on a 7, don’t make the any 7 bet, with a house edge of 16.67%. Instead, divide your bet by 3, and put 1/3 each on the 1-6, 2-5, and 3-4 hop bets. Even at the stingy payoff of 15 to 1, you will still lower the house edge to 11.11%."

While I agree betting on the 7 in general is a terrible bet, I cannot agree with this tip.

The problem I see with this tip is that it does not factor in that in order to win a hopped 7 bet, 2 of your bets are guaranteed losers. In effect, this lowers the house payout to 13 to 1 or "14 FOR 1". The any seven bet would pay 4 to 1 or "5 FOR 1" on a bet 3 times larger for a net of an additional unit (5*3=15 units instead of 14 units).

From a simple example:

If you bet $12 on an any 7 bet paying 4 to 1 and a 7 is rolled, the bet will win you $48, plus you will keep your original wager of $12 for a total of $60 in your pocket after a win.

If you take the same $12 and spread it across the 3 hop bets at $4 a piece and the 2-5 comes out, you will win $60 on the 2-5 hop bet, retain the $4 from the 2-5 hop bet for a total of $64. However, you win also lose the $4 on each of the 1-6 and 3-4 bets, leaving you with a net of $56.

While I agree betting a single hop bet has a better statistical edge, I do not agree that if you intend to play "any 7" it is better to wager on the individual hop bets, because of the guaranteed losses that come with a win.

Please let me know where this logic is incorrect.

Win Any 7: Start with $12, end up with $60.

Win one of Hop bets: Start with $12, end up with $64.

Either way, a non-7 results in a $12 loss.

Your analysis sounds a lot like the bogus explanation in the old story of three men renting a hotel room.

Quote: dcapone

If you take the same $12 and spread it across the 3 hop bets at $4 a piece and the 2-5 comes out, you will win $60 on the 2-5 hop bet, retain the $4 from the 2-5 hop bet for a total of $64. However, you win also lose the $4 on each of the 1-6 and 3-4 bets, leaving you with a net of $56.

In your pocket, you will have $64 by betting the three hops, not $56.

Unless you know for sure that the next roll will be Big Red, a better play would be to put it all on the Pass or Come. If a seven (or eleven) you double up, but if not 2,3,12, you have another shot to double by hitting a point.

Quote: dcapone

If you take the same $12 and spread it across the 3 hop bets at $4 a piece and the 2-5 comes out, you will win $60 on the 2-5 hop bet, retain the $4 from the 2-5 hop bet for a total of $64. However, you win also lose the $4 on each of the 1-6 and 3-4 bets, leaving you with a net of $56.

Please let me know where this logic is incorrect.

You don't net 56, you net 64. it's 12-4-4+60=64.

yep, I am an idiot...wanted to pay the house twice on those losses.

You can tell I have never made this bet or it would have been obvious.

You're not an idiot.

You just needed a slap in the head.

Here ya go:

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