probability of winning at plinko

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galactus

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I have a probability question I was wondering about, if anyone is interested.

Please excuse my lop-sided dots, but I hope it conveys the idea.

I seen an episode of the Price is Right where they played this game called Plinko

The contestant stands at the top and lets a disc randomly fall through the 10X10 grid. The arrow at the bottom is $10,000. They can win up to 5 disks. What is the probability of actually landing in that spot 5 times?.

Mighty slim I would think.

Regardless, the two slots on either side of the arrow is where most land. That's $0, of course. I was just wondering, somewhere along the line someone may have actually calculated this probability somehow.



plinko5fk.gif
 
The contestant can drop anwhere desired, not just from the one spot.
 
i was just goign to post a question regarding price is right as well.

5 6-sided dice, on three sides are a car

you roll all 5 dice three times

what are the chances of getting 5 or more cars after a total of 15 rolls ?
 
mcrae said:
i was just goign to post a question regarding price is right as well.

5 6-sided dice, on three sides are a car

you roll all 5 dice three times

what are the chances of getting 5 or more cars after a total of 15 rolls ?
Click to expand...

What are on the other sides besides the 3 cars?
 
tkhunny said:
The contestant can drop anwhere desired, not just from the one spot.
Click to expand...

Yes, anywhere they choose.

thanks, pka, for the link.
 
Hello, mcrae!

If I interpret the problem correctly, the game is far too easy to win.
Are you sure of those rules?

i was just going to post a question regarding "Price is Right" as well.
Five 6-sided dice, on three sides are a car
You roll all five dice three times.
What are the chances of getting 5 or more cars after a total of 15 rolls ?
Click to expand...
The rolling of the dice is an independent event.
Rolling five dice three times is the same as rolling one die fifteen times.

On any die, three of the six faces have cars.
The probability of getting a Car on a die is \(\displaystyle \,\frac{1}{2} \,=\,0.5\)


The opposite of "5 or more" is "4 or less".
We will compute that probability . . . and subtract from 1.

\(\displaystyle \text{0 Cars: }\;C(15,0)(0.5)^{15}\)
\(\displaystyle \text{1 Car: }\;\;C(15,1)(0.5)^{15}\)
\(\displaystyle \text{2 Cars: }\;C(15,2)(0.5)^{15}\)
\(\displaystyle \text{3 Cars: }\;C(15,3)(0.5)^{15}\)
\(\displaystyle \text{4 Cars: }\;C(15,4)(0.5)^{15}\)

Their sum is: \(\displaystyle \,P(\text{4 or less Cars}) \:=\:1941(0.5)^{15} \:= \;0.059124619 \:\approx\;6\%\)

Therefore: \(\displaystyle \,P(\text{5 or more Cars})\:=\:100\%\,-\,6\%\:=\:94\%\)
 
soroban: yeah, i believe those were the correct rules, it seemed extremely easy to me too, oh well.

joey: the other three sides said $500, $1000, and $1500. if you didnt roll 5 or more cars, you got the total amount of money that you rolled

galactus: sorry for hijacking your thread
 
galactus said:
Here's an interesting link if anyone cares. It was given to me on another site. Thanks pka for the link, but I can not get the applet to work. I must not have Java enabled.
Click to expand...

I did not know that the java was not working even on my setup.
So I went to the following site to test it:
http://www.ies.co.jp/math/indexeng.html
Sure enough I had to download a new version of java: JAVA5.
I have used the demo in a probability class.
If you can get the new version of java to work, it is worth taking a look at the applet.
 
i havent updated firefox in a while, and it displayed the applet fine. not sure if you still need the same version of java if you're using firefox, but just throwing that out there :)
 
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