No result is likely - Part IV (conclusion)

Everything I wrote regarding the 35:1 bet on the 12 (see part II),
where the house expects to make no money, applies equally well to the
odds bets, the difference being that the probabilities of winning and
the payoffs are different, the similarity being that the payoffs
exactly balance the probability advantage. The DP/DC odds are "mirror
images" of the pass/come odds, favoring the player in win probability
but requiring him/her to bet the long end of the odds.

MoneyLA's question:

"When do the dice know that it's time to give the player the win so
that in the overall results, the player will win once for every two
wins for the house? When does that happen?"

could just as well be asked from the point of view of the casino.
Suppose the player gets lucky and wins the first 3 odds bets on the 4
or 10. Do the dice know that the casino is not supposed to lose money
on these bets? Of course not. The casino may never "catch up" with
this player, any more than some other player who loses a few may
"catch up" with the casino. We know, however, that, over a long time
and a very large number of bets, any variation from even is going to
be TINY in comparison to the total amount bet.

Or, take the DP/DC odds bets. The player is expected to win 2 of every
3 laid odds bets against the 4 or 10. How do the dice know the house
is supposed to win 1 of 3? On every bet, the player is more likely to
win; on every bet, if the player wins he gets paid only half of what
he risked.

To sum up:

When you throw one die, every possible result is an "underdog,"
compared to the sum of the other possible results, yet this does not
prevent some face from showing.

When you throw two dice, each face of each die is an "underdog,"
relative to the sum of the other possible results, yet this does not
prevent both dice from landing and coming to a stop with some face
showing.

When you throw two dice, the face showing on one when it comes to rest
does not affect which face shows when the other die comes to rest.

If you decide to combine the number of pips showing on both dice in
some way, and propose to make bets on some combined outcome, you have
to be able to calculate the probabilities of different outcomes, as
defined by whoever is offering the bet.

The joint probability of two independent events is calculated by
MULTIPLYING their individual probabilities. The set of probabilities
so calculated MUST sum to 1.0.

Without using different-colored dice, you can't distinguish between
6,2 and 2,6, so those two possibilities become one distinguishable
outcome. Since they are mutually exclusive, i.e. can't both happen on
the same roll, their combined probability is the SUM of their
individual probabilities. Any other set of outcomes that we decide to
interpret as the same result for some purpose is treated the same. So,
if we decide to use the SUM of the pips showing on both dice as the
"result," then we have to add up the combinations that satisfy that
condition. For example, if the "result" we are interested in is:

SUM of pips = 5

1,4 by multiplication, probability is 1/6 * 1/6 = 1/36
4,1 "
2,3 "
3,2 "
----------
by addition, probability is 1/36 + 1/36 + 1/36 + 1/36 = 4/36

This is the same thing as just adding the number of combinations
satisfying the condition and dividing by the total number of possible
combinations.

If we do this for all the possible sums of pips, we get the familiar
"perfect 36". It is important to note that the seven is still a pretty
big "underdog," relative to ALL the other outcomes, but is MORE LIKELY
than any other individual sum. This principle becomes of great
importance in many of the bets offered by the casinos.

After this, it is just a matter of deciding what combinations to
distinguish and bet on. You could create a game where the product of
the two numbers of pips was the outcome, so 6,2 would be 12, not 8. We
could call it "Multi-Craps." Other combinations resulting in 12 would
be 2,6 , 3,4 , 4,3 , 5,2.4 (whoops! you can't have 2.4 pips showing).
So, if I decided to offer a one-roll bet that the product of the two
faces would be 12, I would know there are 4 possible combinations that
would win and 32 that would lose. What if I offered you this bet,
agreeing to pay $7 for every $1 bet? How would you decide whether to
make such a bet? Is it a "good deal?" What about its "volatility?" Is
there an actual bet offered on the craps table that has the same
probability of winning it and the same payoff?

I hope this series has been of some interest and assistance to those
RCG readers who are not beyond hope and not advanced beyond these
simple principles.
Cheers,
Alan Shank
.

Relevant Pages

  • Re: probability problem
    ... Does "dice" mean 2 six-sided dice? ... Slightly higher since the first player goes first (because this is ... The chances could be equal if you do random selection ... And for B, sum the even throws. ...
    (sci.math)
  • Re: probability problem
    ... >> How do you do a question of the type, 2 players are throwing dice, ... > dice don't change from player to player). ... The chances could be equal if you do random selection ... And for B, sum the even throws. ...
    (sci.math)
  • Re: probability problem
    ... >> How do you do a question of the type, 2 players are throwing dice, ... > dice don't change from player to player). ... The chances could be equal if you do random selection ... And for B, sum the even throws. ...
    (sci.math)
  • Re: Probabilities, random numbers and dice throws
    ... A particular ordered roll of dice occurs with probability 1/z^y, multiply that by the number of ordered rolls, c, that can give a particular sum. ... You have to be careful to not overcount cases where multiple dice contain the same value. ... My problem is with probabilities. ...
    (microsoft.public.excel.worksheet.functions)
  • Re: Chances of (random(0,n) + random(0,n) <= m)
    ... what is the chance of the sum being less than or equal to m? ... the sum of two such picks is n, ... You can get some insight into this by working out the probabilities ... The individual dice are each 1-6 in equal ...
    (sci.math)