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Telepathy test odds

Psiphi

New Blood
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Joined
Apr 10, 2004
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Part of this query has already been discussed, but I need to know the odds of positive trials back to back as in the prelim and formal tests. Maybe you odds-makers and probability buffs can help with this.

The published odds are 2.6 million to one against a chance event of correctly guessing five non-repeating numbers from one to fifty two (preliminary).

The published odds are 135 million to one against a chance event for correctly guessing five non-repeating numbers from one to fifty-two and a wild card from one to fifty-two which may be a repeatable number. (formal test)

What are the odds of telepaths sucessfully completing these two situations back to back? I wish I knew how to calculate these odds, but I do not . Can anyone help.

This scenario seems mind-boggling but would be parlor games for true telepaths. Do you think Randi would accept this protocol?
 
Multiply the two tegether. In probabily when dealing with and and (probabilty of a and b happening) you muliply the two sets of odds together.
 
Psiphi said:
Part of this query has already been discussed, but I need to know the odds of positive trials back to back as in the prelim and formal tests. Maybe you odds-makers and probability buffs can help with this.

The published odds are 2.6 million to one against a chance event of correctly guessing five non-repeating numbers from one to fifty two (preliminary).

The published odds are 135 million to one against a chance event for correctly guessing five non-repeating numbers from one to fifty-two and a wild card from one to fifty-two which may be a repeatable number. (formal test)

What are the odds of telepaths sucessfully completing these two situations back to back? I wish I knew how to calculate these odds, but I do not . Can anyone help.

This scenario seems mind-boggling but would be parlor games for true telepaths. Do you think Randi would accept this protocol?
Click to expand...

1 in 351 trillion assuming no anomalous cognition.
 
Thanks for the input guys, that is mind boggling. I bet Randi would love those odds!
 
The question depends on the protocol, if they are in seperate rooms , under observation by judges. The cards ( I assume cards given 1 to 52) are purchased and under seal by a neutral judge and randomly shuffled. And the sender has an oportunity to see but not to manipulate the cards.

Then there would be little chance of cue-ing or cheating. The only possibility would be a good metal detector scan for radios and off you go.
 
Thanx Dave... that sounds reasonable.

Why we're still on the odds thing, I'd be curious as to how you would arrive at 2.6 million to one odds as in the first situation. Should have gone to the university, huh? I'd appreciate a kind reply.

Thanx.
 
Psiphi said:
Thanx Dave... that sounds reasonable.

Why we're still on the odds thing, I'd be curious as to how you would arrive at 2.6 million to one odds as in the first situation. Should have gone to the university, huh? I'd appreciate a kind reply.

Thanx.
Click to expand...

1/52*1/51*1/10*1/49*1/48 should be how it is done but that gives me odds of 311,875,200 to 1

If we go for 4 picks the odds are 6,497,400 to 1

If we go for 3 picks the odds are 132,600 to 1

I'm not sure where the orignal number came from.
 
Hi Geni, those odds are published by the Massachusetts Lottery Commission. I really would like to know how its arrived at.

Thanx kindly for your reply.
 
I'm assuming you got the numbers from here.
I have no idea how they got those numbers, I get the same as Geni. Even for guessing just the megaball they list odds of 1 in 88, when you are picking from 52 balls. Hmmm. Maybe the rules are not exactly as stated?

Edit: For info on combinations and permutations you can read here.
 
Geni: Your odds are correct IF the order in which the cards are picked is part of the protocol. For lotteries, the order does not matter, so you must divide by the number of ways the five cards can be arranged (5! or 5 X 4 X 3 X 2 X 1 = 120). That will give you the 2.6 million to 1 figure.

I don't know how they get 1 in 88 for the odds of picking one number out of 52. Note that 135 million is approx 2.6 million X 52, so it seems they are using odds of 1 in 52, not 1 in 88, in arriving at that number...
 
Interesting Ian said:
You multiply 5/52 * 4/51 * 3/50 * 2/50 * 1/49 =1/2.6mil

Clearly the order of the balls is not taken into account.
Click to expand...

Souldn't that be 5/52 * 4/51 * 3/50 * 2/49 * 1/48 or is my stats even worse than I thought?
 
patnray said:
Geni: Your odds are correct IF the order in which the cards are picked is part of the protocol. For lotteries, the order does not matter, so you must divide by the number of ways the five cards can be arranged (5! or 5 X 4 X 3 X 2 X 1 = 120). That will give you the 2.6 million to 1 figure.
Click to expand...


Typeing 52C5 into a calculator is quicker:)
 
Probability of getting al five balls=1/(52C5)=1/(2598960)
Probability of not getting the Megaball=51/52.
The 2.6 stated on the page (2,649,920) is 1/(1/(52C5)*(51/52)), so it is the probability of getting all 5, but not the megaball.

Edit: I guess the 1:88 is the probability of getting the Megaball, but none of the other 5. Too lazy to get the actual numbers.
 
Interesting thing about those huge odds is how some luckly swine manages to pick those numbers and win the lottery most weeks . Does such extreme odds prove some kind of physic power ? Of course not , enough people and enough time and you'll get any result you want .
 
Psiphi said:

The published odds are 2.6 million to one against a chance event of correctly guessing five non-repeating numbers from one to fifty two (preliminary).
Click to expand...


This is calculated by: 1/nCr(52,5) = ... = 1/2,598,960.


The published odds are 135 million to one against a chance event for correctly guessing five non-repeating numbers from one to fifty-two and a wild card from one to fifty-two which may be a repeatable number. (formal test)
Click to expand...


And that figure is obtained by taking: (1/2,598,960)* (1/52) = 1/135,145,920.


What are the odds of telepaths sucessfully completing these two situations back to back?
Click to expand...


As mentioned, if the events A and B are independent, then Probability(A and B) = Prob(A) * Prob(B). Clearly A and B aren't independent here, because A has to occur before B can occur. -Unless you are interested in having a psychic satisfy A, then satisfy B with different numbers?

If you are interested in Probability(doing A twice) or Probability(doing B twice), then those probabilities are just Probability(A)*Probability(A), and Probability(B)*Probability(B).
 
I think any odds over 2500 to 1 against chance is a fair demonstration. With the proper protocals of course to prevent any cheating intentional or otherwise.
 
geni said:


1/52*1/51*1/10*1/49*1/48 should be how it is done but that gives me odds of 311,875,200 to 1
Click to expand...

That's in order. If in any order, you have to divide by 5!, which gives 2598960.
 
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