AceHound
Joined: Jun 2, 2010
• Posts: 23
April 20th, 2011 at 12:26:50 PM permalink
Don't know if anyone can figure this out? The game is "match 5".
Rules are: you pick envelopes one at a time revealing an amount. Once you reveal 5 of the same amount you win that amount.
There are:
5 of one amount
6 of one amount
9 of one amount
20 of one amount

Now to throw a twist on an alternate game play. Same concept, match 5 and win the amount but now you have 2 "free play" envelopes. If you choose a "Free play" you hold on to that and can choose to discard an envelope of a smaller amount. The denomination of envelopes would look like this:
5 of one amount
6 of one amount
8 of one amount
19 of one amount
2 of the free play

Ayecarumba
Joined: Nov 17, 2009
• Posts: 6763
April 20th, 2011 at 12:49:06 PM permalink
Quote: AceHound

Don't know if anyone can figure this out? The game is "match 5".
Rules are: you pick envelopes one at a time revealing an amount. Once you reveal 5 of the same amount you win that amount.
There are:
5 of one amount
6 of one amount
9 of one amount
20 of one amount

Odds of getting five in a random draw:

5 - 658008 to one
6 - 109668 to one
9 - 5222.3 to one
20 - 42.4 to one

Odds that you will pick five from these before you pick five from the 20:
5 - 15504 to one
6 - 2584 to one
9 - 123 to one
Simplicity is the ultimate sophistication - Leonardo da Vinci
PapaChubby
Joined: Mar 29, 2010
• Posts: 493
April 20th, 2011 at 1:59:03 PM permalink
Quote: Ayecarumba

Odds that you will pick five from these before you pick five from the 20:
5 - 15504 to one
6 - 2584 to one
9 - 123 to one

I don't immediately know how to do the math, but intuitively those numbers look awfully high to me. Especially the 123 to 1 number. I would've expected a number more in the range of 5 or maybe 10 to 1.
AceHound
Joined: Jun 2, 2010
• Posts: 23
April 20th, 2011 at 2:48:35 PM permalink
Quote: Ayecarumba

Odds of getting five in a random draw:

5 - 658008 to one
6 - 109668 to one
9 - 5222.3 to one
20 - 42.4 to one

Odds that you will pick five from these before you pick five from the 20:
5 - 15504 to one
6 - 2584 to one
9 - 123 to one

I may have mislead the concept. You continue to pick envelopes until you match 5 of the same amount. So you could end up picking a max of 17 envelopes. I find it hard to believe the odds of winning were as high as 658,008 to one.
Ayecarumba
Joined: Nov 17, 2009
• Posts: 6763
April 20th, 2011 at 4:13:36 PM permalink
Quote: PapaChubby

I don't immediately know how to do the math, but intuitively those numbers look awfully high to me. Especially the 123 to 1 number. I would've expected a number more in the range of 5 or maybe 10 to 1.

I just divided the odds of drawing each in five pulls by the odds of getting the common prize. It is very difficult to pull anything other than the common prize in this game as they make up half the possible draws. Imagine that you had a deck of cards with the diamond suit removed. What are the odds that you will pull the cards required to make a Royal Flush in hearts before you draw any five non-picture black cards? That is what you are up against.
Simplicity is the ultimate sophistication - Leonardo da Vinci
PapaChubby
Joined: Mar 29, 2010
• Posts: 493
April 20th, 2011 at 5:24:27 PM permalink
Quote: Ayecarumba

I just divided the odds of drawing each in five pulls by the odds of getting the common prize. It is very difficult to pull anything other than the common prize in this game as they make up half the possible draws. Imagine that you had a deck of cards with the diamond suit removed. What are the odds that you will pull the cards required to make a Royal Flush in hearts before you draw any five non-picture black cards? That is what you are up against.

That sounds like a much more daunting task than that proposed by the original poster. There is only one way to make a royal flush in hearts. I think the problem is more like "what is the probability that you will draw any five red cards before you draw any five black cards (with the diamond suit removed)". Just to make sure we're both evaluating the same problem, if I draw RBBRRBRBR then this is a win for red, as I drew the fifth red before I got to a 5th black.

It appears that you are calculating the probability of getting 5 in a row of each category, then comparing these probabilities to determine the relative likelihood of getting 5 of any category before 5 of another category shows up. It is possible that this yields the correct solution, but I don't see it. I'd think the correct solution would require combinations and permutations out the wazoo.
PapaChubby
Joined: Mar 29, 2010
• Posts: 493
April 20th, 2011 at 5:55:19 PM permalink
I ran an experiment. I created a deck with 9 red cards and 20 black cards. Thoroughly shuffled the deck (I hope). Examined the resulting deck. There are 29 ways I could cut this deck. How many of those would result in drawing 5 red cards before 5 black cards? I repeat this procedure 10 times, for a total of 290 different possible draws (some of which are correlated with each other, but I really didn't want to shuffle and cut 290 times).

Trial 1 resulted in 12 times out of 290 that I would have gotten 5 reds first. About 1 in 24.

Trial 2 resulted in 10 times out of 290. 1 in 29. The results appear fairly consistent.

So I think my intuitive guess of 1 in 5 or 10 was too low. I'll meet you half way?
Ayecarumba
Joined: Nov 17, 2009
• Posts: 6763
April 22nd, 2011 at 11:03:31 AM permalink
Actually, there are 40 total envelopes (not 29) in the original query. Five specific envelopes of the 40 are needed to win the top prize. The odds that you will pull those particular five before pulling five of the 20 commons are:

[(5/40)*(4/39)*(3/38)*(2/37)*(1/36)] divided by [(20/40)*(19/39)*(18/38)*(17/37)*(16/36)] = 1/15504.

In other words, if you did this test 15,504 times you might pull the five card prize once.
Simplicity is the ultimate sophistication - Leonardo da Vinci
miplet
Joined: Dec 1, 2009
• Posts: 1941
April 22nd, 2011 at 2:22:43 PM permalink
I did a quick sim of 1,000,000 trials for 1st one:
5: 1887
6: 8001
9: 69043
20: 916213
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Ayecarumba
Joined: Nov 17, 2009
• Posts: 6763
April 22nd, 2011 at 3:17:25 PM permalink
Quote: miplet

I did a quick sim of 1,000,000 trials for 1st one:
5: 1887
6: 8001
9: 69043
20: 916213

Hmm, I would have expected a figure closer to 64 or 65 for the "perfect 5". Your sim demonstrate a 485.5 - 1 ratio for "Common20" vs. "P5". Granted, my figures are the odds of drawing five of each, in five tries, from a full deck of 40, but wouldn't the ratio of outcomes with more attempts still be proportional to those of a "perfect pull"?
Simplicity is the ultimate sophistication - Leonardo da Vinci