On Sun, 20 Jul 2008 01:51:28 GMT, Odysseus
<odysseus1479-at@[EMAIL PROTECTED]
> wrote:
>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.
Actually, what I did was this. (1-(.95^2))*.05=204 to 1
It was late and I was tired. I see now on my notepade I jotted down
the following:
3c2 * 19...and that's where I left it and went to bed. Almost.
>
>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
I have a question about this formula. My albegra is rusty.
(1/20+19/20)^3 <----Do you take the 1/20 to the 3rd power and add
that to 19/20 to the 3rd power? A little rusty on exponents and
powers. Because it looks like you should add them first, and if you
did that you'd be taking 1 to the third power, which equals 1.
I guess what I'm trying to say is that I followed everything in your
post except for this formula.
>(1/20 + 19/20)^3 = (1 + 19)^3 * (1/20)^3
> = (1^3 + 3*1^2*19 + 3*1*19^2 + 19^3)/8000
To be more specific: I don't see how you get (1 + 19)^3 * (1/20)^3
from (1/20 + 19/20)^3, and therefore I don't see how you get
= (1^3 + 3*1^2*19 + 3*1*19^2 + 19^3)/8000
from (1 + 19)^3 * (1/20)^3
Thanks again for the detailed response.
>
>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. ;)
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