Mac is idly tossing two dice when
he decides to see how many tosses it
would take to go from 1 to 6 in order.
The rules are to toss the two dice
until one or the other or both show
a 1. Then, toss until a 2 shows.
However, if both a 1 and a 2 show at the
same time, both can be used.
He then
tosses for a 3 and similarly for other
numbers. What is the expected number
of tosses to go from 1 to 6?
Find expected value of number of throws to get the target throwing 2 dice.
Prob of hitting the target value is 11/36 is for each throw. Just picture a 6 by 6 grid of the options 1,...,6.
EV of throws to get the target, e.g. a 1, is 36/11. (ignoring whether you also got a 2).
(I used a spreadsheet for this, but the answer is always the reciprocal of the probability of success for each throw).
Define:
S = a single, you got a 1 but not both a 1 and a 2.
D = a double, you got both 1 and 2
H = a final Hit, doesn't matter if it is a single or a double when 6 is the target.
Returning to the 6 by 6 grid, there are 11 ways to get a 1 out of 36 total ways.
1 way is two 1s.
2 ways with a 1 and a 2
8 ways with a 1 and either a 3,4,5,6
So assuming one of the dice was a 1,
p(S | you got a 1) = 9/11
p(D | you got a 1) = 2/11
So on average, a hit comes every 36/11 throws, and if you do get a hit, 9/11 are singles and 2/11 are doubles. The exception is when you have already found a 5 but not a 6, so the target is 6. When the target is 6, 11/11 will be successes.
When the target is 6, just signify success as H, a Hit
List the ways a combination of S and D can get you to 6. When determining combinations, the final hit must be kept fixed, only the hits before the final hit may be shifted around in different orders.
consider S=9/11 D=2/11 and H=1
pattern prob combos
SSSSSH S^5 1 comb(5,1)
SSSSD S^4 D 1 D at the end
SSSDH S^3 D 4 comb(4,1)
SSDD S^2 D^2 3 D at the end
SDDH S D^2 3
DDD D^3 1
pattern
SSSSSH 0.367 1 1 0.367
SSSSD 0.448 0.182 1 0.081
SSSDH 0.548 0.182 4 0.398
SSDD 0.669 0.033 3 0.066
SDDH 0.818 0.033 3 0.081
DDD 1 0.006 1 0.006
1
Then multiply each probability times the number of hits times 37/11 for each
The next spreadsheet columns:
6 3.27272727272727 7.200
5 3.27272727272727 1.333
5 3.27272727272727 6.518
4 3.27272727272727 0.869
4 3.27272727272727 1.062
3 3.27272727272727 0.059
Grand sum: 17.0414047272434
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Posted by Larry
on 2025-01-28 12:12:38 |
Successful analytic solution
