(via CotPi) 2012 --------- (One) Extra coin -------- Akio has one more coin than Bansi. They throw all of their coins and count the number of heads. If all the coins are fair, what is the probability that Akio obtains more heads than Bansi?

omg... this is a Great Problem!
pls tell me about another great problem about Coins or Dice.

___________________________________

------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints to the following.

i throw 2 dice, (let P = Product of the numbers i get)
What is the probability that...
--- P is a multiple of 5 ?
--- P is a multiple of 10 ?

(the same thing with 3 dice)

pls tell me about another (nice) problem about Coins or Dice.

On 01/10/2022 22:33, henh...@gmail.com wrote:
>
> (via CotPi) 2012 --------- (One) Extra coin -------- Akio has one more coin than Bansi. They throw all of their coins and count the number of heads. If all the coins are fair, what is the probability that Akio obtains more heads than Bansi?
>
>
> omg... this is a Great Problem!
> pls tell me about another great problem about Coins or Dice.
>
>

[snip]

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1/2

Assume the first coin thrown by Akio is a head. Then the probability
that Akio throws more heads than Bansi is the probability, P, that Akio
throws at least as many heads as Bansi in the remaining n throws each.

Assume the first coin thrown by Akio is a tail. Then the probability
that Akio throws more heads than Bansi is the probability that Bansi
does not throw at least as many heads as Akio in the remaining n throws
each. As the coins are fair this equals 1-P.

1/2 * P + 1/2 * (1-P) = 1/2

A little Python to illustrate,

import random

def sim(n, its):
tot = 0
for _ in range(its):
A = bin(random.getrandbits(n+1)).count('1')
B = bin(random.getrandbits(n)).count('1')
tot += A > B
return tot / its

sim(1, 1000000)
0.500135
sim(40, 1000000)
0.499144

Duncan

On Sunday, October 2, 2022 at 9:10:50 AM UTC-7, duncan smith wrote:
> On 01/10/2022 22:33, henh...@gmail.com wrote:
> >
> > (via CotPi) 2012 --------- (One) Extra coin -------- Akio has one more coin than Bansi. They throw all of their coins and count the number of heads. If all the coins are fair, what is the probability that Akio obtains more heads than Bansi?
> >
> >
> > omg... this is a Great Problem!
> > pls tell me about another great problem about Coins or Dice.
> >
> >
> [snip]
>
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> 1/2
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>
> Assume the first coin thrown by Akio is a head. Then the probability
> that Akio throws more heads than Bansi is the probability, P, that Akio
> throws at least as many heads as Bansi in the remaining n throws each.
>
> Assume the first coin thrown by Akio is a tail. Then the probability
> that Akio throws more heads than Bansi is the probability that Bansi
> does not throw at least as many heads as Akio in the remaining n throws
> each. As the coins are fair this equals 1-P.
>
> 1/2 * P + 1/2 * (1-P) = 1/2
>
> A little Python to illustrate,
>
>
> import random
>
> def sim(n, its):
> tot = 0
> for _ in range(its):
> A = bin(random.getrandbits(n+1)).count('1')
> B = bin(random.getrandbits(n)).count('1')
> tot += A > B
> return tot / its
>
>
> sim(1, 1000000)
> 0.500135
> sim(40, 1000000)
> 0.499144
>
>
> Duncan

(does [its] stand for something ?)

thanks.... i 'll study your Proof later....
----------------- ( i 'd never seen getrandbits before.)

for 3-sided and 6-sided Dice, i wonder if there's a similar invariant.

-------------- waht if Akio had to beat B by a margin of 1 ?

def dice(N, Sides, simN):
tot = 0
for _ in range(simN):
A,B= 0,0
for __ in range(N+1): A += random.randrange(Sides)
for ___ in range(N): B += random.randrange(Sides)
# tot += A > B
tot += A-1 > B
return tot / simN

Nsim=100000
for n in range(1,11): print('\t', "(2-sided)", n, ": ", dice(n, 2, Nsim))
print()
for n in range(1,11): print('\t', "(3-sided)", n, ": ", dice(n, 3, Nsim))
print()
for n in range(1,11): print('\t', "(6-sided)", n, ": ", dice(n, 6, Nsim))

On Sunday, October 2, 2022 at 12:10:50 PM UTC-4, duncan smith wrote:
> On 01/10/2022 22:33, henh...@gmail.com wrote:
> >
> > (via CotPi) 2012 --------- (One) Extra coin -------- Akio has one more coin than Bansi. They throw all of their coins and count the number of heads. If all the coins are fair, what is the probability that Akio obtains more heads than Bansi?

Nice. Here is another argument (below the space)

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> 1/2

Just flip all of Bansi's coins and ask whether the majority of coins now show heads. This is the same as the answer to the original question (if a=b, then a+(n-b)=n); and the probability is clearly 1/2 (the majority must be either heads or tails, and the post-flip distribution is just as random as the pre-flip).

Jonathan

On 03/10/2022 00:08, henh...@gmail.com wrote:
> On Sunday, October 2, 2022 at 9:10:50 AM UTC-7, duncan smith wrote:
>> On 01/10/2022 22:33, henh...@gmail.com wrote:
>>>
>>> (via CotPi) 2012 --------- (One) Extra coin -------- Akio has one more coin than Bansi. They throw all of their coins and count the number of heads. If all the coins are fair, what is the probability that Akio obtains more heads than Bansi?
>>>
>>>
>>> omg... this is a Great Problem!
>>> pls tell me about another great problem about Coins or Dice.
>>>
>>>
>> [snip]
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>> 1/2
>>
>>
>> Assume the first coin thrown by Akio is a head. Then the probability
>> that Akio throws more heads than Bansi is the probability, P, that Akio
>> throws at least as many heads as Bansi in the remaining n throws each.
>>
>> Assume the first coin thrown by Akio is a tail. Then the probability
>> that Akio throws more heads than Bansi is the probability that Bansi
>> does not throw at least as many heads as Akio in the remaining n throws
>> each. As the coins are fair this equals 1-P.
>>
>> 1/2 * P + 1/2 * (1-P) = 1/2
>>
>> A little Python to illustrate,
>>
>>
>> import random
>>
>> def sim(n, its):
>> tot = 0
>> for _ in range(its):
>> A = bin(random.getrandbits(n+1)).count('1')
>> B = bin(random.getrandbits(n)).count('1')
>> tot += A > B
>> return tot / its
>>
>>
>> sim(1, 1000000)
>> 0.500135
>> sim(40, 1000000)
>> 0.499144
>>
>>
>> Duncan
>
>
> (does [its] stand for something ?)
>

Iterations.

>
> thanks.... i 'll study your Proof later....
> ----------------- ( i 'd never seen getrandbits before.)
>
>
>
> for 3-sided and 6-sided Dice, i wonder if there's a similar invariant.
>
> -------------- waht if Akio had to beat B by a margin of 1 ?
>
>
>
> def dice(N, Sides, simN):
> tot = 0
> for _ in range(simN):
> A,B= 0,0
> for __ in range(N+1): A += random.randrange(Sides)
> for ___ in range(N): B += random.randrange(Sides)
> # tot += A > B
> tot += A-1 > B
> return tot / simN
>
>
> Nsim=100000
> for n in range(1,11): print('\t', "(2-sided)", n, ": ", dice(n, 2, Nsim))
> print()
> for n in range(1,11): print('\t', "(3-sided)", n, ": ", dice(n, 3, Nsim))
> print()
> for n in range(1,11): print('\t', "(6-sided)", n, ": ", dice(n, 6, Nsim))

The original question is set up so that things cancel out nicely. There
are two possible outcomes for Akio's extra coin throw, and the relevant
probabilities conditional on those outcomes involve a probability P that
subsequently cancels.

Looking at your code for the dice example it appears that you're
interested in the sum of the throws (and for dice numbered starting at
zero). So there are six possible outcomes (for the six sided die) which
are equiprobable. So the answer is,

1/6 * P(A_sum > B_sum) +
1/6 * P(A_sum -1 > B_sum) +
1/6 * P(A_sum -2 > B_sum) +
1/6 * P(A_sum -3 > B_sum) +
1/6 * P(A_sum -4 > B_sum) +
1/6 * P(A_sum -5 > B_sum)

where the probabilities are for the sums over N dice rolls.

I don't see any obvious simplifications (apart from with a 2-sided die).

Duncan

In article <cKi_K.289335$9Yp5.27263@fx12.iad>,
duncan smith <duncan@invalid.invalid> wrote:
>On 01/10/2022 22:33, henh...@gmail.com wrote:
>>
>> (via CotPi) 2012 --------- (One) Extra
>coin -------- Akio has one more coin than Bansi. They throw all of
>their coins and count the number of heads. If all the coins are fair,
>what is the probability that Akio obtains more heads than Bansi?
>>
>>
>> omg... this is a Great Problem!
>> pls tell me about another great problem about Coins or Dice.
>>
>>
>
>[snip]
>
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>1/2
>
>
> Assume the first coin thrown by Akio is a head. Then the probability
> that Akio throws more heads than Bansi is the probability, P, that Akio
> throws at least as many heads as Bansi in the remaining n throws each.
>
> Assume the first coin thrown by Akio is a tail. Then the probability
> that Akio throws more heads than Bansi is the probability that Bansi
> does not throw at least as many heads as Akio in the remaining n throws
> each. As the coins are fair this equals 1-P.
>
> 1/2 * P + 1/2 * (1-P) = 1/2
>

We can omit P. By symmetry, the probability A wins on heads equals the
probability A wins on tails. And A must win on one because a tie is
impossible. So both probabilities are 1/2.

Gareth