In article <otm484d9he4i3rv3bth42hfoa93i77smi5@[EMAIL PROTECTED]
>,
Paul <Msr33@[EMAIL PROTECTED]
> wrote:
> You have three wheels with 20 numbers on each wheel.
> You get to spin each wheel one time.
>
> Three 7s pays $100,000
> Two 7s pays $10,000
> 1 7 pays $1000.
The amounts payable are irrelevant to the odds: they don't come into the
picture unless you want to calculate the average or expected yield.
> Please correct me if I'm wrong.
>
> 1. Odds against hitting all three 7s=7999-1 (.000125)
Yes: the probability of success is (1/20)^3 = 1/8000.
> 2. Odss of getting two 7s out of three=204-1 (.004875)
Where # represents any number other than 7, a pair of 7s must occur as
one of 7-7-#, 7-#-7, or #-7-7. Since each of these can be made 19 ways,
we have p = (3 * 19)/8000 = 0.007125; the odds against will be 7943:57
or about 139:1.
> 3. Odds of getting one 7 out of three=6.01-1 (.1426)
Yes, if you mean getting *at least* one 7. Be sure that's really the
statistic you want! Getting at least a certain number (> 1) of hits and
getting exactly that many will have different odds -- see below.
> I'm pretty sure about 1. and 3., but I'm a little sketchy on #2.
Venturing to extrapolate from experiences of my own, I'm inclined to
guess that your method was sound enough but, in a momentary lapse, you
subtracted 7 from 20 when you actually intended to exclude the (single)
successful spin of 7 from the sample space, writing a 13 instead of a 19.
In general problems like this (with independent events) are easiest
solved with the "binomial theorem". The probabilities can be broken out
of the expansion of (p_success + p_failure)^n_trials -- which sums to 1
-- in this case
(1/20 + 19/20)^3 = (1 + 19)^3 * (1/20)^3
= (1^3 + 3*1^2*19 + 3*1*19^2 + 19^3)/8000
From the terms in parentheses, with their ***ulative sums:
| Ways to get | Ways to get
| exactly n Odds | at least n Odds
n | successes Prob. against | successes Prob. against
----------------------------------------------------------------
3 | 1 .000125 7999:1 | 1 .000125 7999:1
2 | 57 .007125 ~ 139:1 | 58 .00725 ~ 137:1
1 | 1083 .135375 ~6.39:1 | 1141 .142625 ~6.01:1
0 | 6859 .857375 ~0.17:1 | 8000 1 --
Factoring in the payout schedule in your preamble, I'd pay up to $219 to
take a spin -- that is, had I nothing better to do with the money I
might. It would be a more attractive proposition were I allowed to
choose my own target number. ;)
--
Odysseus
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