

stempleton Inner circle 1339 Posts 
Working on a type of Ping Pong matching trick as found in Mark Wilson's book. In my case, I have 2 identical sets of 5 different colored balls poured into a bag. I and the volunteer blindly choose 5 balls in rotating order and make 2 individual stacks of balls. What is the probability that both stacks will be ordered in the same color combination at the reveal. For example:
YBRPG YBRPG I know the number of combinations each person has for stacking: 120, but the chance of us BOTH choosing the same 1/120 combo eludes me. And then I get bogged down knowing that there is a possibility that two balls of the SAME color could be chosen by the same participant. Any help appreciated. 
stempleton Inner circle 1339 Posts 
I have contacted a professor of mathematics/probability and statistics from a local University for help. Until then, place your bets

FrankFindley Veteran user 328 Posts 
Are there two bags each with five balls? Or is there one bag with ten balls?
It the first case it is 1 in 120. In the second case I need to calculate it out. There are 10 factorial arrangements of the ten balls (ie 3,628,800). From this we simply need to divide out the numbers where pairs alternate, if I understand the process. So YYBBRRPPGG, YYRRBBPPGG, etc. However, for any given pattern, the balls of same color are interchangeable. So this needs to be taken into account. 
stempleton Inner circle 1339 Posts 
Frank, there is only one bag... all ten balls (2 of each color) are poured into the bag and mixed, then I and the volunteer take turns blindly reaching in and placing whatever ball is chosen into the inner, unseen tube. BTW, this is a twist on the routine explained by Mark Wilson... there he uses two bags, and the whole thing is structured as a joke on the spectator. I am using one bag and playing it as a serious bit.

FrankFindley Veteran user 328 Posts 
Off the top of my head, I come to 1 in 945 chance.
What I did is assigned a letter to each ball. For each color one ball will be a capital letter and one a small, So we have the following ten balls: Y,R,B,P,G,y,r,b,p,g These ten balls can come out of the bag in 3,628,800 different combinations. So what we want to know is how many of these combinations are in alternating pairs like Y,y,R,r,B,b,P,p,G,g; Y,y,R,r,B,b,G,g,P,p, etc. This would be the 120 alternations of the five colors multiplied by 2 for each color because we don't care if capital or small comes out first. For example Y,y is equally good as y,Y. This gives us 120x2x2x2x2x2= 3840 So it is 3,840 out of 3,628,800 which simplifies to 1 in 945. At least that is the thinking off the top of my head. 
FrankFindley Veteran user 328 Posts 
Just checked Mark Wilson's Complete Course in Magic. That is a funny routine.
In it they replace the balls after a pair is pulled out. Is that how you plan on doing it? The 1 in 945 was the probability of pulling out five matching pairs without replacement until all ten balls were removed. So, for example, Yy then Rr then Bb then Gg then Pp would be pulled from bag. The probability of pulling a match with a single pull is 1 in 9. And that actually verifies the 1 in 945. Because the probability to pull first pair is 1 in 9, the probability of pulling second pair is 1 in 7, probability of pulling third pair is 1 in 5, probability of pulling fourth pair is 1 in 3, and, of course, the fifth must match so it is 1 in 1. By multiplying these together we get the probability of the sequence: 1/9 X 1/7 X 1/5 X 1/3 X 1/1 = 1/945 
stempleton Inner circle 1339 Posts 
Quote:
On Oct 1, 2019, FrankFindley wrote: No, the balls are replaced only after the trick is over in order to do it again. Also, it only uses 3 pairs, I'm using 5, and I'm playing it as magic, not as comedy, because I want to avoid the scenario of the spectator being the only one not in on the joke. I think Paper Balls over Head is much better for something like that. No, the way I perform is listed above. One bag, 5 pairs of balls inside, spectator and magician alternate removing a ball, and at the end both stacks are matched exactly. Still haven't heard back from the professor I contacted... I guess a magician is not academic enough for his time 
FrankFindley Veteran user 328 Posts 
Quote:
On Oct 2, 2019, stempleton wrote: Very cool. This would be 1 in 945 chance. Think of it this way. There are 10 balls in the bag. Spectator chooses 1 (say Yellow) leaving nine in the bag. The magician then has a 1 in nine chance of choosing the other yellow one. Then the spectator takes another ball (say blue) leaving 7 balls in the bag. So now magician has a one in seven chance of picking the other blue one. Then spectator chooses another ball (say red) leaving 5 in bag. So magician has a 1 in 5 chance of picking the other red one. Then spectator chooses another ball (say purple) leaving 3 in the bag. So now magician has a 1 in 3 chance of choosing the remaining purple one. This leaves the remaining two which must match, that is a 1 in 1 chance. Collectively this gives a 1 in 945 (9x7x5x3x1 = 945) chance of such a sequence playing out. Or, if you prefer, such an occurance would happen just over 0.1% of the time (1/945 = 0.1058%). Not very likely at all! 
stempleton Inner circle 1339 Posts 
I follow that
A lot of info to wrap one's head around, but your steps seem logical. Many thanks for that 
Sealegs Inner circle The UK, Portsmouth 2567 Posts 
Working out probabilities can be very tricky and confusing. In Franks working out above he seems to be assuming that after each round of the spec and the performer removing a ball there will still be a matching colour ball in the bag to any ball that is removed 1st on the next round. What he seems to have not factored in are the (many likely) occasions where there is no match available in the bag to the 1st of the two balls that are removed due to it already having been removed in a previous round.
Neal Austin
"The golden rule is that there are no golden rules." G.B. Shaw 
FrankFindley Veteran user 328 Posts 
Quote:
On Nov 30, 2019, Sealegs wrote: That is a funny double entendre. Quote:
On Nov 30, 2019, Sealegs wrote: In both calculation methods cited above this is not assumed. Are you referring to the second method? If so, it doesn't assume that. But there is an inherent pattern in that the only way to get a situation where there aren't matching pairs in the bag for a draw is if an earlier removed pair was nonmatching. For example, to get R,B,P,G,y,b,p,g in the bag for the second round requires that the nonmatching Y,r had been pulled in the first round. So, that sequence was eliminated in the first step (that is, it is one of the 8 in 9 that didn't match). In other words, to be included in the 'all match' category requires that for each earlier pull there was a match and therefore only pairs are in the bag after each round. 
Sealegs Inner circle The UK, Portsmouth 2567 Posts 
Thanks Frank. Like I said, tricky and confusing.
Neal Austin
"The golden rule is that there are no golden rules." G.B. Shaw 
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